# The Grey Labyrinth is a collection of puzzles, riddles, mind games, paradoxes and other intellectually challenging diversions. Related topics: puzzle games, logic puzzles, lateral thinking puzzles, philosophy, mind benders, brain teasers, word problems, conundrums, 3d puzzles, spatial reasoning, intelligence tests, mathematical diversions, paradoxes, physics problems, reasoning, math, science.

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bonanova
Daedalian Member

 Posted: Wed Sep 22, 2010 8:56 pm    Post subject: 1 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
Vote: Zag

 Posted: Wed Sep 22, 2010 10:16 pm    Post subject: 2 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
Daedalian Member

 Posted: Wed Sep 22, 2010 11:50 pm    Post subject: 3 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
Daedalian Member

Posted: Thu Sep 23, 2010 12:54 am    Post subject: 4

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

 ralphmerridew wrote: Possibility C: After tossing the dice, the magician reveals the larger number. Possibility D: ...
I'd like to contribute Possibility D: After tossing the dice, the magician reveals the first die...
... thus converting it to the "older son" case of the two children, and a 1/6 probability.
HAMMOS
Icarian Member

 Posted: Thu Sep 23, 2010 2:43 pm    Post subject: 5 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
Tired of his old title

 Posted: Thu Sep 23, 2010 3:15 pm    Post subject: 6 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
Threefold Repetition

 Posted: Thu Sep 23, 2010 11:27 pm    Post subject: 7 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
Daedalian Member

Posted: Fri Sep 24, 2010 3:09 am    Post subject: 8

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

 zag wrote: If the contestant were to ask, "Is there at least one 6?" and got a positive response, his correct answer would be 2/11.

I put this puzzle [in another forum] before some of the 1/2 vs 1/3 protagonists [one of my kids is a boy] to have them see that an answer can't be certain unless we know how the informant chose what information to give. I wish I had had Zag's insight. The puzzle solver most likely goes astray because he doesn't realize the difference between asking and being told.

Thanks both. Very clearly put.

I also asked [in the other forum] if there was a best or most reasonable algorithm to assume for the informant in puzzles of this type. Didn't get a response. Owing to the number and types of available statements, even an assumption that his statement was random doesn't seem helpful.
_________________
Vidi, vici, veni.
bonanova
Daedalian Member

Posted: Fri Sep 24, 2010 4:48 am    Post subject: 9

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

Assuming he said I have two children,
I asked, is one a girl born on Friday?
Then I'd say .481... [repeating]

Because...
Two children gives a 196-case sample space

B[m-f] B[m-f] 49 cases
B[m-f] G[m-f] 49 cases
G[m-f] B[m-f] 49 cases
G[m-f] G[m-f] 49 cases

A Friday girl exists reduces the sample space to 27 cases:

bb: 0 cases
bg: B[m-f] G[f] 7 cases
gb: G[f] B[m-f] 7 cases
gg: G[f] G[m-f] + G[m-f]G[f] - G[f]G[f] [double counted] 13 cases.

gg/(bg+gb+gg) = 13/27 = .481...

_________________
Vidi, vici, veni.
THUDandBLUNDER
Threefold Repetition

 Posted: Fri Sep 24, 2010 9:21 am    Post subject: 10 : An example of why hard scientists chose science, not math? http://5g2c.sl.pt :
referee
June 21st, 2004 Member

 Posted: Fri Sep 24, 2010 2:53 pm    Post subject: 11 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._________________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. Be the Ultimate Ninja! Play Billy Vs. SNAKEMAN today!
Trojan Horse
Daedalian Member

 Posted: Fri Sep 24, 2010 2:56 pm    Post subject: 12 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
No Place Like Home

 Posted: Fri Sep 24, 2010 4:29 pm    Post subject: 13 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
Daedalian Member

 Posted: Fri Sep 24, 2010 4:44 pm    Post subject: 14 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
No Place Like Home

 Posted: Fri Sep 24, 2010 4:48 pm    Post subject: 15 For a brief moment, you did. I was about to call it 50-50
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