# Riddles



## Fort Fun Indiana (Dec 12, 2019)

*First one:

You meet a girl. She tells you she has one sibling. What are the odds her sibling is also a girl?*


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## Toddsterpatriot (Dec 12, 2019)

50/50


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## OldLady (Dec 12, 2019)

Fort Fun Indiana said:


> *First one:
> 
> You meet a girl. She tells you she has one sibling. What are the odds her sibling is also a girl?*


48%?
But that's not a riddle.


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> 50/50


No.


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## Fort Fun Indiana (Dec 12, 2019)

OldLady said:


> Fort Fun Indiana said:
> 
> 
> > *First one:
> ...


No. And yes, it's a riddle. You have to approach it correctly to get the correct answer.


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## OldLady (Dec 12, 2019)

This calls for Treeshepherd.


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## depotoo (Dec 12, 2019)

Boy or Girl paradox - Wikipedia


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## Fort Fun Indiana (Dec 12, 2019)

depotoo said:


> Boy or Girl paradox - Wikipedia


depotoo  had to look it up. He is awarded zero points.  Let's see if someone can get it on their own.


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > 50/50
> ...



Yes.


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## OldLady (Dec 12, 2019)

depotoo said:


> Boy or Girl paradox - Wikipedia


I was just looking at that, but the answer is still 1/2.

Oh lord, is this going to be another "The answer is 1" thread?


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## OldLady (Dec 12, 2019)

Fort Fun Indiana said:


> depotoo said:
> 
> 
> > Boy or Girl paradox - Wikipedia
> ...


Maybe I didn't read far enough, but my article didn't have an answer.


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## depotoo (Dec 12, 2019)

Fort Fun Indiana said:


> depotoo said:
> 
> 
> > Boy or Girl paradox - Wikipedia
> ...


I had heard it before.


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> Fort Fun Indiana said:
> 
> 
> > Toddsterpatriot said:
> ...


No, sorry. I will explain, eventually.


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## Fort Fun Indiana (Dec 12, 2019)

OldLady said:


> depotoo said:
> 
> 
> > Boy or Girl paradox - Wikipedia
> ...


No.


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## Crepitus (Dec 12, 2019)

50/50 I would think


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## OldLady (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
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> 
> > Fort Fun Indiana said:
> ...


You need to move along with the answer.  We're going to die of old age.  Anyone who wants to play can try the next riddle.


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## Fort Fun Indiana (Dec 12, 2019)

OldLady said:


> Fort Fun Indiana said:
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> 
> > Toddsterpatriot said:
> ...


Ha! Okay, will do.


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
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> 
> > Fort Fun Indiana said:
> ...



Unless your explanation is 50/50, you're wrong.


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## Fort Fun Indiana (Dec 12, 2019)

OldLady 

So, the answer is 1/3. (depotoo  knew this already)

Why?

There are four equally likely possibilities (in this case, permutations) for two siblings, oldest listed first:

BB
BG
GB
GG

We have met one sibling, and she is a girl. So now, there exist three equally likely possibilities:

BG
GB
GG

As these are equally likely, the probability of each possible permutation is now 1/3. So, the probability that the other sibling is also a girl is 1/3.

Had you been given the information of whether the girl we met was the older or younger sibling, the answer would have been 1/2.


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## Fort Fun Indiana (Dec 12, 2019)

Crepitus said:


> 50/50 I would think


Right, that's the immediately intuitive answer. But it is not correct.


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> Unless your explanation is 50/50, you're wrong.


Now that the explanation has been posted, do you still think it is wrong?


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> OldLady
> 
> So, the answer is 1/3. (depotoo  knew this already)
> 
> ...



*We have met one sibling, and she is a girl. So now, there exist three equally likely possibilities:*

*BG
GB
GG
*
Nope.
*
You meet a girl. She tells you she has one sibling. What are the odds her sibling is also a girl?*

G<-- girl you met  ------> sibling B
G<-- girl you met  ------> sibling G

Two possibilities.
*
*​


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > Unless your explanation is 50/50, you're wrong.
> ...



Your 1/3 answer is wrong.


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> Fort Fun Indiana said:
> 
> 
> > OldLady
> ...


No, that is incorrect. There are, in fact, four permutations of two children and two genders, not three. There are three combinations. But four permutations.

BB
BG
GB
GG

...all equally likely. As you can see, the probability of having one girl and one boy is 50%, if you have two children.  But, given we have at least one girl, our sample space has been reduced to three permutations.

BG
GB
GG

...all equally likely. Now that we know at least one is a girl, the probability that there is one girl and one boy is 2/3.

1/3 is correct.


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## Fort Fun Indiana (Dec 12, 2019)

Same riddle, reiterated:

2 quarters in a cup. You shake the cup and turn it over on a table. You reach under and slide out one quarter; it shows "heads". 

What is the probability the other quarter will also show '"heads"?

1/3.


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > Fort Fun Indiana said:
> ...



*As you can see, the probability of having one girl and one boy is 50%, *

Exactly. One girl......50% chance her sibling is a boy, 50% chance her sibling is a girl.

*There are, in fact, four permutations of two children*

We aren't looking for permutations of two children, because one was already determined.


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Same riddle, reiterated:
> 
> 2 quarters in a cup. You shake the cup and turn it over on a table. You reach under and slide out one quarter; it shows "heads".
> 
> ...



*2 quarters in a cup. You shake the cup and turn it over on a table. You reach under and slide out one quarter; it shows "heads". 

What is the probability the other quarter will also show '"heads"?*

50%.


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> Exactly. One girl......50% chance her sibling is a boy, 50% chance her sibling is a girl.


False. 



Toddsterpatriot said:


> We aren't looking for permutations of two children, because one was already determined.


Which eliminates only one of the four, equally likely permutations. You are left with three, equally likely permutations. Yes, the correct method is to consider the permutations.


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > Exactly. One girl......50% chance her sibling is a boy, 50% chance her sibling is a girl.
> ...



*False. *

LOL! 

*Which eliminates only one of the four, equally likely permutations.*

Nope. Girl eliminates half the permutations. Only two left.

*Yes, the correct method is to consider the permutations.*

Sure.

How many permutations are there for the second, unmet sibling?


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> *2 quarters in a cup. You shake the cup and turn it over on a table. You reach under and slide out one quarter; it shows "heads".
> 
> What is the probability the other quarter will also show '"heads"?*
> 
> 50%.


Incorrect.


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> Nope. Girl eliminates half the permutations. Only two left.


Obviously false. The following 3 permutations remain:

BG
GB
GG

You have, again, confused the concepts of "permutation" and "combination". And even then you are still wrong, as there are 3 possible combinations. Knowing one of the two children is a girl would only eliminate one of these 3 combinations, so not half.


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > Nope. Girl eliminates half the permutations. Only two left.
> ...



Let's look at the possibilities, taking birth order into account.

We meet a girl, G.

She has either a younger or older brother (b) or a younger or older sister (s)

Gb
bG
Gs
sG

Any possibilities I missed?


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> Fort Fun Indiana said:
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> > Toddsterpatriot said:
> ...


No need to complicate things and confuse yourself with odd annotation. We will just stick to the annotation used thus far:

BB
BG
GB
GG

These are, indeed, all four possible permutations. Keep in mind: this is your sample space before you know one of the children is a girl.


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > Fort Fun Indiana said:
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*No need to complicate things and confuse yourself with odd annotation. *

I'm not confused at all by my clear annotation.

* this is your sample space before you know one of the children is a girl. *

*BB
BG
GB
GG*

This is your sample space after you know one of the children is a girl. 

BG

GG​


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> This is your sample space after you know one of the children is a girl.
> 
> BG
> 
> GG


False. Those are the possible combinations, not permutations. And they are not equally likely. This is the source of your errors.


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > This is your sample space after you know one of the children is a girl.
> ...



You didn't like the real permutations.

Gb
bG
Gs
sG


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## Fort Fun Indiana (Dec 12, 2019)

Another riddle, somewhat similar:

A dealer has an infinite deck of playing cards. He pulls one off the top and places it to his left. He pulls another card off the top and places it to his right. He flips the card to his left. It is red. What is the probability that the card to his right is also red?

50%

Now he pulls two cards from the top of the  deck  and shuffles the two cards. He places them to his left and right. He flips the one to his left, and it is red. What is the probability the other card is also red?

50%

Now he pulls two cards from the deck and, without shuffling them, holds them behind his back. He shows you one of the cards, and it is red. What is the probability the other card is also red?

33%


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> Fort Fun Indiana said:
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> 
> > Toddsterpatriot said:
> ...


That annotation gives no extra information and use 3 symbols instead of 2. So no, that is not correct. The third symbol is unnecessary.

And it obviously causes you to confuse yourself, as "Gs" and "sG" are not actually unique permutations. You have listed a permutation twice, which is an error.

If you insist on these being unique data points in the sample space, then you must modify their probability and reduce the probability of each by half. In which case, you will still arrive at the correct answer: 1/3.


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
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> 
> > Fort Fun Indiana said:
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*That annotatiin give no extra information *

It gave birth order info.

*And it obviously causes you to confuse yourself, as "Gs" and "sG" are not actually unique permutations.*

They are unique. In the first case, the girl we met (G) is the older sister, in the second case she is the younger.

*You have listed a permutation twice, which is an error.*

Like you did.....

BB
BG <here
GB <and here
GG


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Another riddle, somewhat similar:
> 
> A dealer has an infinite deck of playing cards. He pulls one off the top and places it to his left. He pulls another card off the top and places it to his right. He flips the card to his left. It is red. What is the probability that the card to his right is also red?
> 
> ...



*Now he pulls two cards from the deck and, without shuffling them, holds them behind his back. He shows you one of the cards, and it is red. What is the probability the other card is also red?

