Dice probability paradox

Factual Questions
21

I want to point out that, if the magician had rolled the dice until he got at least one of some particular number N, then there aren’t thirty-six possible dice landings, there are only eleven possible ones. In that case, even though the magician never said anything to the last person, there would still be a 2/11 chance of a seven.

If the magician just rolled the dice once, and picked a number that was showing, then there would be a 1/6 chance of getting a seven. In this case, Fantome there’s your answer as to why the probabilities are different. The setup is different.

22

Am I understanding this correctly, that you are saying the odds of at least 1 die showing a six is 11/36?

As in:
If (Die_One==6 || Die_Two==6) at_least_one_six_count++;

23

I think so. Here’s the code I used:


CLS
RANDOMIZE TIMER
FOR i = 1 TO 5000000
x = INT (RND * 6) + 1
y = INT (RND * 6) + 1
IF x = 6 OR y = 6 THEN total = total +1
NEXT i
PRINT total / 5000000
24

Huh. I think I posted in haste earlier.

25

Of course all the statements about possible combination are a complete nonsense and the trick to the whole exercise is to simply ignore them. If you could take the magician’s spiel at face value there is a 2 in 11 chance of getting a 7 if one of the dice is a 1, 2, 3, 4, 5, or 6.

In reality if you are handed two dice and told that one is any integer, then since the other dice can be any one of 1, 2, 3, 4, 5 or 6. There is one chance in 6 that you are holding a 7.

Do you think I can find anyone dumb enough to not see this and take the 9/2 odds about the result. I don’t have a lot of money but I won’t need much.

26

Looks ok, unless the QBASIC pseudo-random number generator repeats prior to 10 million??? (2 rands per iteration)

Because I’m pretty sure there are 12 different ways to roll 2 dice to get a six out of a total of 36 combos

27

I address exactly this erroneous counter argument in the rest of the post you quoted.

The fact that the magician gives different numbers to each of six different people suggests that you are intending us to take his sentences to mean: “I have chosen one of the dice at random and it shows the number N.” This is the critical point. If his algorithm makes any number special, then the boy-girl problem becomes applicable. (In that problem, “girl” is special.) An example here would be if the number six were special, with his algorithm being: “Look at the dice and report 6 if a six is present; otherwise report the value from one of the dice chosen at random”.

This is related to what Omphaloskeptic is saying, as the conditional probability of the other die matching depends on the algorithm used by the magician. It seems like you are intending the first algorithm I mentioned above, as there is no indication of a special number.

In summary: Yes, [4,4] comes up exactly as often as [1,4] and exactly as often as [4,1]. You are correct not to double count the degenerate case when enumerating the space of dice rolls. But, the magician subsequently chooses a particular die to report, and in the following list of (equiprobable) outcomes:
[4,4] [1,4] [4,1]
the number “4” appears twice as often in the [4,4] outcome as it does in the others, inflating that roll’s conditional probability by a factor of two as soon as the magician says “4”. (This is [not coincidentally] equal to the factor you took care not to introduce when counting the space of dice rolls.)

28

I actually had to do it myself and even go so far as to capture the count of each pair before before I finally realized that there is only 1 spot to count a 6 and 6 pair. Sometimes I stun myself.

29

If the magician rolls two dice and tells you the number on a randomly selected die then it does not allow you to infer the outcome of the second die. The odds remain 1/6 of it making a number to total to 7 with the first. It’s true that there are 11 possible outcomes with at least one number 6, but he is also twice as likely to report ‘I see one six’ when selecting a die at random in the case of 6 - 6 compared to 1 - 6. (Which is what Larry was getting at I think)

For the odds to be 2/11 he would need to say ‘im going to roll dice until I see at least one 6, what are the odds of them totaling seven?’. In that case you are picking randomly from the set of equally probable outcomes with one 6 so your odds are 2/11.
It’s the same for the boy/girl problem. If you report the sex of a random child, the other one is 50/50 boy/girl. The sex of the first child doesn’t matter because they are discrete events. If instead you select a family randomly from the set of permutations [GG BG GB BB] until you get one with at least one G, then yes the second child is more likely to be B. (The way this question was worded in the Monty Haul Dope I really think it is asking for the former and not the latter.)
I think that generally the point of these kinds of riddles is to imply the former but truly ask for the latter through very tricky wording (or vice versa). If you could somehow ask both questions at once then I can see how it could become a paradox.

