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bonanova

Question

A magician blindfolded seven subjects, then for each of them he rolled a

pair of fair dice and asked them the probability of their dice totaling 7.

He said to the first, here's a hint: truthfully, at least one of your dice shows a 6.

The subject counted 11 cases of at least one 6, two of which, 1-6 and 6-1, total 7.

So he answered, my chances of having 7 are 2/11. Very good, said the magician.

He then said to the second, I've looked at your dice, and at least one of them is a 5.

This subject counted 11 cases of at least one 5, of which 2-5 and 5-2 made 7.

So he answered 2/11. Very good, said the magician.

He told the third subject, I see at least one 4 on your dice.

That subject also found 11 cases of 4, of which 3-4 and 4-3 made 7.

So he answered 2/11. Very good, said the magician.

The next subject was told at least one of his dice was a 3.

Like the others, he found 11 cases, and of them only 4-3 and 3-4 were favorable.

So he answered, my chances of having 7 are 2/11. Very good said the magician.

The next two were told their dice showed at least one 2 and one 1, respectively.

They found 5-2, 2-5 for one, and 6-1, 1-6 for the other, among 11 cases gave 7.

They both answered their chances of 7 were 2/11. Very good said the magician.

The seventh subject had been listening to all of this. And before the magician

could speak, he said, I don't need a hint. I know that you're going to tell me

some number appears on at least one of my dice. And you've already confirmed

what the right answer is in each case. So whatever you were going to say,

I know most certainly what the odds probability of my dice totaling 7 are is.

My answer is 2/11.

But we know his odds are probability is 1/6, right?

Edited by bonanova
Scraff's observation

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I have two possible solutions (not sure which if any is correct):

1. You have to count the dice possibilities of there being a 1 and a 1 twice, as they can be in each others positions, meaning the magician was wrong or being deceitful by saying "very good". The probability is 1/6.

2. If at least one of your dice is a 1, we then know that the other die is from 1 to 6. Only the 6 works, so the probability is 1/6. The magician was still wrong or deceitful.

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The probability should be 1/6. But I'm not sure why it isn't...I believe that you have to count the double numbers (like 5 and 5) twice...

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Of course, another way to look at it is...

Consider the two dice A and B. And at least one of them is a 6.

The probability of A being a 6 is 1/2, and then the probability of making a 7 (B being a 1) is 1/6. So (1/2)(1/6)=1/12

Now the probability of B being a 6 is 1/2, and then the probability of making a 7 (A being a 1) is 1/6. So (1/2)(1/6)=1/12

Now the probability of A and B BOTH being a 6 is probably 1/6 (since one of them must be a 6), but regardless of the initial odds, the probability of that making a 7 is 0. So (1/6)(0)=0.

Now 1/12+1/12+0=1/6, which is the actual probability of having a 7.

EDIT: Clarified "probability"

Edited by rossbeemer

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But, doesn't counting 6-6 twice [to get 12 cases instead of 11] say that

[1] Die A=6 and die B=6

[2] Die B=6 and die A=6

are somehow different?

Subject #1 [at least one of whose dice, call them A and B, was a 6] reasoned these were the possible cases:

A-B = 6-1, 6-2, 6-3, 6-4, 6-5, 6-6, 5-6, 4-6, 3-6, 2-6 and 1-6; of which the first and last total 7.

Similarly for Subjects 2-6.

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But, doesn't counting 6-6 twice [to get 12 cases instead of 11] say that

[1] Die A=6 and die B=6

[2] Die B=6 and die A=6

are somehow different?

Subject #1 [at least one of whose dice, call them A and B, was a 6] reasoned these were the possible cases:

A-B = 6-1, 6-2, 6-3, 6-4, 6-5, 6-6, 5-6, 4-6, 3-6, 2-6 and 1-6; of which the first and last total 7.

Similarly for Subjects 2-6.

given that you have one six... doesn't that increase the probability that you are looking at the pair 6-6 compared to the other combinations? Methinks the maths is similar to that problem where we have two cards, one with two black sides and one with one black and one white...

Or is it similar to monty hall where the magician's knowledge of the two dice has a counterintuitive effect on the odds?..

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Scraff's second point was the closest I've seen to the solution to this mystery. While it is true that there are 11 possible combinations of two dice where one of the dice shows a particular number, the relevant information is that one of those dice is locked and therefore there are only 6 possibilities for the other die.

