Riddles

[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?

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.

Yep!

 

[Toddsterpatriot]

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.

Which are not equally likely possibilities.

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

 

[Dogmaphobe]

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.

you need to take a class in probability, kid.

 

[Toddsterpatriot]

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.

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.

 

[Fort Fun Indiana]

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)

 

[Fort Fun Indiana]

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.

 

[Toddsterpatriot]

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)

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

Rain isn't siblings.

 

[Toddsterpatriot]

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.

How many initial permutations in your world?

 

[Fort Fun Indiana]

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.

How many initial permutations in your world?

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.

 

[Toddsterpatriot]

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.

How many initial permutations in your world?

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.

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

 

[Fort Fun Indiana]

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.

How many initial permutations in your world?

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.

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

Also already explained. Several times.

 

[Toddsterpatriot]

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.

How many initial permutations in your world?

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.

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

Also already explained. Several times.

And I already explained your error, several times.

You're still confused though.

 

[RandomPoster]

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.

 

[Fort Fun Indiana]

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

How many initial permutations in your world?

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.

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

Also already explained. Several times.

And I already explained your error, several times.

You're still confused though.

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.

 

[Toddsterpatriot]

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.

How many initial permutations in your world?

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.

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

Also already explained. Several times.

G1 already met me.

BG isn't possible anymore.

 

[Fort Fun Indiana]

G1 already met me.

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

 

[Toddsterpatriot]

G1 already met me.

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

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

 

[Fort Fun Indiana]

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.

 

[Toddsterpatriot]

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.

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.

 

[Fort Fun Indiana]

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.