33% *

50%


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> It gave birth order info.


Which is unnecessary, and causes you to make yet another error. As the probability of these two permutations is 1/2 the probability of "GG", you still arrive at 1/3 as the correct answer and have not actually added any information. You have just forced yourself to multiply by 2 then multiply by 1/2. No net change.


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> *Now he pulls two cards from the deck and, without shuffling them, holds them behind his back. He shows you one of the cards, and it is red. What is the probability the other card is also red?
> 
> 33% *
> 
> 50%


Wrong. For all the same reasons. That's why the riddle is called a "paradox"... Even though it actually isn't. When you don't fully understand the problem -- as you clearly do not -- it appears as a paradox.


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > It gave birth order info.
> ...



*Which is unnecessary*

You didn't want permutations?


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > *Now he pulls two cards from the deck and, without shuffling them, holds them behind his back. He shows you one of the cards, and it is red. What is the probability the other card is also red?
> ...



Your confusion doesn't make these a paradox.


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## G.T. (Dec 12, 2019)

Theres 3 doors. One has a prize.

You get to choose a door.

Before it's opened, the host (who knows where the prize is), opens one of the doors that has no prize.

He asks you if youd like to switch to the other door left, before the reveal.

Should you switch?


Yepp, odds on your door = 33%
Odds on winning if you switch = 66%


Most initially intuit that its 50/50. Its not.


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## RandomPoster (Dec 12, 2019)

I came up with approximately 33.6%.  There is a .8% chance that her unidentified sibling is an identical twin and this slightly increases the likelihood of the sex being the same.  This also may change slightly across different populations.


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## Fort Fun Indiana (Dec 12, 2019)

RandomPoster said:


> I came up with approximately 33.6%.  There is a .8% chance that her unidentified sibling is an identical twin and this slightly increases the likelihood of the sex being the same.  This also may change slightly across different populations.


Ha! I will make it clear next time that they are not identical twins.


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## Fort Fun Indiana (Dec 12, 2019)

G.T. said:


> Theres 3 doors. One has a prize.
> 
> You get to choose a door.
> 
> ...


Love it.


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## RandomPoster (Dec 12, 2019)

Fort Fun Indiana said:


> RandomPoster said:
> 
> 
> > I came up with approximately 33.6%.  There is a .8% chance that her unidentified sibling is an identical twin and this slightly increases the likelihood of the sex being the same.  This also may change slightly across different populations.
> ...



  I'm not certain, except it might be more clear to also specify whether she is the older or younger sibling, anything that clearly designates which one she is, as opposed to simply a sibling.


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## luchitociencia (Dec 12, 2019)

This is a riddle that needs Android 101 fonts. No other way to find the answer. You can download it from 

Android 101 Font | dafont.com

This is a riddle from the 60s when Texas Instrument watches were in the market like crazy. If you are old like me, you might know the answer already, but please give it using the Android 101 fonts.


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## Fort Fun Indiana (Dec 12, 2019)

RandomPoster said:


> I'm not certain, except it might be more clear to also specify whether she is the older or younger sibling,


That would add more information and change the answer to 1/2.


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## RandomPoster (Dec 12, 2019)

Fort Fun Indiana said:


> RandomPoster said:
> 
> 
> > I'm not certain, except it might be more clear to also specify whether she is the older or younger sibling,
> ...



  Yep.  I see.


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## Toddsterpatriot (Dec 12, 2019)

RandomPoster said:


> Fort Fun Indiana said:
> 
> 
> > RandomPoster said:
> ...



With no birth order, there are two combinations
1) she has a brother  
2) she has a sister.

With birth order, there are four permutations, 
1) she has an older brother
2) she has a younger brother
3) she has an older sister
4) she has an younger sister

Either way, there is a 50% chance she has a sister.​


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> With birth order, there are four permutations,
> 1) she has an older brother
> 2) she has a younger brother
> 3) she has an older sister
> ...



First, those are not four distinct permutations. Sorry. You were wrong the first time you said it, and you are still wrong.

Furthermore, those four states of the girl you met are not equally likely. This is easily shown by proof by contradiction.

1 and 2 each have a probability of 33.33%, while 3 and 4 each have a probability of 16.67%.


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## MAGAman (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
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> 
> > Fort Fun Indiana said:
> ...


No.

If she's older its 50/50

If she's younger its 50/50

BG
GG


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > With birth order, there are four permutations,
> ...



*First, those are not four distinct permutations*

Why not?

*This is easily shown by proof by contradiction.*

Okay, prove it.

*1 and 2 each have a probability of 33.33%*

Your confusion still doesn't prove your claim.


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## Fort Fun Indiana (Dec 12, 2019)

MAGAman said:


> If she's older its 50/50
> 
> If she's younger its 50/50


Correct, if you have that information at the start. But you don't. The answer is 1/3.


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## MAGAman (Dec 12, 2019)

But thanks for the riddle

BG
And 
GB 

Each have 1/2 the chance of GG

So it's really

BG
GG
GG
GB


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## Fort Fun Indiana (Dec 12, 2019)

MAGAman said:


> So it's really
> 
> BG
> GG
> ...


Wrong. GG is one permutation, no matter whether you meet the younger or older sibling.


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> MAGAman said:
> 
> 
> > So it's really
> ...



*Wrong. GG is one permutation*






In that case, BG and GB is also one permutation


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> In that case, BG and GB is also one permutation


False. You are confusing permutations and combinations again.


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## MAGAman (Dec 12, 2019)

Fort Fun Indiana said:


> MAGAman said:
> 
> 
> > So it's really
> ...


Then the chance of 
BG
and 
GB
are each 1/2 as likely as
GG

Either way, it's 50/50


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > In that case, BG and GB is also one permutation
> ...



*You are confusing permutations and combinations again.*

You did, when you said, "*GG is one permutation*"

With no birth order, there are two combinations
1) she has a brother
2) she has a sister.

With birth order, there are four permutations,
1) she has an older brother
2) she has a younger brother
3) she has an older sister
4) she has an younger sister

Either way, there is a 50% chance she has a sister.


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## Fort Fun Indiana (Dec 12, 2019)

Maybe this will help. Back to the card riddle.

Two cards are drawn from the infinite deck. They are laid in front of you. The dealer tells you that he knows at least one of the cards is red (he has xray vision). What are the odds both cards are red? 3:1. The probability is 1/3.

When you have two cards laid in front of you, you remove one third of the sample space by choosing one and flipping it to see it is red. You have now ordered your choice. If you do not know which card was flipped, the sample space is not reduced.

When reaching under the cup to grab a quarter or meeting the girl, you have not ordered your choice. When you dont know which card has been flipped, but only know at least one of the two cards is red, you have not ordered your choice.


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> l"*GG is one permutation*"


Correct, it is. Describing the girl you have met to be the older or the younger describes the state of the girl. It does not describe unique permutations. It is just two ordered choices of whether the girl you met is the older or younger of the one GG permutation. And, to maintain the absolute fact that the sample space contains three equally likely permutations (BG, GB, GG) your descriptions of the girl's state each have probabilities that, taken together, sum to the probability of the permutation GG. There is no getting around that.


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## Fort Fun Indiana (Dec 12, 2019)

MAGAman said:


> Then the chance of
> BG
> and
> GB
> ...


Which is obviously absurd. You would then have to say that you are equally likely to have mixed sublings as you are to have two girls, should you have two children. Clearly you see that is false., assuming each birth has a 50/50 chance for boy or girl.

Surely you agree that, should you flip a coin twice, you are twice as likely to end up with one head and one tail (in any order) than you are two tails.


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > l"*GG is one permutation*"
> ...



* It does not describe unique permutations.
*
She's older or she's younger. Unique permutations.

*the sample space contains three equally likely permutations (BG, GB, GG, ) *

If you don't care about birth order for two girls, why do you use two permutations with a brother?


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> She's older or she's younger. Unique permutations.


False. Only one permutation exists: GG.

Maybe it would help you if we added (unnecessary) subscripts: x1x2, x1 is the older sibling.

Whether the girl you meet is G1 or G2 does not have any bearing on the fact that G1G2 is a unique permutation. Same for B1G2 and G1B2. These are your three possible permutations. There are no others. There is no "G2G1" permutation. Your choice has not been ordered. 

And, these three permutations, which are the entire sample space:

B1G2
G1B2
G1G2

...are equally likely. That is a fact that you cannot get around. Whether you meet the older or younger sibling (you dont know which, so the only information you have is "G"), this fact remains. Do you agree?


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > She's older or she's younger. Unique permutations.
> ...



*Only one permutation exists: GG.*





How can you have a permutation which disregards order?

*And, these three permutations, which are the entire sample space:*​​*B1G2*​*G1B2*​*G1G2*​​Nah.​Every permutation has to include the girl(G) we met in post #1.

That's why this works.......

Gb
bG
Gs
sG


​


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## Fort Fun Indiana (Dec 12, 2019)

Another reiteration:

So, let's say two siblngs are in a house with 2 rooms, a hallway, and one front door. This is all the information you have.

If you go inside and open one of the doors to find a girl, what is the probability that there is a girl in the other room?

50%.

Now, instead, you wait out front, one of the siblings is sent outside, and it is a girl. Now, what is the probability that the remaining sibling in the house is a girl?

1/3


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## Fort Fun Indiana (Dec 12, 2019)

Toddsterpatriot said:


> *B1G2 G1B2 G1G2*
> 
> Nah.
> Every permutation has to include the girl(G) we met in post #1.​


Which all of those do. Who did you meet... G3? Doesn't matter if you met G1 or G100.... as far as you know, you only met G.

And no, that weird annotation doesn't "work" in that the four possibilities you listed are not equally likely. Do you get that?