Your 11 combinations are as follows:

1 X

2 X

3 X

4 X

5 X

6 X

X 1

X 2

X 3

X 4

X 5

X 6

and one of the above has to be removed because it is a duplicate, but all 12 cases are equally likely rolls.

Now, you can look at this in two ways:

1. As you can see in my listing of 12 equally likely cases above, X X is twice as likely to appear as any single combination X Y or Y X. If you considered X Y and Y X to be equivalent to each other, the number of cases reduces to 6, of which only one has a sum of 7. So your probability is 1/6.

2. A better way to look at is like this: The selection of one die to be X means that all the variability is in the other die. It doesn't matter if X is showing on die 1 or die 2 - in order to get a total of 7, the thing that matters is the value on the "other" die. So the problem reduces to this: given that one die shows X, what is the probability that the other die displays a value such that the sum of the two is 7? With that as the problem, you can see that there are 6 possible values for the "other" die and the probability of rolling a combined 7 is 1/6, which is consistent with the overall probability of rolling a 7 (6 possibilities / 36 cases).

The fact that there are 11 cases where X appears on one of the two dice is misleading at best, and lying with statistics at worst (lies, d--n lies, & statistics!). All you really need to do to show the irrelevance of that fact is to see that there are only 36 possible combinations of dice rolled, not 6 * 11 = 66.

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Scraff's second point was the closest I've seen to the solution to this mystery. While it is true that there are 11 possible combinations of two dice where one of the dice shows a particular number, the relevant information is that one of those dice is locked and therefore there are only 6 possibilities for the other die.

Your 11 combinations are as follows:

1 X

2 X

3 X

4 X

5 X

6 X

X 1

X 2

X 3

X 4

X 5

X 6

and one of the above has to be removed because it is a duplicate, but all 12 cases are equally likely rolls.

Now, you can look at this in two ways:

1. As you can see in my listing of 12 equally likely cases above, X X is twice as likely to appear as any single combination X Y or Y X. If you considered X Y and Y X to be equivalent to each other, the number of cases reduces to 6, of which only one has a sum of 7. So your probability is 1/6.

2. A better way to look at is like this: The selection of one die to be X means that all the variability is in the other die. It doesn't matter if X is showing on die 1 or die 2 - in order to get a total of 7, the thing that matters is the value on the "other" die. So the problem reduces to this: given that one die shows X, what is the probability that the other die displays a value such that the sum of the two is 7? With that as the problem, you can see that there are 6 possible values for the "other" die and the probability of rolling a combined 7 is 1/6, which is consistent with the overall probability of rolling a 7 (6 possibilities / 36 cases).

The fact that there are 11 cases where X appears on one of the two dice is misleading at best, and lying with statistics at worst (lies, d--n lies, & statistics!). All you really need to do to show the irrelevance of that fact is to see that there are only 36 possible combinations of dice rolled, not 6 * 11 = 66.

Well said.

But, doesn't counting 6-6 twice [to get 12 cases instead of 11] say that

[1] Die A=6 and die B=6

[2] Die B=6 and die A=6

are somehow different?

Subject #1 [at least one of whose dice, call them A and B, was a 6] reasoned these were the possible cases:

A-B = 6-1, 6-2, 6-3, 6-4, 6-5, 6-6, 5-6, 4-6, 3-6, 2-6 and 1-6; of which the first and last total 7.

Similarly for Subjects 2-6.

counting 6-1 and 1-6 as two separate possibilities is also flawed, as well as 6-2 & 2-6, etc. It matters not which of the two dice is labeled as A and B after the roll, so let's say that die A is always the one on which the magician calls out the number that is showing. Then, using A-B nomenclature, 6-1 is a possibility, but 1-6 is not. Conversely, if die B is always called out, 1-6 is a possibility, but 6-1 is not. Good thinker of a problem though.

Edited by lazboy

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Well said.
counting 6-1 and 1-6 as two separate possibilities is also flawed, as well as 6-2 & 2-6, etc. It matters not which of the two dice is labeled as A and B after the roll, so let's say that die A is always the one on which the magician calls out the number that is showing. Then, using A-B nomenclature, 6-1 is a possibility, but 1-6 is not. Conversely, if die B is always called out, 1-6 is a possibility, but 6-1 is not. Good thinker of a problem though.

Assume your premise that for example 1-6 and 6-1 are not separate cases.

What do you then calculate for the probability of Subject 7's dice totaling 7?

Assume there are 36 possible outcomes of rolling 2 dice.