30

I don’t think the OP is trying to imply this, though I may be reading too much into it. In my experience of probability word problems, the “at least one” language usually is intended to imply just a probability conditioned over the subspace of outcomes satisfying the constraint.

Here’s one way to make the probabilities agree with the story in the OP:

The magician’s nightly routine is to call seven volunteers up and blindfold them. For the first volunteer he rolls two dice, and reports (truthfully) either “At least one 1 is showing” (probability 11/36) or “There are no 1s showing” (probability 25/36)–note that this is a partition of the event space, so he doesn’t have any choice on what to report; then he asks the probability that the dice sum to 7 (2/11 or 4/25, respectively). For the second, he does the same thing, except that he reports on the presence of a 2; and so on for the first six volunteers. I don’t know what he plans to do for the seventh; maybe he reports on the number of 7s showing.

This doesn’t seem very magical to me; maybe he’s not a very good magician, I don’t know. But about one night in 1200, he can truthfully report on each of his six rolls that his chosen number is showing. This probably doesn’t impress anyone either, but at least it may have the benefit of puzzling the seventh volunteer and any credulous statisticians in the crowd. Fantome just happened to show up that one night, so he got an unusual story to tell.

31

Which reminds me of a cool brainteaser (from memory, so I’ll probably cock this up):

Three prisoners are put in separate rooms, and have either a white or a black hat placed on their head, randomly with equal likelihood. They can’t see what colour their own hats are, but can see the other prisoners’ hats by video camera. The prisoners have no way of communicating with each other.

On the stroke of midday, each prisoner must either guess the colour of his own hat or say nothing.

If nobody says anything, or any of the prisoners makes an incorrect guess, the prisoners are all executed.
Otherwise, i.e. if at least one of the prisoners guesses correctly and nobody guesses incorrectly, they are freed.

The prisoners are allowed to confer the day before this procedure, and are of course all highly intelligent mathematicians.

What is the probability that the prisoners go free?

32

  1. P(7) = 6/36 = 1/6.

  2. Here’s the situation, as described by the OP:

P(7|at least one die = 1) = 2/11
P(7|at least one die = 2) = 2/11
P(7|at least one die = 3) = 2/11
P(7|at least one die = 4) = 2/11
P(7|at least one die = 5) = 2/11
P(7|at least one die = 6) = 2/11

  1. #2 can be summarized as follows:

P(7|at least one die = n where n is an element of {1,2,3,4,5,6}) = 2/11

  1. But the condition in #3 is always true when a pair of dice is rolled, so that collapses to:

P(7|we rolled a pair of dice) = 2/11

  1. So the ‘condition’ of this conditional probability is that we performed the experiment. So #4 collapses to:

P(7) = 2/11

which we know is wrong.

Conclusion: the problem lies with the conditional probability of 2/11.

I know what all the textbooks say, but I think the problem is that the classic description of the condition - that at least one die is a 6, but we don’t know what die it is - doesn’t fit reality very well. Sure, you can design an experiment where this is so, but you’d better be careful.

When the magician tells one person that at least one of his dice shows a 1, the magician sees a specific die showing a 1. Maybe he doesn’t tell the blindfolded person that it’s the red die or the green die that shows a 1, but nonetheless, that’s what he sees. (If the dice can’t be distinguished in some manner, it doesn’t matter: they’re still distinct dice.) So like it or not, he’s really collapsing the sample space to “die X shows a 1, and I’m not telling you about die Y.” And the probability that die Y is a 6, so the pair of dice totals 7, is 1/6.

Same with the children; in fact, it’s more obvious there. “Diane has two kids, and at least one is a boy.” Well, how did you know? You saw the toddler with the dirty blond hair in the yard? That’s a specific kid; the likelihood that the other’s a boy is 1/2, not 1/3. You know one of her children is named Justin? Justin’s almost certainly a boy; her non-Justin child has a 50-50 chance of being a boy.

ETA: Thanks, Fantome, for sharing this great paradox. It really required some thought, and I’m still not 100% sure I’m right - I just can’t see any other way out.

33

So, to summarize, the problem with the OP is simply that (x, x, magician reports x) is twice as likely as either (x, i, magician reports x) or (i, x, magician reports x), so that the assumption that ‘all eleven possibilities have the same probability’ is wrong. Right?