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > *B1G2 G1B2 G1G2*
> ...



*Which all of those do.* 

_B1G2 G1B2 G1G2_

G1 and G2 are different, they can't both exist with a brother in your "riddle".

*And no, that weird annotation doesn't "work"*

It works. Every permutation has the girl we met (G)
You can clearly see two permutations with a brother and two with a sister.

*the four possibilities you listed are not equally likely. *

Prove it.


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## Fort Fun Indiana (Dec 12, 2019)

*3rd riddle:

*


Toddsterpatriot said:


> G1 and G2 are different, they can't both exist with a brother in your "riddle".


And, unfortunately for you who must solve the problem, you have no idea which one you have met. So you cannot eliminate anything from the sample space.You only know that at least one sibling is a girl. So, the fact remains that the siblings are twice as likely to be mixed as to be all girls.


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## Toddsterpatriot (Dec 12, 2019)

Fort Fun Indiana said:


> *3rd riddle:
> 
> *
> 
> ...



*And, unfortunately for you who must solve the problem, you have no idea which one you have met. *

You're lying.

_You meet a girl. She tells you she has one sibling. What are the odds her sibling is also a girl?_


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## luchitociencia (Dec 12, 2019)

> Permutations GED basic 1, Combinations PE QS 2600 2x 76R, 2x 77R QS Value: 2502 to find if *she* was a boy or a girl...



And I thought solving riddles was fun and like a poster says "with simple steps". Like Batman movies... sigh*


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## OldLady (Dec 13, 2019)

Fort Fun Indiana said:


> OldLady
> 
> So, the answer is 1/3. (depotoo  knew this already)
> 
> ...


Interesting.  I would have thought with each child the probability would start from a clean slate.  Good question, Indiana.  Not a riddle, though.


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## Toddsterpatriot (Dec 13, 2019)

OldLady said:


> Fort Fun Indiana said:
> 
> 
> > OldLady
> ...



* I would have thought with each child the probability would start from a clean slate.*

You would be correct.


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## OldLady (Dec 13, 2019)

Toddsterpatriot said:


> OldLady said:
> 
> 
> > Fort Fun Indiana said:
> ...


You'll never win against these math minds, Todd.  I Googled permutation, still have no idea what the diff is between that and the combinations, and you know what?  Since I'm not a gambler in Vegas, I can get on with my life perfectly well NOT knowing.
I will remember the answer, 1/3, but this was a cheap trick to pull us into a statistics class.


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## Toddsterpatriot (Dec 13, 2019)

OldLady said:


> Toddsterpatriot said:
> 
> 
> > OldLady said:
> ...



*You'll never win against these math minds, Todd.*

FF is far from a math mind.

*I will remember the answer, 1/3,*

Then you're remembering it wrong.

With no birth order, there are two combinations
1) she has a brother
2) she has a sister.

With birth order, there are four permutations,
1) she has an older brother
2) she has a younger brother
3) she has an older sister
4) she has an younger sister

Either way, there is a 50% chance she has a sister.


----------



## K9Buck (Dec 13, 2019)

Every time you flip a coin, the odds of it coming up on one particular side is ALWAYS going to be 50-50, regardless of how it came up previously.  This is simple logic, folks.


----------



## OldLady (Dec 13, 2019)

And so, this has become another "The Answer is One" thread.

Math is a bunch of clever contortions of numbers to torture English majors.  It may be why we can have suspension bridges and skyscrapers, space capsules hurtling toward Neptune with cameras, but it still sucks.


----------



## Toddsterpatriot (Dec 13, 2019)

OldLady said:


> And so, this has become another "The Answer is One" thread.
> 
> Math is a bunch of clever contortions of numbers to torture English majors.  It may be why we can have suspension bridges and skyscrapers, space capsules hurtling toward Neptune with cameras, but it still sucks.



Not familiar with that thread.


----------



## OldLady (Dec 13, 2019)

Toddsterpatriot said:


> OldLady said:
> 
> 
> > And so, this has become another "The Answer is One" thread.
> ...


It's in the FZ, I believe.  It was too long ago for me to bother searching for, though, sorry.  It was an algebra problem that can be solved in two ways, depending on what rules regarding PEMDAS you learned as a nipper.

The answer was either 1 or 16, depending.  The argument raged for days.
IIRC, I was in the 16 camp.


----------



## Fort Fun Indiana (Dec 13, 2019)

K9Buck said:


> Every time you flip a coin, the odds of it coming up on one particular side is ALWAYS going to be 50-50, regardless of how it came up previously.  This is simple logic, folks.


No, sorry, it does not follow that the answer is 50%


----------



## K9Buck (Dec 13, 2019)

Fort Fun Indiana said:


> K9Buck said:
> 
> 
> > Every time you flip a coin, the odds of it coming up on one particular side is ALWAYS going to be 50-50, regardless of how it came up previously.  This is simple logic, folks.
> ...



If I flip a coin, what are the odds of it being heads or tails?


----------



## Fort Fun Indiana (Dec 13, 2019)

K9Buck said:


> Fort Fun Indiana said:
> 
> 
> > K9Buck said:
> ...


Just make your point. By the way, your answer is wrong, and you can look it up.


----------



## K9Buck (Dec 13, 2019)

Fort Fun Indiana said:


> K9Buck said:
> 
> 
> > Fort Fun Indiana said:
> ...



There are two possibilities, heads or tails.  The odds of either turning up are 50-50.  Do you dispute this?


----------



## Fort Fun Indiana (Dec 13, 2019)

K9Buck said:


> There are two possibilities, heads or tails.


Wrong. You need to be considering 4 possibilities, as we are considering two "coin flips", not one. Then we are given the information that at least one flip resulted in tails, reducing this to 3 possibilities. The probability that both flips resulted in tails then becomes 1/3.


----------



## K9Buck (Dec 13, 2019)

Fort Fun Indiana said:


> K9Buck said:
> 
> 
> > There are two possibilities, heads or tails.
> ...



I was talking about a single flip of the coin.  This discussion is obviously a waste of time.  Carry on.


----------



## Fort Fun Indiana (Dec 13, 2019)

K9Buck said:


> Fort Fun Indiana said:
> 
> 
> > K9Buck said:
> ...


Yes, you were contibuting the wisom that the probability of getting tails in a coin flip is 50%. Thank you.


----------



## Toddsterpatriot (Dec 13, 2019)

Fort Fun Indiana said:


> K9Buck said:
> 
> 
> > Fort Fun Indiana said:
> ...



*No, sorry, it does not follow that the answer is 50% *

And you were disagreeing.


----------



## Fort Fun Indiana (Dec 13, 2019)

Toddsterpatriot said:


> Fort Fun Indiana said:
> 
> 
> > K9Buck said:
> ...


False. I did not disagree on the probability of a coin flip.


----------



## Toddsterpatriot (Dec 13, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > Fort Fun Indiana said:
> ...




*False. I did not disagree on the probability of a coin flip.*






LOL!


----------



## Fort Fun Indiana (Dec 13, 2019)

Toddsterpatriot said:


> Fort Fun Indiana said:
> 
> 
> > Toddsterpatriot said:
> ...


Which is not what I did there. Obviously, I meant that it does not follow from his statements that the answer to the riddle is 50%. Settle down, you will never grasp this riddle when you are in that state.


----------



## Wuwei (Dec 13, 2019)

Fort Fun Indiana said:


> Another reiteration:
> 
> So, let's say two siblngs are in a house with 2 rooms, a hallway, and one front door. This is all the information you have.
> 
> ...


One way to look at it is to go in the house and stand in the hallway. The probabilities are 50% that there is a G in one room, and 50% that there is a G in the other room because there are only 2 possibilities in each room. (Excluding the 70 other genders that Facebook has.)

The four possible combinations of both room all have equal probabilities.

As soon as you open one door, and see a G you have eliminated the BB possibility.
if you see a B you have eliminated the GG possibility. 

The three combinations left still still have have equal probabilities.

That leaves three possibilities, each with the same probability. Namely 1/3.


----------



## Fort Fun Indiana (Dec 13, 2019)

Wuwei said:


> The four possible combinations of both room all have equal probabilities.


*permutations, not combinations. There are only three combinations: BB, BG, GG, with BG starting with twice the probability of each of the other two combinations. 


Wuwei said:


> As soon as you open one door, and see a G you have eliminated the BB possibility.


You have also eliminated half of the BG combinations, specifically, all of the BG permutations (where the first element is the gender in the first room you enter).

Because of this, if you open the first door and see a girl, the probability you will find a girl behind the second door is 1/2.


----------



## Fort Fun Indiana (Dec 13, 2019)

Wuwei 

However, if you sent someone else into the house who opened a door (you don't know which door) and shouted out that he sees a girl, the probability, from your perspective, that he finds a girl in the other room as well is 1/3.


----------



## Toddsterpatriot (Dec 13, 2019)

Fort Fun Indiana said:


> Wuwei said:
> 
> 
> > The four possible combinations of both room all have equal probabilities.
> ...



*You have also eliminated half of the BG combinations, specifically, all of the BG permutations (where the first element is the gender in the first room you enter).*​​Hmmmmm….where have I heard that before?​
_Nope. Girl eliminates half the permutations. Only two left._

Riddles

Oh, yeah.


----------



## Fort Fun Indiana (Dec 13, 2019)

Toddsterpatriot said:


> Hmmmmm….where have I heard that before?


i believe you said it while incorrcetly using the term "permutations". And also notice you are off base anyway, as i spoke of eliminating half of the BG combinations, while you spoke of eliminating half of the permutations. If you read your own post slowly, you can see this error on your part.