Write them out on a piece of paper.

Cross out the cases that are eliminated by the statement "at least one of the dice is a 6."

How many cases reamain?

How do you explain that?

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Assume your premise that for example 1-6 and 6-1 are not separate cases.

What do you then calculate for the probability of Subject 7's dice totaling 7?

Assume there are 36 possible outcomes of rolling 2 dice.

Write them out on a piece of paper.

Cross out the cases that are eliminated by the statement "at least one of the dice is a 6."

How many cases reamain?

How do you explain that?

The key is in "at least one". Eleven cases remain, but in one of those cases six must appear twice (boxcars). so a six shows up 12 times out of 72 or 1/6 of the time.

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I majored in mathematics at college and the most common mathematics/statistics "stumper" puzzle is one where probaility is misused to create a false conflict (whether intentional or not). Sadly, this case is no different. Here is the probability of any number being rolled when 2 dice are rolled simultaneously.

2 - 1/36

1 then 1

3 - 2/36 or 1/18

1 then 2 or 2 then 1

4 - 3/36 or 1/12

1 then 3, 2 then 2, or 3 then 1

5 - 4/36 or 1/9

1 then 4, 2 then 3, 3 then 2, or 4 then 1

6 - 5/36

1 then 5, 2 then 4, 3 then 3, 4 then 2, or 5 then 1

7 - 6/36 or 1/6

1 then 6, 2 then 5, 3 then 4, 4 then 3, 5 then 2, or 6 then 1

8 - 5/36

2 then 6, 3 then 5, 4 then 4, 5 then 3, or 6 then 2

9 - 4/36 or 1/9

3 then 6, 4 then 5, 5 then 4, or 6 then 3

10 - 3/36 or 1/12

4 then 6, 5 then 5, or 6 then 4

11 - 2/36 or 1/18

5 then 6, or 6 then 5

12 - 1/36

6 then 6

If you add up all the probability you see you have covered all your bases (36/36)

BTW, 1 then 1, 2 then 2, 3 then 3, etc. are only used once because it doesn't matter which comes first (that's the way it works).

Now, the probability changes if you roll the dice separately and then calculate probability (which is the equivalent of what the magician did by telling them the number on "at least one" of the dice). The reason for this is because the events become independent instead of dependent. The overall probability of the die roll from the beginning doesn't change, but the probability of having a certain total, once the one of the dice is know, is different. This can also be seen if you watch texas hold-em poker on TV and they show the probability of a particular playing winning the hand. The probability changes as the flop, turn, and river cards are shown. Anyhow, here's the probability for getting a certain total, if the first die is known.

If the total you are seeking is less than the die known, the probability is obviously 0. If the total you are seeking is more than 6 away from the die that is known, the probability is obviously 0. This leaves, no matter what, only 6 possible totals that can be made. This is true, regardless of what the known die shows. Since there are only 6 totals, and all are equally likely, the probabilities are all 1/6. Now, the only total that this is true for, in all cases, is 7, because it's the only total that will always be greater than the known number and will never be more than 6 greater than the known number.

If 8 were the desired total, the probability of getting an 8 if the known number was 1 would be 0, since there is no number you can roll to get 8. If the total you are seeking is 6 and the known number is 6, the probability is 0, since there is no number you can roll to get a total of 6.

Since we know we are talking about a total of 7, let's look at that

Known number X Potential other number

1X1

1X2

1X3

1X4

1X5

1X6

2X1

2X2

2X3

2X4

2X5

2X6

... you get the idea

No matter what you start out knowing, there are always 6 possible rolls (not 11 like the magician and other solvers have stated) that can be made, only one of which will yield a 7, so the proability is 1/6.

The importance of the wording of the question saying that "at least one" of the dice was a "x" is to say that duplicates ARE allowed. If the question said, "one of your dice is a "x", then you would have a probability of 1/5 instead, since you wouldn't be allowed to use duplicates.

Therefore, the answer to the question "How can this be", is that the magician was either lieing to the first 6 contestants in order to trick the 7th, or he doesn't know statistics and probability very well. In the case of a total of 7, the probability is ALWAYS 1/6, whether you know one of the dice or not.

Edited by BeastMaster

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The probability changes as the flop, turn, and river cards are shown. Anyhow, here's the probability for getting a certain total, if the first die is known.

