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I'm looking at this problem in a bayesian stats book:

A card game is played with 52 cards divided equally between 4 players, North, South, East and West, all arrangements equally likely. 13 of the cards are referred to as trumps. If you know North and South have ten trumps between them, what is the probability all 3 remaining trumps are in the same hand?

Would this reduce to the problem of 'there are three trump cards in 26 cards, what is the probability that all 3 are in a set of 13 when divided into two sets of 13 (East & West)?' ? or does the 10 cards in the North, South have influence and so you have to use conditional probability?

Also just out of interest is the answer to this assumption (even if wrong) a $\frac{1}{2} $ ? As the equally likely possibilities are {3,0}, {2,1}, {1,2}, {0,3} and the last two are winners, where {2,1} is the event that there are two trump cards in the first 13 and 1 trump card in the second 13.


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I'm not quite sure you should read this as N and S having exactly ten trumps between them. – Kevin Carlson Oct 11 '12 at 10:45
@KevinCarlson I think it must mean that the hands of N and S contain exactly ten trumps between them. So for example if N has 7 trumps then S would have 3. – Sam Forbes Oct 11 '12 at 10:58
Well, it is a conditional probability problem, and could be attacked that way. However, the natural (and correct) approach is the one you suggest. – André Nicolas Oct 11 '12 at 11:10

The first part of your approach is correct; no need to use conditional probabilities here.

The second part is wrong, though; the three possibilities aren't equally likely. The probability that all $3$ remaining trumps are in a particular one of the two sets of $13$ is

$$ \frac{\binom{13}3}{\binom{26}3}=\frac{11}{100}=0.11\;, $$

so the probability that they're all in either of the two sets is $2\cdot0.11=0.22$.

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Thank you. I'm very bad at doing 'counting problems' like this. – Sam Forbes Oct 11 '12 at 11:49

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