----------



## Toddsterpatriot (Dec 13, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > Hmmmmm….where have I heard that before?
> ...



*i believe you said it while incorrcetly using the term "permutations".*

If you've already picked the first sibling, the girl you met, there are only combinations.

Girl-brother
Girl-sister

You've eliminated the permutations, brother-brother and brother-sister.


----------



## Fort Fun Indiana (Dec 13, 2019)

Toddsterpatriot said:


> If you've already picked the first sibling, the girl you met, there are only combinations.


You did not "pick" anything in the riddle, though. That changes the riddle. And "picking" actually eliminates half of the BG combinations. So this reduction in the sample space is what leads to your second "pick", in your scenario, being, indeed, 50% likely to be a girl.


----------



## Wuwei (Dec 13, 2019)

Fort Fun Indiana said:


> Wuwei
> 
> However, if you sent someone else into the house who opened a door (you don't know which door) and shouted out that he sees a girl, the probability, from your perspective, that he finds a girl in the other room as well is 1/3.



Why does the particular door matter? The problem seems to be symmetric.


----------



## Toddsterpatriot (Dec 13, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > If you've already picked the first sibling, the girl you met, there are only combinations.
> ...



*You did not "pick" anything in the riddle, though*

The first sibling was picked for you in post #1.


You meet a girl. She tells you she has one sibling.


----------



## Fort Fun Indiana (Dec 13, 2019)

Wuwei said:


> Why does the particular door matter?


Because, when you know which door has been opened, you have "ordered" your choice. Before opening the door, you were faced with a sample space of 4, equally likely permutations (you can look at them as the first element being behind the first door you open, with the second element being behind the second door you open):

BB
BG
GB
GG

When you open the first door and see a girl, you immediately eliminate half the permutations from your sample space: BB and BG.

Now, you are left with a sample space of two equally likely permutations. The probability of finding a girl behind the second door is 1/2.

Now, if it is shouted outside the house to you that somene inside has opened a door and found a girl, but you don't know which door, this changes. You can only eliminate one permutation from this sample space -- BB -- and are left with THREE, equally likely permutations. The choice has not been "ordered".  Now, to you, the probability that the other room contains a girl is only 1/3.


----------



## cnm (Dec 13, 2019)

Fort Fun Indiana said:


> We have met one sibling, and she is a girl. So now, there exist three equally likely possibilities:
> 
> BG
> GB
> GG


No. We are told a girl has one sibling. The possibilities are

GG
GB

Age is not mentioned in the given information.


----------



## cnm (Dec 13, 2019)

Fort Fun Indiana said:


> OldLady
> 
> So, the answer is 1/3. (depotoo  knew this already)
> 
> ...


You framed the question incorrectly for that solution to apply.


----------



## luchitociencia (Dec 13, 2019)

K9Buck said:


> There are two possibilities, heads or tails.  The odds of either turning up are 50-50.  Do you dispute this?



I do.

I dropped several times five coins until this happened


----------



## Fort Fun Indiana (Dec 13, 2019)

So yes, if you want to scam someone and take some of their money, make a 4 card deck from two red and two black cards. Only allow bets when at least one card is red. Pull two cards. Keep them  hidden from the mark, but peek at them. If at least one card is red, tell the mark this. Give him even money odds on betting that there are two red cards. Your expectation will be 0.33. If you start with $300, you will end up with $400. Your mark, who started with $300, will end up with $200.


----------



## Fort Fun Indiana (Dec 13, 2019)

cnm said:


> You framed the question incorrectly for that solution to apply.


I don't think so.


----------



## Wuwei (Dec 13, 2019)

Fort Fun Indiana said:


> Because, when you know which door has been opened, you have "ordered" your choice. Before opening the door, you were faced with a sample space of 4, equally likely permutations (you can look at them as the first element being behind the first door you open, with the secnd element being behind the second door you open):
> 
> BB
> BG
> ...


I think my objection is that I don't think it is an ordering problem.


----------



## CrusaderFrank (Dec 13, 2019)

Fort Fun Indiana said:


> *First one:
> 
> You meet a girl. She tells you she has one sibling. What are the odds her sibling is also a girl?*


100%


----------



## luchitociencia (Dec 13, 2019)

Fort Fun Indiana said:


> So yes, if you want to scam someone and take some of their money, make a 4 card deck from two red and two black cards. Only allow bets when one card is red. Pull two cards. Keep them  hidden from the mark, but peek at them. If at least one card is red, tell the mark this. Give him even money odds on betting that there are two red cards. Your expectation will be 0.25. If you start with $300, you will end up with $375. Your mark, who started with $300, will end up with $225.



Your riddle was OK until you started with your nonsense. Now your riddle sucks.


----------



## luchitociencia (Dec 13, 2019)

CrusaderFrank said:


> 100%


 That is correct, because being a girl she will say straight she has an older brother.


----------



## Toddsterpatriot (Dec 13, 2019)

Fort Fun Indiana said:


> Wuwei said:
> 
> 
> > Why does the particular door matter?
> ...



*Before opening the door, you were faced with a sample space of 4, equally likely permutations (you can look at them as the first element being behind the first door you open, with the second element being behind the second door you open):

BB
BG
GB
GG*

There are two choices for each room. 

Room 1, boy (B1) or
Room 1, girl (G1)

Room 2, boy (B2) or
Room 2, girl (G2)

Doesn't matter if your friend opens door one and finds G1 or door two and finds G2.
In the first case, the odds of a girl behind door two is 50%.
In the second case, the odds of a girl behind door one is 50%.​


----------



## cnm (Dec 13, 2019)

Toddsterpatriot said:


> *You did not "pick" anything in the riddle, though*
> 
> The first sibling was picked for you in post #1.
> 
> ...


Yes.

The chances of throwing 7 with two dice are 6 in 36

If it is given one die shows 6, for example, the chance of totalling 7 with the second die is one in 6. The permutations are rendered moot.


----------



## Toddsterpatriot (Dec 13, 2019)

cnm said:


> Toddsterpatriot said:
> 
> 
> > *You did not "pick" anything in the riddle, though*
> ...



*If it is given one die shows 6, for example, the chance of totalling 7 with the second die is one in 36. *

No. If the first die shows 6, there is a one-in-six chance the second die will show a 1.


----------



## cnm (Dec 13, 2019)

Fort Fun Indiana said:


> cnm said:
> 
> 
> > You framed the question incorrectly for that solution to apply.
> ...


Stating the gender of one of the children killed it stone dead.


----------



## cnm (Dec 13, 2019)

Toddsterpatriot said:


> cnm said:
> 
> 
> > Toddsterpatriot said:
> ...


Already corrected.


----------



## Wuwei (Dec 13, 2019)

Fort Fun Indiana said:


> BB
> BG
> GB
> GG
> ...



This is the way I look at it. There are just two kids each having one of two properties: B or G.
These are all the possibilities of properties:
BB, BG, GG
BB and GG each have probabilities of 1/4
BG has a probability of 2/4
(Note: BG and GB  are two different ways of saying the same thing: "one of each" That assumes ordering is not involved. I will just use the first label, BG).

If either kid is identified as B, the BB and BG possibilities remain.  
Probabilities BG = 1/2, BB = 1/4.
If either kid is identified as G, the GG and BG possibilities remain.  
Probabilities BG = 1/2, GG = 1/4.

If you normalize the two cases after one kid is identified, you get 2/3 and 1/3.

That is the analysis if ordering is not an issue.
(Maybe.)
.


----------



## Fort Fun Indiana (Dec 13, 2019)

Wuwei said:


> These are all the possibilities of properties:
> BB, BG, GG


Yes, the three combinations. And they have the following probabilities:

BB: 25%
BG: 50%
GG: 25%

Half of the BG combinations are the permutation BG, and half are the permutation GB. 

When you "order" your choice by opening a door, you eliminate all the BB combinations and half the BG combinations (specifically, all the BG permutations). 

Not knowing which door was opened keeps your choice unordered, and you can only eliminate the BB combinations from your sample space. This leaves you with a 66% chance of a BG combination and a 33% chance of a GG combination.


----------



## Fort Fun Indiana (Dec 13, 2019)

cnm said:


> Stating the gender of one of the children killed it stone dead.


Knowing at least one child is a girl is the key to the riddle. If you think that, given only the information at least one child is a girl yields a 50% chance of the other child also being a girl, you are wrong. It is 33%.


----------



## Fort Fun Indiana (Dec 13, 2019)

Toddsterpatriot said:


> You meet a girl.


That is not you "picking" anything, any more than the dealer pulling one card from behind his back and showing it to you is "picking".

Now, if you had the two siblngs in two containers in front of you, then picked one and "met" a girl, that would be different.


----------



## Fort Fun Indiana (Dec 13, 2019)

cnm said:


> If it is given one die shows 6, for example, the chance of totalling 7 with the second die is one in 6. The permutations are rendered moot.



If you roll a die and get 6, you are right that the permutations are then "moot", because there no longer are any permutations or combinations to consider. There is only a sample space of 6 equally likely elements: 1-6.   And yes, the odds of then rolling a total of 7 are one in 6.

Thats the riddle, really. While it must be accurately solved by counting permutations, this is unnecessary work, when you "open one of two doors and find a girl". At this point, you have ordered your two choices, and now all that remains is a sample space of two, equally likely elements: boy or girl. 

*Now consider:

If I roll two dice in a cup and look at them and tell you at least one is a 6 (thus "giving" you the information that one die shows 6), what are the odds the dice total 7? You have to count permutations.

