Probability paradox

This crazy problem arose during a discussion at a bridge game. I’ll reduce it to its simplest terms.

Take a packet of four cards: the 33,5 and 8 of hearts and the ace of spades. Randomly give two cards each to players A and B face down. Ask each to place one heart on the table face up. The player who has two hearts may apply either one at random. A plays the 3, and B plays the 5.

One can argue that A’s 3 is probably a singleton as follows: On only 50% of the hands where he holds 8,3 he plays the 3. But if he has A,3 he plays the 3 100% of the time Therefore, the odds are 2 to 1 that he holds the ace of spades. One could make the same case that B has a 2/3 probability of holding the ace.

What’s wrong with this picture? *****

For bridge players: You are south, declarer. Dummy holds the A, J, 10 of clubs, and you hold 3 small clubs. You lead toward dummy and play the 10. East wins with the Q. There is a 2/3 probability that west hold the K. If east held the K,Q there is a 50% probability that he would have played the K. But if the Q is his only honor, he has no choice.

How does this differ from the above example?

Bill Ryan

First, the possible distributions (order is not important here, just division between A and :

A B 35 8A 38 5A 3A 58 58 3A 5A 38 8A 35

So there are only six combinations. At present, assuming that A plays randomly except when he has the ace, he is equally likely to play the three, five, or eight. (p = 0.3r for each card; I could prove that if need be.) B likewise.

Now, we are given some extra information which reduces the number of possible combinations. A has played the 3, and B has played the 5. Therefore, we can eliminate all combinations where A doesn't have a 3 or B doesn't have a 5, as follows:

A B 38 5A 3A 58

Only two possibilities remain. They are equally likely, and in one case A holds the ace, and in the other case B holds the ace.

Conclusion: your wording was misleading. No doubt deliberately so.

I wonder if this is a cleverly disguised variant of the 'missing pound' puzzle?

) Only two possibilities remain. They are equally likely, and in one case ) A holds the ace, and in the other case B holds the ace.

Duh.

The question was _what_ is wrong with this. The puzzle is to determine what exactly is the flaw in the argument. Giving a completely different argument to refute it is pointless.

In other words: This is a 'what's wrong with this proof' puzzle, or more precisely, a 'which law of probability calculus was broken' one.

Oh, sorry.

Here's the argument again:

'One can argue that A’s 3 is probably a singleton as follows: On only 50% of the hands where he holds 8,3 he plays the 3. But if he has A,3 he plays the 3 100% of the time Therefore, the odds are 2 to 1 that he holds the ace of spades. One could make the same case that B has a 2/3 probability of holding the ace.'

'On only 50% of the hands where he holds 8, 3 he plays the three' is flawed, because in 100% of the cases under consideration he has in fact played the three. Because we know he has played the three (and that B has played the five), we can reduce the situation to the one I outlined earlier.

That's it. Because that first part is flawed, the rest falls. The 'therefore' simply doesn't apply.

Please post in ASCII. That's 'I'll'. There seem to be some extra 3's in the quoted text, which I'll delete as I go.

I assume this means 'must play'.

Okay.

You're using part of the information from B's play to think about A's hand, but not all of it.

Look at A's play alone, supposing that A plays first. There are 6 equally probable hands in this 'game': A-8, A-5, A-3, 8-5, 8-3, 5-3. The probability that A plays the 3 is 1/3, made up as follows: 1/6 that it's the 3 from A-3, 1/12 that it's the 3 from 8-3, and 1/12 that it's the 3 from 5-3. Of these, half the cases (1/6) include the ace, and half don't, so the probability is still 1/2 that he had the ace. The same argument applies to B.

Now look at all the information from both plays. The possible deals are now A-3 / 8-5 and 8-3 / A-5, both of which were equally likely originally. In the first case, there is a probability of 1/2 that the 3 and 5 are played. In the second case, there is also a prob- ability of 1/2 that the 3 and 5 are played. The situation is still symmetrical, and so the probability of A holding the A is still 1/2.

Rule of Restricted Choice, yes.

The first example was asymmetrical

It ignores the fact that A is twice as likely to have a doubleton as a singleton in the first place.

Jonathan Dushoff

I don't think it does ignore that fact. I think that the error is the mis-application of the extra information that A does not hold the 5.

I am not convinced that anybody has really pinpointed the error in the above argument, which I think is quite a subtle one. People are generally just replacing it with a correct argument, rather than exposing the flaw.

The incorrect argument uses the fact that B playing 5 tells us that A does not hold 5.

So the question is apparently:

What is the probability that A holds the ace of spades, given that A has played 3 and that A does not hold 5.

The answer to that question really is 2/3. So what is going wrong?

It has to be that the 'given that A does not hold 5' part of the information is being mis-applied. As Mark Brader (I think) said, we are making use of some of the information available from the fact that B played 5, but not all of it. The moral is that when you make use of extra information in a probability computation, you have to take into account where that information comes from.

This is very similar to the old chestnut 'A man has two children. What is the probability that that they are both girls given that at least one is a girl?' The answer to that is well-known to be 1/3. But suppose instead that we know that the man has two children, and we ask him to make a true statement of the form 'At least one of my children is a girl/boy', where, if he happens to have one of each, he must choose boy or girl at random. Suppose he now says 'At least one of my children is a girl'. Then the probability of two girls becomes 1/2, even though we apparently have the same information as in the original problem.

Derek Holt.

I see what you are getting at. In fact, what this is is the Monty Hall problem. Since B will definitely play 5 or 8, and since the apparent information content of either is the same, in fact we get no information from B's play. My earlier answer was implicitly assuming this, and thus ignoring the 'information' from the 5.

To put it another way: the ace is one of the four cards that we can't see. We can think of the four card backs as four doors. A and B each control two doors, and they sequentially open one to show us that it is not the ace. Since they always can do this, and always will do this, neither of these two actions change the probability of the statement 'the ace is behind either door 1 or door two', equivalent to 'the ace is in A's hand.

Jonathan Dushoff