Constrained probability

[bonanova]

I made up this puzzle a while ago while moderating a debate of the "one of my children is a boy" chestnut. When the debate was over, during which I championed the 1/3 answer, I converted to the position that even when the puzzle is stated "clearly" the probability of two sons could be 1/2. That's the background, here's the puzzle. I'm fairly sure I haven't already posted it ...

A magician blindfolded 100 subjects, then for each of them he rolled a
pair of fair dice and asked the probability that the total of the dice was 7.

He said to the first, here's a hint: truthfully, at least one of your dice shows a 6.
Of the 36 equally likely cases, the subject counted 11 that show at least one 6.
Two of them, 1-6 and 6-1, total 7. He answered, the probability that they total 7 is 2/11.

To the second he said, I've looked at your dice. At least one of them is a 4.
This subject also found 11 cases that fit the clue, of which only 3-4 and 4-3 made 7.
He also answered, 2/11.

The magician continued this, each time naming a number from 1-6 that
appeared on at least one of the subject's dice. In each case, the subject
deduced that p[7] was 2/11. By the time he got to the 100th subject, he'd
identified each of the numbers 1-6 at least once.

As it turns out, the 100th subject had no grasp of probability. But he had
listened to all that preceded. Before the magician could speak, he said,
I don't need a hint. I know that you're going to tell me that some number 1-6
appears on at least one of my dice. Well, I've already heard the calculated
answer in each case. So for whatever number you would have said, I already
know what p[7] is. My answer is 2/11.

But we know the unconstrained probability is 1/6, right?
_________________
_{Vidi, vici, veni.}

[Chaz]

Not if there are two dice and we know that we'll know one of the dice.

It's just Monty Hall Logic in a new outfit--a very sexy outfit too.
_________________
The enemy's base is down.

[ralphmerridew]

The problem is that the subjects are giving the answer to P(sum = 7 | at least one X) when they're being asked P(sum = 7 | told at least one X). The two events are not necessarily the same.

The correct answer depends on how the magician decides what hint to give.

Possibility A: Before tossing the dice, the magician picks a number 1-6. After rolling, the magician tells the subject whether or not that number showed on either die.

Possibility B: After tossing the dice, the magician picks one of the two numbers and reveals that number to the subject.

Possibility C: After tossing the dice, the magician reveals the larger number.

Possibility D: ...

Under possibility A, the two events are the same, and 2/11 is be the correct answer for the first 99 people. (It's very unlikely that the magician would get through 99 people without having a fail result at any time, though.) The final person's answer is incorrect because he can't be certain that he will get a "At least one die shows a ...."

But the two events are not the same under other rules.
For example, under possibility B with X=6:
6/36 times, it will be true that at least one die shows a 6 and the subject will be told that at least one die shows 6.
25/36 times, it will be false that at least one die shows a 6, and the subject will not be told that at least one die shows a 6.
5/36 times, it will be true that at least one die is a 6, but the subject will not be told that fact.

[L'lanmal]

/agree with rm. Whenever information is revealed by an agent, the behavior of the revealer comes into question.

[HAMMOS]

When doing the count that leads to a 2/11 odds calculation, the mistake being made is that 6-6 should be counted twice.

The statement "You have at least one six" means one of the following is true:
The dice referred to is the first dice, and you have 6-1
The dice referred to is the first dice, and you have 6-2
The dice referred to is the first dice, and you have 6-3
The dice referred to is the first dice, and you have 6-4
The dice referred to is the first dice, and you have 6-5
The dice referred to is the first dice, and you have 6-6
The dice referred to is the second dice, and you have 1-6
The dice referred to is the second dice, and you have 2-6
The dice referred to is the second dice, and you have 3-6
The dice referred to is the second dice, and you have 4-6
The dice referred to is the second dice, and you have 5-6
The dice referred to is the second dice, and you have 6-6

Thus, the probability of a 7 total is 2/12 = 1/6 and the mystery is solved.

[Zag]

HAMMOS, this is also incorrect. You are also making an assumption about how the the magician decides what hint to give. It might be a better assumption than the 99 people made, but it is still an assumption.

Specifically, you are assuming that he rolls the dice, picks one of the two dice at random, and tells the value of that die. With that assumption, you are correct. But it is still an assumption. It is probably the best one you can make, under the circumstances, but it is not guaranteed.

What if, rather than being the 100th person, you were the 1,000,000th person. You carefully kept track of the numbers that the magician revealed, and you found that

11/36 of the time it was 6.
1/4 of the time it was 5.
7/36 of the time it was 4.
5/36 of the time it was 3.
1/12 of the time it was 2.
1/36 of the time it was 1.

---------------------

bonanova, this difference between this case and the puppy chestnut puzzle (which, BTW, I posted in our chestnuts thread), is that the magician offered the information rather than being asked for it. If the contestant were to ask, "Is there at least one 6?" and got a positive response, his correct answer would be 2/11. (And if he got a negative response, his correct answer would be 4/25.) However, since the magician offered the information, we have to know his algorithm for what information to offer, or we can't draw conclusions from it. Probably the best assumption is the one HAMMOS made, which leads to the result 1/6.

[THUDandBLUNDER]

In similar vein:

a) I have two children. One is a girl. What is the probability that I have two girls?

b) I have two children. One is a girl born on a Friday. What is the probability that I have two girls?
:

[bonanova]

[bonanova]

[THUDandBLUNDER]

:
An example of why hard scientists chose science, not math? http://5g2c.sl.pt
:

[referee]

No, it works like this.

P(a 6 appears) = 1 - (5/6)^2 = 1 - 25/36 = 11/36
P(sum is 7 | a 6 appears) = 2/11 (Correctly calculated by 99 people)

P(A 6 appears AND sum is 7) = 11/36 * 2/11 = 2/36 (1-6 and 6-1, indeed)

So what are we being asked about?

P (A ^ B) = P (A) * P (B | A)
P (A ^ B ^ C) = P (A) * P (B | A) * P (C | [A ^ B])
So on.
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Jan 21st, 2008: The pillaging continues.
Mar 4th, 2008: Rest in Peace, Gary Gygax. May your dice always roll a natural 20 wherever you are.

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[Trojan Horse]

A variation of THUDandBLUNDER's:

I come from a two child family myself; it's me and one sibling. (I'm a boy, by the way, in case anyone here didn't know that.) I'm going to give you a chance to make a wager on the gender of my sibling. If my sibling is a girl, you win. If my sibling is a boy, I win.

What odds would I have to offer you to make this a fair bet?

[groza528]

Assuming you would not want to lose the bet, and also assuming that you know your sibling, the only way I could win this bet is if you thought your sister was in fact a boy. Based on your regular and semi-respected () presence here we can throw out the probability of mental instability on your part. Now, there is a disorder that can cause a male to appear female (http://en.wikipedia.org/wiki/Androgen_insensitivity_syndrome), but I don't think anything exists in the reverse, so any odds you could offer would have to cover my penalty in the event that I were to get caught tampering with the genetic testing results.

[Trojan Horse]

Hee hee. I just can't fool you guys, can I? (Yes, I do have a brother, and I wouldn't have offered this bet if I had a sister.)

[groza528]

For a brief moment, you did. I was about to call it 50-50