The odds are 11:2. Do you know why?*


----------



## Fort Fun Indiana (Dec 13, 2019)

Fort Fun Indiana said:


> So yes, if you want to scam someone and take some of their money, make a 4 card deck from two red and two black cards. Only allow bets when at least one card is red. Pull two cards. Keep them  hidden from the mark, but peek at them. If at least one card is red, tell the mark this. Give him even money odds on betting that there are two red cards. Your expectation will be 0.33. If you start with $300, you will end up with $400. Your mark, who started with $300, will end up with $200.


Anyone? Anyone up for a game of cards? I'll deal. I'll give you 3:2, instead of even money.


----------



## Fort Fun Indiana (Dec 13, 2019)

*Now consider:

If I roll two dice in a cup and look at them and tell you at least one is a 6 (thus "giving" you the information that one die shows 6), what are the odds the dice total 7? You have to count permutations.

The odds are 11:2. Do you know why?
*
Worth noting:

In the sibling scenario, the odds got worse,  when all the information you had was that at least one is a girl, compared to opening a door to find one (a dice roll).

In this scenario, you end up with better odds if you _don't_ roll a die and instead are divined that at least one of two dice rolled is a 6. 11:2 are betters odds than 6:1.

Let there be six genders instead of two in the sibling riddle, and you will get the same results as the above dice riddle.


----------



## Toddsterpatriot (Dec 13, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > You meet a girl.
> ...



*That is not you "picking" anything,*

She can't be a boy, the choice is made.

That eliminates permutations B1-B2 and B1-G2.
Only two left, G1-B2 and G1-G2.


----------



## Toddsterpatriot (Dec 13, 2019)

Fort Fun Indiana said:


> cnm said:
> 
> 
> > If it is given one die shows 6, for example, the chance of totalling 7 with the second die is one in 6. The permutations are rendered moot.
> ...



*If you roll a die and get 6, you are right that the permutations are then "moot", because there no longer are any permutations or combinations to consider.  There is only a sample space of 6 equally likely elements: 1-6.*

If you meet the first sibling and get a girl, you are right that the permutations are then "moot", because there no longer are any permutations or combinations to consider.  There is only a sample space of 2 equally likely elements: boy or girl.


----------



## Fort Fun Indiana (Dec 13, 2019)

Toddsterpatriot said:


> If you meet the first sibling and get a girl,  you are right that the permutations are then "moot"


False. Because you dont know which sibling you have met. Just as when the dealer holds the two cards behind his back, then hands you one red card. You never got to choose. You didn't order your choice. If the dealer draws two cards (from an infinite deck, for the hall monitors), holds them behind his back, then hands you one red card, the probability that the other card is also red is 1/3.

Now you have retreated to a semantic argument. So it appears that you DO understand that, if you are only given the information that at least one sibling is a girl, the probability of the other also being a girl is 1/3. Else you wouldn't be  focused on calling this a "pick", or a card flip, using the word "meet"..

So, good on you, you do understand. However, "meeting" the girl doesn't mean you have ordered your choice. As I will show in the next post.


----------



## Fort Fun Indiana (Dec 13, 2019)

Two siblings are in a room behind a door, with Toddsterpatriot  waiting outside . One is sent out, and it is a girl. Toddsterpatriot  gets to *meet* her and shake her hand and exchange pleasantries. They don't just *meet*, they go on to get married and have 11 kids.

What is the probability that there remains a girl behind the door?

Correct answer: 1/3.


----------



## Fort Fun Indiana (Dec 14, 2019)

Now, Toddsterpatriot walks into a house with two rooms and a hallway. Two siblings are in the house, one in each room. He opens one door. He *meets* a girl and she spurns him; she plays for the wrong team, or something. What is the probability a girl is behind the other door?

1/2


----------



## Fort Fun Indiana (Dec 14, 2019)

*Now consider:

If I roll two dice in a cup and look at them and tell you at least one is a 6 (thus "giving" you the information that one die shows 6), what are the odds the dice total 7? You have to count permutations.

The odds are 11:2. Do you know why?*


----------



## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > If you meet the first sibling and get a girl,  you are right that the permutations are then "moot"
> ...



*False. Because you dont know which sibling you have met. 
*
I met a girl sibling.

You meet a girl. 

What are the other possible siblings?

*However, "meeting" the girl doesn't mean you have ordered your choice. *

I don't have to order my choice, because I didn't make one.


----------



## Fort Fun Indiana (Dec 14, 2019)

Two siblings are in a room behind a door, with Toddsterpatriot waiting outside . One is sent out, and it is a girl. Toddsterpatriot gets to *meet* her and shake her hand and exchange pleasantries. They don't just *meet*, they go on to get married and have 11 kids.

What is the probability that there remains a girl behind the door?

Correct answer: 1/3.

Now, Toddsterpatriot walks into a house with two rooms and a hallway. Two siblings are in the house, one in each room. He opens one door. He *meets* a girl and she spurns him; she plays for the wrong team, or something. What is the probability a girl is behind the other door?

1/2


----------



## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Two siblings are in a room behind a door, with Toddsterpatriot  waiting outside . One is sent out, and it is a girl. Toddsterpatriot  gets to *meet* her and shake her hand and exchange pleasantries. They don't just *meet*, they go on to get married and have 11 kids.
> 
> What is the probability that there remains a girl behind the door?
> 
> Correct answer: 1/3.



*Two siblings are in a room behind a door, *

Excellent!
Let's look at the possibilities. 
Sibling one is either B1 (boy) or G1 (girl). Sibling two is either B2 (boy) or G2 (girl)

*One is sent out, and it is a girl. *

Excellent!
Let's look at the possibilities. 
She's either G1 or G2, the only two possible girl siblings.

If she's G1, the remaining sibling is either B2 (boy) or G2 (girl).

If she's G2, the remaining sibling is either B1 (boy) or G1 (girl).
​*Correct answer: 1/3.*​
There aren't 3 siblings remaining behind the door, there aren't 3 choices. There aren't 3 possibilities.

One sibling left, 2 possibilities. Still 1/2.​


----------



## Fort Fun Indiana (Dec 14, 2019)

Toddsterpatriot said:


> If she's G1, the remaining sibling is either B2 (boy) or G2 (girl).
> 
> If she's G2, the remaining sibling is either B1 (boy) or G1 (girl).


Yep, three equally possible permutations. 

BG
GB
GG

You're just not getting this.


----------



## Wuwei (Dec 14, 2019)

Fort Fun Indiana said:


> Two siblings are in a room behind a door, with Toddsterpatriot waiting outside . One is sent out, and it is a girl. Toddsterpatriot gets to meet her and shake her hand and exchange pleasantries. They don't just meet, they go on to get married and have 11 kids.
> 
> What is the probability that there remains a girl behind the door?
> 
> Correct answer: 1/3.


Probability = 0. It is neither a boy or girl. It's either a really ticked off woman or a man.... Waiting in a room for a couple of decades is going to do that to you while Todd is out having fun overpopulating the world with your sister who never writes to you.



Fort Fun Indiana said:


> Now, Toddsterpatriot walks into a house with two rooms and a hallway. Two siblings are in the house, one in each room. He opens one door. He meets a girl and she spurns him; she plays for the wrong team, or something. What is the probability a girl is behind the other door?
> 
> 1/2


But seriously folks, I think we are overthinking it. I don't think ordering plays a role.

At the start of this puzzle, the premise is that the genders of the two kids are statistically independent, namely 50:50 for each. Suppose you went through life meeting thousands of parents with their two kids visible or known. You will compile these statistics.
(Notation: OE logically means BG ∪ GB or *O*ne of *E*ach. The order doesn't matter)

BB 25%
OE 50%
GG 25%

So, if for the first time you encounter one kid, a G, and you don't know the gender of the other kid, There can be many circumstances (The first kid is discovered behind a door, walking in the street..., the second kid is locked in a room, or in jail...) You will eliminate the BB possibility and resort to your statistics. You will see that the OE possibility is twice the GG possibility. That means the chances of a second G is half that for OE. So the chance is 1/3 that the second is a G and 2/3 that its an OE (in this case, a boy).

I think.
.


----------



## Fort Fun Indiana (Dec 14, 2019)

Wuwei said:


> But seriously folks, I think we are overthinking it. I don't think ordering plays a role.


And you are wrong. By rolling a die or opening a door, you have ordered your choice of the two elements which constitute the permutation. It's a simple mathematical fact that you cannot get around.

I think tackling the dice riddle i just posted might give you some clarity on this. Note the difference in probability when you roll a die and get 6, versus the probability when someone tells you that at least one of two rolled dice shows a 6.


----------



## Dogmaphobe (Dec 14, 2019)

Damn, and here I thought Fort Fun was a right, proper lefty.  Is he actually implying there are just two genders?

 Shouldn't we, like, try to get him fired from his job or something?


----------



## Fort Fun Indiana (Dec 14, 2019)

Dogmaphobe said:


> Damn, and here I thought Fort Fun was a right, proper lefty.  Is he actually implying there are just two genders?
> 
> Shouldn't we, like, try to get him fired from his job or something?


For the sake of the math, there are two, equally likely genders. If you feel left out, I can add a third, just to include you. Androgynous, or hermaphrodite? You tell me.


----------



## Dogmaphobe (Dec 14, 2019)

Fort Fun Indiana said:


> Dogmaphobe said:
> 
> 
> > Damn, and here I thought Fort Fun was a right, proper lefty.  Is he actually implying there are just two genders?
> ...


I'm selecting hermaphrodite.

It makes things a lot more fun when you tell me to go fuck myself.


----------



## Fort Fun Indiana (Dec 14, 2019)

Dogmaphobe said:


> Fort Fun Indiana said:
> 
> 
> > Dogmaphobe said:
> ...


But less fun for me, when you are going to do it anyway.


----------



## Dogmaphobe (Dec 14, 2019)

Fort Fun Indiana said:


> Dogmaphobe said:
> 
> 
> > Fort Fun Indiana said:
> ...


 how did you know I was hung like a horse and could pull it off?

have people been talking?