You're making the same mistake many have made in this thread. We don't know what the first die shows. We just know what one of them is in each case. If we knew what one specific die shows, then the probability of both adding up to seven is 1/6. In the first case, since we know that at least one shows a 6, the probability of both adding up to seven is 2/11. In the riddle I linked to, if we knew the first child was a girl, then the probability of them both being girls is 1/2. Since we are told that at least one is a girl, then the probability is 1/3 that they both are.

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I'm likely reiterating in different words what others have already said:

The magician (and all his subjects) were incorrect in asserting the 2/11 probability.

It's true that there are 11 outcomes possible if "At least one die is x", however they do not all have the same probability.

For any number y != x, the probability that the roll is (x,y) [not (y,x)] is (1/2)*(1/6) = 1/12. The (1/2) takes care of the x being first.

However for the outcome of (x,x) has a probability of (1/2)*(1/6)+(1/2)*(1/6)=1/6, as the "at least one" could be either die in this case.

Thus the probability of rolling a 7 is ((1/12)+(1/12))/(10*1/12+1/6) = 1/6.

In other words, if at least one die is a 6, there the odds that the other die is 6 are the same as the odds that the other die is 5.

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You're making the same mistake many have made in this thread. We don't know what the first die shows. We just know what one of them is in each case. If we knew what one specific die shows, then the probability of both adding up to seven is 1/6. In the first case, since we know that at least one shows a 6, the probability of both adding up to seven is 2/11. In the riddle I linked to, if we knew the first child was a girl, then the probability of them both being girls is 1/2. Since we are told that at least one is a girl, then the probability is 1/3 that they both are.

No, no, no, no, no. The puzzle you linked to has a wrong solution as well. The reason the probability changes for poker hands is because there is a finite number of cards, out of which the next card comes. So say there are 5 players playing Texas Hold 'Em, each has 2 down cards, so that's 10 total and leaves 42 in the deck. Then one is burned and 3 are flopped, leaving 38 in the deck, so the probability of getting any specific card on the turn rises from 1/42 to 1/38, and subsequently 1/36 for the river. For dice, the "second" die is completely uncoupled from the result of the "first" die, much as the gender of the "second" kid is completely unrelated to the gender of the "first" kid. It doesn't matter which kid or which die you label as first and second. Boy-Girl and Girl-Boy are the SAME result (the puzzle states ONE of the kids is a girl, it doesn't specify if it's the older or younger one, and if the other child is, in fact, a boy, then how can you count that as two possibilities?), just as 6-1 and 1-6 are exactly the same when you are rolling two dice.

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It's true that there are 11 outcomes possible if "At least one die is x", however they do not all have the same probability.

Sure they do. If at least one die shows a six, the following results all have an equal possibility:

6-1, 6-2, 6-3, 6-4, 6-5, 6-6, 5-6, 4-6, 3-6, 2-6

Each of the above has just as much of a chance as being rolled as any other.

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You're making the same mistake many have made in this thread. We don't know what the first die shows. We just know what one of them is in each case. If we knew what one specific die shows, then the probability of both adding up to seven is 1/6. In the first case, since we know that at least one shows a 6, the probability of both adding up to seven is 2/11. In the riddle I linked to, if we knew the first child was a girl, then the probability of them both being girls is 1/2. Since we are told that at least one is a girl, then the probability is 1/3 that they both are.

If you know that at least one is a certain number, let's say a 3, then it doesn't matter WHICH is a 3, what matters is that the OTHER one can be one of six things. That makes the probability 1/6. The girl riddle is also wrong (on the basis of statistics). The probability of both being girls is 1/2. If you know one of the children is a girl, it doesn't matter WHICH is a girl, the OTHER can only be one of two things, hence a probability of 1/2. Both of these riddles are based on false logic and false statistics. Let me try to explain differently.

All possible combinations (36 total)

1 x 1

1 x 2

1 x 3

1 x 4

1 x 5

1 x 6

2 x 1

2 x 2

2 x 3

2 x 4

2 x 5

2 x 6

3 x 1

3 x 2

3 x 3

3 x 4

3 x 5

3 x 6

4 x 1

4 x 2

4 x 3

4 x 4

4 x 5

4 x 6

5 x 1

5 x 2

5 x 3

5 x 4

5 x 5

5 x 6

6 x 1

6 x 2

6 x 3

6 x 4

6 x 5

6 x 6

If you know that “at least one” of the dice is a 3, then your options are

3 x 1

3 x 2

3 x 3

3 x 4

3 x 5

3 x 6

That’s it, only 6. In this case you do not repeat by adding…

1 x 3

2 x 3

4 x 3

5 x 3

6 x 3

Because if you already know that one of them is a 3, that die’s probability is removed from the equation, so you CANNOT duplicate the combinations. Once one of the numbers is known, you have to only concern yourself with the second number (1 x 3 is the same as 3 x 1)

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No, no, no, no, no. The puzzle you linked to has a wrong solution as well.

