Riddles

[Fort Fun Indiana]

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.

I'm selecting hermaphrodite.

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

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

 

[Dogmaphobe]

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.

I'm selecting hermaphrodite.

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

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

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

have people been talking?

 

[Fort Fun Indiana]

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.

I'm selecting hermaphrodite.

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

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

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

have people been talking?

Just the neighsayers.

 

[Dogmaphobe]

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.

I'm selecting hermaphrodite.

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

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

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

have people been talking?

Just the neighsayers.

horses do say neigh, after all..

 

[Toddsterpatriot]

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.

BG​

GB​

Why are you repeating a combination here?​

 

[Toddsterpatriot]

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.

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 One of Each. 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.
.

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.

 

[OldLady]

I still say 16.

 

[Dogmaphobe]

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.

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 One of Each. 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.
.

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.

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.

 

[RandomPoster]

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

​

 

[Toddsterpatriot]

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

​

Ummmmmmm……..

 

[Fort Fun Indiana]

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.

BG
GB

Why are you repeating a combination here?​

I'm not.

 

[Fort Fun Indiana]

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

​

Correct!

This guy gets it.

 

[Fort Fun Indiana]

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.

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 One of Each. 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.
.

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.

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.

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

 

[Fort Fun Indiana]

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?

 

[RandomPoster]

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.

 

[Dogmaphobe]

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.

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 One of Each. 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.
.

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.

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.

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

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?

 

[Toddsterpatriot]

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.

BG
GB

Why are you repeating a combination here?​

I'm not.

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.

 

[Fort Fun Indiana]

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.

 

[Fort Fun Indiana]

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.

 

[Fort Fun Indiana]

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.