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## Fort Fun Indiana (Dec 14, 2019)

Dogmaphobe said:


> Fort Fun Indiana said:
> 
> 
> > Dogmaphobe said:
> ...


Just the neighsayers.


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## Dogmaphobe (Dec 14, 2019)

Fort Fun Indiana said:


> Dogmaphobe said:
> 
> 
> > Fort Fun Indiana said:
> ...


horses do say neigh, after all..


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > If she's G1, the remaining sibling is either B2 (boy) or G2 (girl).
> ...




*BG*​*GB*​
Why are you repeating a combination here?​


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## Toddsterpatriot (Dec 14, 2019)

Wuwei said:


> Fort Fun Indiana said:
> 
> 
> > Two siblings are in a room behind a door, with Toddsterpatriot waiting outside . One is sent out, and it is a girl. Toddsterpatriot gets to meet her and shake her hand and exchange pleasantries. They don't just meet, they go on to get married and have 11 kids.
> ...


*At the start of this puzzle, the premise is that the genders of the two kids are statistically independent, namely 50:50 for each. *

And that's your answer. 50%.

*So, if for the first time you encounter one kid, a G, and you don't know the gender of the other kid,*

_the genders of the two kids are statistically independent_

You have two independent choices. 50/50.

*You will eliminate the BB possibility and resort to your statistics.*

Before you met the one kid, your choices were:

B1-G2
B1-B2
G1-B2
G1-G2

The one you meet can be B1 (boy first) or G1 (girl first).

When you meet a girl first, you eliminate B1-G2 AND B1-B2, or 50% of your original 4 possibilities.

Only 2 possibilities left......G1-G2 and G1-B2.
The odds that the second sibling is a girl is 1 out of 2.

The only way Fort Fun's riddle works out to 1/3 is if you have 4 children, two boys and two girls.

One girl leaves the house to meet you.
Now you have 2 boys and 1 girl remaining.
In this case, the odds of picking another girl is 1/3.


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## OldLady (Dec 14, 2019)

I still say 16.


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## Dogmaphobe (Dec 14, 2019)

Toddsterpatriot said:


> Wuwei said:
> 
> 
> > Fort Fun Indiana said:
> ...


Yep.

It's the first thing you learn when studying probability. If you flip a coin 4 times and it comes up heads each time, there is no reason to believe it will come out tails the next time to make up for the fact three previous 4 were heads. Ultimately, the odds will even out over time, assuming there are no factors affecting the outcome. In the short term, however, there are still even odds.

As to this specific question, one would have to determine first that there are no factors from the first pregnancy affecting the second like hormonal changes caused by the gender of the first child.


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## RandomPoster (Dec 14, 2019)

A 12 year old girl has a 9 year old sibling.  What are the odds the sibling is a girl?  1 in 2

  A 12 year old girl has a 15 year old sibling.  What are the odds the sibling is a girl?  1 in 2

  A 12 year old girl has a sibling of unknown age.  What are the odds the sibling is a girl?  1 in 3

​


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## Toddsterpatriot (Dec 14, 2019)

RandomPoster said:


> A 12 year old girl has a 9 year old sibling.  What are the odds the sibling is a girl?  1 in 2
> 
> A 12 year old girl has a 15 year old sibling.  What are the odds the sibling is a girl?  1 in 2
> 
> ...



Ummmmmmm……..


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## Fort Fun Indiana (Dec 14, 2019)

Toddsterpatriot said:


> Fort Fun Indiana said:
> 
> 
> > Toddsterpatriot said:
> ...


I'm not.


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## Fort Fun Indiana (Dec 14, 2019)

RandomPoster said:


> A 12 year old girl has a 9 year old sibling.  What are the odds the sibling is a girl?  1 in 2
> 
> A 12 year old girl has a 15 year old sibling.  What are the odds the sibling is a girl?  1 in 2
> 
> ...


Correct!

This guy gets it.


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## Fort Fun Indiana (Dec 14, 2019)

Dogmaphobe said:


> Toddsterpatriot said:
> 
> 
> > Wuwei said:
> ...


That's not the correct way of looking at the problem


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## Fort Fun Indiana (Dec 14, 2019)

*If I roll two dice in a cup and look at them and tell you at least one is a 6 (thus "giving" you the information that one die shows 6), what are the odds the dice total 7? You have to count permutations.

The odds are 11:2. Do you know why?*


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## RandomPoster (Dec 14, 2019)

Fort Fun Indiana said:


> *If I roll two dice in a cup and look at them and tell you at least one is a 6 (thus "giving" you the information that one die shows 6), what are the odds the dice total 7? You have to count permutations.
> 
> The odds are 11:2. Do you know why?*



  There are 36 different permutations of 2 six sided dice being rolled.  Specifying that at least 1 is a 6 eliminates 25 of them.  Of the remaining 11 permutations, 2 permutations (specifically 1 + 6 and 6 + 1), will equal 7.


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## Dogmaphobe (Dec 14, 2019)

Fort Fun Indiana said:


> Dogmaphobe said:
> 
> 
> > Toddsterpatriot said:
> ...


I am dogmaphobe.

the way I look at every problem is the correct way.

now, the question you should be asking is this:   given that we have seen a twenty seven fold increase in the number of genders in just a few years, and given our current and projected population expansion, in what year will we all be able to enjoy the myriad benefits of having a unique gender all to our own?


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > Fort Fun Indiana said:
> ...



The girl already walked out. the only possibilities are girl left inside or boy left inside.

BG and GB is one combination, before she walks out. GG is the other.


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## Fort Fun Indiana (Dec 14, 2019)

Toddsterpatriot said:


> The girl already walked out. the only possibilities are girl left inside or boy left inside.


Which are not equally likely possibilities. You are stuck on the same error you have been stuck on since square one.


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## Fort Fun Indiana (Dec 14, 2019)

Toddsterpatriot said:


> BG and GB is one combination,


Right, with the number of those two, distinct permutations totalling the number of BG combinations in the sample space.


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## Fort Fun Indiana (Dec 14, 2019)

Dogmaphobe said:


> I am dogmaphobe.
> 
> the way I look at every problem is the correct way.


Then getting an F on the discrete math test is "the right way", where you are from.


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## Fort Fun Indiana (Dec 14, 2019)

RandomPoster said:


> Fort Fun Indiana said:
> 
> 
> > *If I roll two dice in a cup and look at them and tell you at least one is a 6 (thus "giving" you the information that one die shows 6), what are the odds the dice total 7? You have to count permutations.
> ...


Yep!


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > The girl already walked out. the only possibilities are girl left inside or boy left inside.
> ...



*Which are not equally likely possibilities.*

Two potential siblings, two potential possibilities. Equally likely possibilities.


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## Dogmaphobe (Dec 14, 2019)

Fort Fun Indiana said:


> Dogmaphobe said:
> 
> 
> > I am dogmaphobe.
> ...


you need to take a class in probability, kid.


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > BG and GB is one combination,
> ...



*Right, with the number of those two, distinct permutations*

I already explained the 4 initial permutations collapse to two possibilities once you met the girl.

B1-B2
B1-G2
G1-B2
G1-G2

Once you meet a girl (G1), the two permutations involving B1 evaporate.

Who is left? B2 or G2.


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## Fort Fun Indiana (Dec 14, 2019)

Toddsterpatriot said:


> Two potential siblings, two potential possibilities


Which does not mean they are equal. Again, same error since square one.

Toddsterpatriot  , the weatherman:

"Only two possibilities exist today: it will rain, or it will not rain. Therefore, the chance of rain is 50%."

(He was fired that afternoon)


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## Fort Fun Indiana (Dec 14, 2019)

Toddsterpatriot said:


> I already explained the 4 initial permutations collapse to two possibilities once you met the girl.


And you are just as wrong now as you were the first time. Repeating it does not make it more correct.


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > Two potential siblings, two potential possibilities
> ...



*"Only two possibilities exist today: it will rain, or it will not rain. Therefore, the chance of rain is 50%."*

Rain isn't siblings.


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > I already explained the 4 initial permutations collapse to two possibilities once you met the girl.
> ...



How many initial permutations in your world?


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## Fort Fun Indiana (Dec 14, 2019)

Toddsterpatriot said:


> Fort Fun Indiana said:
> 
> 
> > Toddsterpatriot said:
> ...


Already answered. Several times, actually. Instead of asking others to repeat themselves for you, you should just go back to page one and start reading again.


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > Fort Fun Indiana said:
> ...



Why does meeting the girl only eliminate one of your permutations?


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## Fort Fun Indiana (Dec 14, 2019)

Toddsterpatriot said:


> Fort Fun Indiana said:
> 
> 
> > Toddsterpatriot said:
> ...


Also already explained. Several times.


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > Fort Fun Indiana said:
> ...



And I already explained your error, several times. 

You're still confused though.


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## RandomPoster (Dec 14, 2019)

*  The Three Prisoners Problem*

Three Prisoners problem - Wikipedia

*Problem*
Three prisoners, A, B, and C, are in separate cells and sentenced to death. The governor has selected one of them at random to be pardoned. The warden knows which one is pardoned, but is not allowed to tell. Prisoner A begs the warden to let him know the identity of one of the two who are going to be executed. "If B is to be pardoned, give me C's name. If C is to be pardoned, give me B's name. And if I'm to be pardoned, secretly flip a coin to decide whether to name B or C."

The warden tells A that B is to be executed. Prisoner A is pleased because he believes that his probability of surviving has gone up from 1/3 to 1/2, as it is now between him and C. Prisoner A secretly tells C the news, who reasons that A's chance of being pardoned is unchanged at 1/3, but he is pleased because his own chance has gone up to 2/3. Which prisoner is correct?