No, it doesn't. Both moderators on this site agree that the correct solution has been given and so do many cites given by math experts. Please don't re-open that can of worms here.

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All possible combinations (36 total)

1 x 1

1 x 2

1 x 3

1 x 4

1 x 5

1 x 6

2 x 1

2 x 2

2 x 3

2 x 4

2 x 5

2 x 6

3 x 1

3 x 2

3 x 3

3 x 4

3 x 5

3 x 6

4 x 1

4 x 2

4 x 3

4 x 4

4 x 5

4 x 6

5 x 1

5 x 2

5 x 3

5 x 4

5 x 5

5 x 6

6 x 1

6 x 2

6 x 3

6 x 4

6 x 5

6 x 6

If you know that “at least one” of the dice is a 3, then your options are

3 x 1

3 x 2

3 x 3

3 x 4

3 x 5

3 x 6

You list all the 36 possible choices, then for some reason when including all that have at least one 3, you have neglected to actually list all of the possibilities that have at least one 3. You claimed some didn't count for some reason. They do. Of your 36 equal possibilities, 11 of them have at least one 3.

If you're going to list both "1 x 3" and "3 x 1" in your first list, then you have to do the same in the second. There's no reason both are relevant in the first and not in the second.

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No, it doesn't. Both moderators on this site agree that the correct solution has been given and so do many cites given by math experts. Please don't re-open that can of worms here.

Ohhhh OK, after realizing there were 32 pages of replies to that puzzle and reading bonanova's explanation, that puzzle makes sense to me now based on the probability of a family of 4 having two girls vs. a boy and a girl, in either order. And I think I now know how to translate that logic to the dice problem, even as both dice are rolled simultaneously. If we physically label one as A and the other as B (say A is akin to the older child and B to the younger in the boy/girl problem), and the magician randomly chooses which one to call out, then we would have 11 possible combinations using the number that is called, 2 of which sum to 7. The last contestant, however, IS NOT TOLD what either dice says, and therefore has a probability of 1/6 of having them sum to 7. Say the boy/girl problem did not tell you that one child was a girl, then the probability of having two girls is 1/4, but since it says one is a girl, that eliminates one of the possibilities. Same with the magician telling you what one of the dice says, it eliminates 25 of the 36 possible combinations, whereas the last contestant still has all 36 options available, 6 of which sum to 7. I feel like an idiot for contradicting my previous post, but I believe it makes sense to me now.

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You pretty much got it, lazboy, but I have one nitpick:

If we physically label one as A and the other as B (say A is akin to the older child and B to the younger in the boy/girl problem), and the magician randomly chooses which one to call out, then we would have 11 possible combinations using the number that is called, 2 of which sum to 7.

The magician didn't randomly choose to call one out. If he did, the probability would be 1/6. What he did is say "at least one of your dice shows a...", which is why the probability is 2/11.

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You list all the 36 possible choices, then for some reason when including all that have at least one 3, you have neglected to actually list all of the possibilities that have at least one 3. You claimed some didn't count for some reason. They do. Of your 36 equal possibilities, 11 of them have at least one 3.

If you're going to list both "1 x 3" and "3 x 1" in your first list, then you have to do the same in the second. There's no reason both are relevant in the first and not in the second.

I had a long response, but realised this can be explained quickly. I knew you would ask that. The reason is because the "x" in "1 x 3" is translated as "1 then 3". This is a result of the fact that rolling 2 dice is considered 2 separate statistical events which are not influenced or affected by each other. This is was causes 1 x 3 to be different from 3 x 1, and it's the fact that they are different that causes you to not list both when you know you have at least one 3. The moment that the magician tells the contestant that he has at least one "3" he has told him to eliminate ALL potential rolls except the ones that start "3 then...". This would exclude...

1 x 3

2 x 3

4 x 3

5 x 3

6 x 3

because all of these are scenarios where the 3 is rolled AFTER the other die (ie. "1 then 3" or "4 then 3")

This leaves you with only

3 x 1

3 x 2

3 x 3

3 x 4

3 x 5

3 x 6

It might be easier to understand or see if you think about it this way.