34

I’m not sure how to solve your version, but I’ve heard a few different version of these hat problems. There’s the classic three-hat problem, of course, and then there’s this one (modified to use your syntax):

Some number (let’s say n) prisoners are put in separate rooms, and have either a white or a black hat placed on their head, randomly with equal likelihood. They can’t see what colour their own hats are, but can see the other prisoners’ hats by video camera.

Starting on the stroke of midday, each prisoner (one at a time, in random order) must guess the colour of his own hat. The prisoner may not say anything other than “black” or “white,” and no other form of communication is allowed.

If the prisoner makes an incorrect guess, he is immediately executed; if his guess ir correct, he’s immediately freed. After that, it’s the next prisoner’s turn. All of this–the guess and the execution/freeing-- is broadcast to the other prisoners (in order to heighten the tension, or course).

The prisoners are allowed to confer the day before this procedure, and are of course all highly intelligent mathematicians.

What is the expected number of prisoners let free?

35

Here, just for fun, let’s rework my simplified problem to make it more obvious:

Suppose a magician has four marbles: one black, one white, and two gray ones. Just for fun, again, he inscribes “1-1” on the black marble, “2-2” on the white marble, “1-2” on one of the gray marbles, and and “2-1” on the other gray marble.

The magician places the marble in a bag and shakes it up. He blindfolds three people, then for each of them he pulls out a marble and asks them the probability of their marble being gray. (He puts each marble back before going on to the next person of course, so he’s always choosing from four marbles.)

He said to the first, here’s a hint: truthfully, your marble is either gray or black–it’s definitely not white. The subject knows one possibility is eliminated, so he answered, my chances of having a gray marble are 2/3. Very good, said the magician.

He then said to the second, I’ve looked at your marble, and it’s either gray or white–it’s definitely not black. The subject reasons the same way, so he answered, 2/3. Very good, said the magician.

The third 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 “the marble is either gray or white,” or you’re going to tell me “the marble is either gray or black.” 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 probability of having a gray marble is. My answer is 2/3.

The third subject is wrong.

I think the reasoning why the third subject is wrong is much clearer here. (As has been noted above, the “correctness” of the 2/3 answer given by subjects one and two depends on your interpretation of how the magician chooses his words. So this devolves into the same kind of argument that the boy-girl problem devolves into; namely, how to interpret English phrasing as probability statements.)

36

Depends on your assumptions. Mine is that what’s actually on the dice doesn’t matter, because once these hypothetical people are unblindfolded, what they see isn’t going to prove anything anyway. The dice, the blindfolds, are all just props for the thought experiment. It’s not like most people would say “2/11” in the first place, after all, unless this event’s happening at the Joint Statistical Meetings.

37

I’m not sure I get what you’re getting at – my thinking is simply: there are 12 cases in which the magician will report x – namely the five cases (x, i, magician reports die 1), the five cases (i, x, magician reports die 2), and the two cases (x, x, magician reports die 1) and (x, x, magician reports die 2) --, two of which sum to seven; thus, (reasonably) assuming equal probability for all twelve cases, a probability of 1/6 that the two dice sum to seven.

38

Okay, you’re claiming that the first six subjects didn’t in fact have a 2/11 probability, but a 1/6 probability.

Let’s look at the example zut supplied:

There are an equal amount of 1s and 2s.

After performing zut’s experiment we gather two groups of 100 people.

In the first group, everyone waits to hear what side at least one of their coins are showing. Out of those 100, about 67 will have two coins that equal 3.

In the second group, no one waits to hear what side at least one of their coins are showing. Out of those 100, about 50 will have two coins that equal 3.

Would you make the same claim in this example also? That each person in the first group didn’t have a 2/3 probability of having two coins that total 3 after hearing the magician, but a 1/2 probability?

39

For the coins, the cases where the magician reports 1 are:

(1, 2, magician reports coin 1)
(2, 1, magician reports coin 2)
(1, 1, magician reports coin 1)
(1, 1, magician reports coin 2)

Two of these cases sum to three – the probability is 50%. Am I missing something?

40

You don’t count 1,1 twice. It counts once just as zut wrote:

Same in the Boy/Girl problem. GG and BB count once (see this thread).