*Solution*
The answer is that prisoner A did not gain any information about his own fate, since he already knew that the warden would give him the name of someone else. Prisoner A, prior to hearing from the warden, estimates his chances of being pardoned as 1/3, the same as both B and C. As the warden says B will be executed, it is either because C will be pardoned (1/3 chance), or A will be pardoned (1/3 chance) _and_ the B/C coin the warden flipped came up B (1/2 chance; for a total of a 1/6 chance B was named because A will be pardoned). Hence, after hearing that B will be executed, the estimate of A's chance of being pardoned is half that of C. This means his chances of being pardoned, now knowing B is not, again are 1/3, but C has a 2/3 chance of being pardoned.

*Explanation*
  "Prisoner A only has a 1/3 chance of pardon. Knowing whether "B" or "C" will be executed does not change his chance. After he hears B will be executed, Prisoner A realizes that if he will not get the pardon himself it must only be going to C. That means there is a 2/3 chance for C to get a pardon"

  This raises the question.  What if C had asked the warden instead of A.  Would the answer then be?

  "Prisoner C only has a 1/3 chance of pardon. Knowing whether "B" or "A" will be executed does not change his chance. After he hears B will be executed, Prisoner C realizes that if he will not get the pardon himself it must only be going to A. That means there is a 2/3 chance for A to get a pardon"

  If you ever find yourself in that position, I guess you shouldn't be in a hurry to ask the warden any questions.


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## Fort Fun Indiana (Dec 14, 2019)

Toddsterpatriot said:


> Fort Fun Indiana said:
> 
> 
> > Toddsterpatriot said:
> ...


Sorry Todd, you got an f on that quiz.

So, see if you can avoid making the same, elementary errors with the dice riddle. I doubt it.


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > Fort Fun Indiana said:
> ...







G1 already met me.

BG isn't possible anymore.


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## Fort Fun Indiana (Dec 14, 2019)

Toddsterpatriot said:


> G1 already met me.


False. You don't have all that info. Already explained.


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > G1 already met me.
> ...







Well, if G2 already met me, GB isn't possible. Already explained.


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## Fort Fun Indiana (Dec 14, 2019)

Toddsterpatriot said:


> Well, if G2 already met me,


Sorry, you dont have that much info, either. As has been explained. You are stuck on the same error since square one.

Sorry Todd, all the information you need to learn that you are wrong is already in the thread. You will have to just live with being wrong. I'm not going to respond further to you regardiing the first riddle.

Try out the dice riddle. See if you can arrive at the correct answer.


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > Well, if G2 already met me,
> ...



*Sorry, you dont have that much info, either.*

I don't need it....since the chances of either are 50/50.

Only two possibilities for the sibling you meet only two for the unmet sibling.


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## Fort Fun Indiana (Dec 14, 2019)

Dogmaphobe said:


> you need to take a class in probability, kid.


Actually, i helped teach them....kid. And,unlike the other classes I helped teach or took myself, these subjects were the only ones wherein students would continue to insist they were correct, despite being shown and told they were wrong by both the textbooks and the professors and the assistant professors. I did not see this phenomenon in the other math subjects, at least to the same degree or with the same fervor.

I remember one poor kid in my office threatening to take his case to the faculty council. I offered to buy him lunch and drive him there. It was over a similar counting problem that caused him to get a poor test grade. His fundamental error was similar to the one being made by those who are getting the wrong answer to the riddle.

So the intransigence we see in this thread is not new or special.


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > Well, if G2 already met me,
> ...



*Try out the dice riddle. See if you can arrive at the correct answer. *

_If I roll two dice in a cup and look at them and tell you at least one is a 6 (thus "giving" you the information that one die shows 6), what are the odds the dice total 7? You have to count permutations.

The odds are 11:2. Do you know why?_

I roll two dice in a cup. I pull one out, it is a 6.
What are the odds the other is a 6?


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## Fort Fun Indiana (Dec 14, 2019)

Toddsterpatriot said:


> I roll two dice in a cup. I pull one out, it is a 6.
> What are the odds the other is a 6?


No, you're confused. I ask the riddle. You answer it. 

Did you get 11:2?


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Toddsterpatriot said:
> 
> 
> > I roll two dice in a cup. I pull one out, it is a 6.
> ...



*No, you're confused. I ask the riddle. You answer it. *

When the girl met me, she was "pulled out of the cup"


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## Fort Fun Indiana (Dec 14, 2019)

I think todd might be afraid to answer the dice riddle.


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> I think todd might be afraid to answer the dice riddle.



No more afraid than you are to admit your sibling error.


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## Fort Fun Indiana (Dec 14, 2019)

Like i said....


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Like i said....



When you met the girl, there is only one "die" left to roll.


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## Fort Fun Indiana (Dec 14, 2019)

Anyone else?


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## Toddsterpatriot (Dec 14, 2019)

Fort Fun Indiana said:


> Anyone else?








One other sibling. Two possibilities.


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## RandomPoster (Dec 14, 2019)

I kind of like this one.

A woman who is sitting on a red couch has an adopted sibling who is sitting on a green couch.  What are the odds the sibling is a female?  1 out of 2
A woman who is sitting on a red couch has an adopted sibling who is sitting on a red couch.  What are the odds the sibling is a female?  1 out of 3


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## Fort Fun Indiana (Dec 14, 2019)

RandomPoster said:


> I kind of like this one.
> 
> A woman who is sitting on a red couch has an adopted sibling who is sitting on a green couch.  What are the odds the sibling is a female?  1 out of 2
> A woman who is sitting on a red couch has an adopted sibling who is sitting on a red couch.  What are the odds the sibling is a female?  1 out of 3


Ha! What you are demonstrating is actually very important. There are many different ways that the sibling choice can become "ordered". Age, which door was opened first, the color of the couch, for example. All have the same effect.


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## RandomPoster (Dec 14, 2019)

Fort Fun Indiana said:


> RandomPoster said:
> 
> 
> > I kind of like this one.
> ...



  Of course you have to be careful not to specify anything that could affect the likelihood of gender, such as height, weight, hair color (girls tend to have lighter hair), getting braces, losing an eye in a childhood accident (on account of boys being more likely to engage in dangerous behavior) etc.


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## Fort Fun Indiana (Dec 14, 2019)

RandomPoster said:


> Actually, the odds of the original question are 33.858 percent when you factor in the likelihood of adoption when calculating the likelihood of the sibling being an identical twin.  .8 percent of births are identical twins
> 
> 
> Fort Fun Indiana said:
> ...


Right, but it is assumed that the probability of getting a girl or boy at each birth or adoption event is 50% each. I felt it extraneous to state this, but i clarified later. Similarly, I have occassionally included "infinite deck" in the card problems.


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## RandomPoster (Dec 14, 2019)

Fort Fun Indiana said:


> RandomPoster said:
> 
> 
> > Actually, the odds of the original question are 33.858 percent when you factor in the likelihood of adoption when calculating the likelihood of the sibling being an identical twin.  .8 percent of births are identical twins
> ...



  I believe you are correct about the adoption not affecting things unless you factor in the microscopic likelihood of a child having an identical twin who died and was replaced by an adopted sibling possibly lowering the percentage chance of the child having a living sibling who is an identical twin from .8% to slightly less, which would require at a minimum knowing the child's age and nationality among any number of other factors.


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## Fort Fun Indiana (Dec 14, 2019)

Assume all dogs are either white, brown, or black, with equal probability of each at each birth, regardless of the color of the parents. 

The odds of all 3 puppies in a litter being white are 27:1.

Someone hands you a white puppy. You are asked what the odds are that his two littermates are also both white.

Answer: 19:1

This time, instead, you go into a pet store and see three occupied cages with dogs hiding under blankets. You pick one at random, and out from under the blanket comes a white puppy. What are the odds that the other two puppies are both also white?

9:1


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## Dogmaphobe (Dec 14, 2019)

Fort Fun Indiana said:


> Dogmaphobe said:
> 
> 
> > you need to take a class in probability, kid.
> ...


Glad to meet you, "professor".

I am a nuclear physicist by day and a Chippendales dancer by night, myself.

the fact remains that there is a roughly 50/50 chance of either sex with all bi

I won't confuse you by mentioning there are actually 1.08 boys born for every girl since you are playing make believe and all.


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## Dogmaphobe (Dec 14, 2019)

Fort Fun Indiana said:


> Assume all dogs are either white, brown, or black, with equal probability of each at each birth, regardless of the color of the parents.
> 
> The odds of all 3 puppies in a litter being white are 27:1.
> 
> ...


In a couple of years when you are a bit older, you should ask the teacher to explain dominant and recessive genes.


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## Fort Fun Indiana (Dec 14, 2019)

Dogmaphobe said:


> the fact remains that there is a roughly 50/50 chance of either sex


...at birth. Yes, thank you, that is one of the assumptions before the riddle even begins. Welcome to square one. We dont need you to clarify the starting conditions. That has already been done. But an honest attempt by you to understand why you are ass backwards wrong about the answer to the riddle would be appreciated.


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## Fort Fun Indiana (Dec 14, 2019)

Dogmaphobe said:


> Fort Fun Indiana said:
> 
> 
> > Assume all dogs are either white, brown, or black, with equal probability of each at each birth, regardless of the color of the parents.
> ...


Yes, often students will try to distract or cut jokes, when asked to answer a question. This is always a sure sign they do not understand the material.


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## Dogmaphobe (Dec 14, 2019)

Fort Fun Indiana said:


> Dogmaphobe said:
> 
> 
> > Fort Fun Indiana said:
> ...