The magician telling the contestant they have at least one 3 is the same as the magician starting them out with at 3 and then giving them a second die and asking them, "what is the probability you will roll a 4". I think everyone would agree, the probability of rolling a 4 on a single, standard die, is 1/6.

The moral of the story - you HAVE to separate the events because they are separate statistical occurances. "1 then 3" is different than "3 then 1", so if the magician tells you that you have at least one 3, then "1 then 3" is no longer a possibility because that means you're first number was a 1, which he didn't tell you it was.

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I'm not trying to reopen the can of worms of the child puzzle, but the probability is 1/2, for the same reason the dice probability is 1/6. Once you've established one is a girl, you remove her individual probability from the picture and you are simply asking the question, "what is the probability a child is either a boy or a girl". It's as simple as that. When you have no kids to start with, your probability of having 2 girls is 1/4, but once you've had a girl, the probability is 1/2 because you eliminate the options ("boy then boy" and "boy then girl"). All you're left with is "girl then boy" and "girl then girl". 2 options means a probabilty of 1/2.

I'm not trying to get heated, so I hope the moderators don't get upset at me, but I really dislike when math is used incorrectly. This is a pet peve of mine.

The fact that there are previous riddles and other people who agree with false logic doesn't make it any more true.

You must look at the math, not people's opinions.

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I'm not trying to reopen the can of worms of the child puzzle, but the probability is 1/2, for the same reason the dice probability is 1/6. Once you've established one is a girl, you remove her individual probability from the picture and you are simply asking the question, "what is the probability a child is either a boy or a girl". It's as simple as that. When you have no kids to start with, your probability of having 2 girls is 1/4, but once you've had a girl, the probability is 1/2 because you eliminate the options ("boy then boy" and "boy then girl"). All you're left with is "girl then boy" and "girl then girl". 2 options means a probabilty of 1/2.

I'm not trying to get heated, so I hope the moderators don't get upset at me, but I really dislike when math is used incorrectly. This is a pet peve of mine.

The fact that there are previous riddles and other people who agree with false logic doesn't make it any more true.

You must look at the math, not people's opinions.

But you've assumed we were given the gender of the older child. All we're told is that there are already two children, and at least one of them is a girl, which means the younger child could be the girl and the older the boy, leaving "boy then girl" as still a viable option.

edit: clarifying

Edited by lazboy

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The moment that the magician tells the contestant that he has at least one "3" he has told him to eliminate ALL potential rolls except the ones that start "3 then...".

Wrong. There are 36 possible ways to roll 2 dice. How do you conclude how many contain at least on 3? Simple. You already listed all 36 possibilities- just count how many have at least one 3. I don't know why you think it should be any more difficult than that.

I'm not trying to reopen the can of worms of the child puzzle...

Yes, you are. See the following web-sites and read the posts in that thread.

http://mathforum.org/dr.math/faq/faq.boy.girl.html

http://en.wikipedia.org/wiki/Boy_or_Girl

While looking for cites regarding the boy/girl problem, I found one that describes the dice problem in this riddle. As you will see, there are 11 possibilities of having at least one of any number.

http://www.mathpages.com/home/kmath036.htm

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But you've assumed we were given the gender of the older child. All we're told is that there are already two children, and at least one of them is a girl, which means the younger child could be the girl and the older the boy, leaving "boy then girl" as still a viable option.

edit: clarifying

Still not a probability of 1/3. Once you know that one is a girl, take that girl and put her out of your mind. What is the probability another child is a boy or a girl. You only have 2 options, boy or girl, so the probability cannot be 1/3. They are 2 distinct statistical events and have to be considered separate. Once you know the gender of one the order no longer matters, I just explained it that way for simplicity. Here's the full picture.

If the older child is a girl, the younger child could be a boy or girl

If the younger child is a girl, the older child could be a boy or girl

Either way you look at it, the probability is 1/2.

Let me return to the dice though. You all have brought up that you should label them A and B. That's fine. Here's how that'd work.

Assume you're the contestant the magician told you had at least one 3.

If A=3

Then B=1, 2, 3, 4, 5, or 6

and your probability of getting 7 total is 1/6

SEPARATE PROBABILITY EVENT

If B=3

Then A=1, 2, 3, 4, 5, or 6

and your probability of getting 7 total is 1/6

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