I doubt that you are fooling too many people here.

the fact remains that when one flips a coin, there is a 50 % chance it will come up heads. If you happen to flip tails, the chance that the next one will be heads is still 50%.


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## Fort Fun Indiana (Dec 14, 2019)

Dogmaphobe said:


> Fort Fun Indiana said:
> 
> 
> > Dogmaphobe said:
> ...



Which nobody has disputed. You are adding zero new information, and you are not arguing that your incorrect answer to the riddle is correct. No, your opinion of me does not make your wrong answer correct.

Just for laughs: have a crack at the dice riddle.


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## Dogmaphobe (Dec 14, 2019)

Fort Fun Indiana said:


> Dogmaphobe said:
> 
> 
> > Fort Fun Indiana said:
> ...


my answer is not incorrect.

given a 50% probability of either gender at birth and assuming there are to genders, if the first birth is a girl, her sibling can STILL only be either a boy or a girl and as we have already established a 50 probability at birth, the answer is undeniably a 50 % probability.

there would have to be three genders, not two to have a 33 % chance.

you are not a professor of anything.  you are just making shit up to assuage some sort of inferiority complex.


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## Fort Fun Indiana (Dec 14, 2019)

Dogmaphobe said:


> my answer is not incorrect.


Wrong.  If your answer to the riddle is not 1/3, it is incorrect.


Dogmaphobe said:


> given a 50% probability of either gender at birth and assuming there are to genders, if the first birth is a girl, her sibling can STILL only be either a boy or a girl and as we have already established a 50 probability at birth, the answer is undeniably a 50 % probability.
> 
> there would have to be three genders, not two to have a 33 % chance.


Wrong, as already explained, in depth, in this thread.  That's why the riddle was posed in the first place: Because I knew so many would get it wrong. I also knew almost all of those who got it wrong would continue to insist they were correct, as these people generally won't have good mathematics or logic backgrounds. You didn't disappoint.


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## RandomPoster (Dec 14, 2019)

A 20 year old man has one sibling.  Both he and his sibling are adopted.  He tells you he was born on a Monday.

If he is telling the truth what are the odds the sibling is male?
6/14 + 1/21 = 140 / 294 = 70/147 about equal to 47.6%

If he is lying, what are the odds the sibling is male?
1/3 about equal to 33.3%


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## RandomPoster (Dec 14, 2019)

Here's something weird.

 A 20 year old man has one sibling. Both he and his sibling are adopted.  Learning his Social Security Number will decrease the likelihood of being able to correctly guess his sibling's gender.


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## OldLady (Dec 15, 2019)

RandomPoster said:


> Here's something weird.
> 
> A 20 year old man has one sibling. Both he and his sibling are adopted.  Learning his Social Security Number will decrease the likelihood of being able to correctly guess his sibling's gender.


Because you'll be in federal prison?


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## Fort Fun Indiana (Dec 15, 2019)

RandomPoster said:


> A 20 year old man has one sibling. Both he and his sibling are adopted. He tells you he was born on a Monday.
> 
> If he is telling the truth what are the odds the sibling is male?
> 6/14 + 1/21 = 140 / 294 = 70/147 about equal to 47.6%


RandomPoster

I think you may have miscounted, here.

There are 27 possible permutations that include a male born on a Monday. Of these 27 permutations, 13 include two males. This yields a probability of 48.15%.




RandomPoster said:


> If he is lying, what are the odds the sibling is male?
> 1/3 about equal to 33.3%



If he is lying -- taken to mean he was NOT born on a Monday -- then there are 132 permutations that include at least one male NOT born on a monday. Of these,  48 include two males. This yields a probability of 36.36%.

*For everyone else:* the age of the man and the fact that he and his sibling are adopted are extraneous bits of info. This is assuming that the probability of a given gender at birth and at time of adoption are both 50%.


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## Frannie (Dec 15, 2019)

Fort Fun Indiana said:


> *First one:
> 
> You meet a girl. She tells you she has one sibling. What are the odds her sibling is also a girl?*


Let me guess the sibling is a liberal who identifies as a toaster


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## Fort Fun Indiana (Dec 15, 2019)

Frannie said:


> Fort Fun Indiana said:
> 
> 
> > *First one:
> ...


I'm pretty sure that's as close as you will ever get to the correct answer. So yeah, have fun with it!


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## Frannie (Dec 15, 2019)

Fort Fun Indiana said:


> Frannie said:
> 
> 
> > Fort Fun Indiana said:
> ...


Which came first, the supermarket, the truck, the farm, the chicken or the egg

Enjoy and neither chickens or eggs grow in ponds unlike your family of salamanders


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## Fort Fun Indiana (Dec 15, 2019)

Frannie said:


> Which came first, the supermarket, the truck, the farm, the chicken or the egg


Simple: the egg. Eggs appeared on Earth long before there were chickens.  Got any other softball questions?


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## Frannie (Dec 15, 2019)

Fort Fun Indiana said:


> Frannie said:
> 
> 
> > Which came first, the supermarket, the truck, the farm, the chicken or the egg
> ...


Yawn, ok I will be more specific because only retards know the answer so you clearly will.

Which came first, the supermarket, the truck, the farm, the chicken or the egg the first chicken hatched from.

Be the town retard and tell us you know


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## Fort Fun Indiana (Dec 15, 2019)

Frannie said:


> Which came first, the supermarket, the truck, the farm, the chicken or the egg the chicken hatched from.


Well, that depends. In what year did the chicken hatch?

I don't think posing riddles is a strong suit for you. Maybe you should just stop for now and try to tackle one of the riddles that has already been posted.


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## ding (Dec 16, 2019)

Fort Fun Indiana said:


> Frannie said:
> 
> 
> > Which came first, the supermarket, the truck, the farm, the chicken or the egg the chicken hatched from.
> ...


You have a glass of apple juice and a glass of water. You take a teaspoon of the apple juice and mix it into the water.  Then you take a teaspoon from the water glass and mix it into the apple juice glass.

Which has more of the other liquid in it?  Does the apple juice glass have more water in it or does the water glass have more apple juice in it?

Ignore the fact that the apple juice contains water for this thought experiment.


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## RandomPoster (Dec 16, 2019)

ding said:


> Fort Fun Indiana said:
> 
> 
> > Frannie said:
> ...



  I believe they would both have the same.

  Let's say they were large glasses with 100 teaspoons in each.  At the end, the apple juice would contain 99 + 1/101 teaspoons of apple juice and 100/101 teaspoons of water.  The glass of water would contain 99 + 1/101 teaspoons of water and 100/101 teaspoons of apple juice.


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## Fort Fun Indiana (Dec 16, 2019)

ding said:


> Fort Fun Indiana said:
> 
> 
> > Frannie said:
> ...


Well, trying to decide if it is a semantics riddle.


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## Fort Fun Indiana (Dec 16, 2019)

RandomPoster said:


> I believe they would both have the same.


Read it again. Word trick, maybe...


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## ding (Dec 16, 2019)

RandomPoster said:


> ding said:
> 
> 
> > Fort Fun Indiana said:
> ...


That is correct. Well done. 

That’s a very nice material balance. 

What if the glasses did not contain the same starting volumes?  What would the answer be then?


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## ding (Dec 16, 2019)

Fort Fun Indiana said:


> ding said:
> 
> 
> > Fort Fun Indiana said:
> ...


It’s not.

While you were deciding someone else solved it.


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## ding (Dec 16, 2019)

Fort Fun Indiana said:


> RandomPoster said:
> 
> 
> > I believe they would both have the same.
> ...


Nope.


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## Fort Fun Indiana (Dec 16, 2019)

ding said:


> It’s not.


Well,I had the mathematical solution

(RandomPoster  : dont make it so hard in yourself! Just start with only one teaspoon of liquid in each glass.  )

BUT

You moved a teaspoon of liquid from the first glass. Then, you only moved a teaspoon from the second glass. Not a teaspoon of liquid. Just the spoon itself.

So, an alternative, correct answer would be that the water glass has more apple juice than the apple juice has water.

Hey, don't blame me. Blame all the people posting semantics riddles.


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## RandomPoster (Dec 16, 2019)

ding said:


> RandomPoster said:
> 
> 
> > ding said:
> ...



  I believe the same.  I am running the numbers to verify, except is looking that way for now.


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## ding (Dec 16, 2019)

RandomPoster said:


> ding said:
> 
> 
> > RandomPoster said:
> ...


Ok, I’ll wait for your final answer before commenting.


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## Fort Fun Indiana (Dec 16, 2019)

RandomPoster said:


> I believe the same.


Chyep,if the answer is expressed as a percentage.

If expressed an an absolute amount...no, not the same.

EDIT: wait,yes it is

Good one!


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## RandomPoster (Dec 16, 2019)

Fort Fun Indiana said:


> ding said:
> 
> 
> > It’s not.
> ...



  I started with 0 and 1, then 1 and 2, then 2 and 3, etc. to see if it holds.  It appears to.  I wrote a quick function and ran variations of it.


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## RandomPoster (Dec 16, 2019)

NNN


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## Toddsterpatriot (Jan 11, 2022)

Fort Fun Indiana said:


> No, that is incorrect. There are, in fact, four permutations of two children and two genders, not three. There are three combinations. But four permutations.
> 
> BB
> BG
> ...



_But you are correct that the "turned the card to your left" messes with the answer and makes it 50%, in scenarios 3 and 4 . But, this is for a slightly different (yet also similar) reason than the answers to 1 and 2 being 50%.

I should have said, "I tell you at least one card is red". Then, the odds both are red are 1/3. This would have made scenarios 3 and 4 analogous to rhe sibling riddle, which was my original intent._

*But you are correct that the "you meet a girl" messes with the answer and makes it 50%.*

Glad  to clear up your confusion.


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