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"In the card game bridge, the $52$ cards are dealt out equally to 4 players—called East,West, North, and South. If North and South have a total of $8$ spades among them, what is the probability that East has $3$ of the remaining $5$ spades?"

The book "A First Course in Probability" by Sheldon Ross states in its solution to the above question that the probability of South-North having a total of $8$ spades among them is $C(26,13)$. I can not understand where it came from. Any help would be appreciated.

N. F. Taussig
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mali1234
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  • It must be a typo. C(26,13) is clearly greater than one and therefore cannot be a probability. – cr001 Dec 24 '20 at 20:40
  • It is likely that the given probability is actually just the number of cases of North and South having 8 spades. The true probability would be calculated by diving by the total number of cases. – Tbw Dec 24 '20 at 20:44
  • So the question is actually "Why is the number of cases where N and S having a total of $8$ spades $26 \choose 13$?" Personally I highly doubt this is correct as well, because $8$ should be involved somewhere. – cr001 Dec 24 '20 at 20:47
  • Yes, that is the question precisely – mali1234 Dec 24 '20 at 20:50

1 Answers1

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The total number of cases where N and S having a total of $8$ spades is not used in the solution, because it is independent from what the question is asking.

$26\choose 13$ is the number of ways to distribute the remaining $26$ cards between E and W. Out of all those distributions, ${5\choose3}$ ways for the E player to choose from the spades and ${21\choose 10}$ ways to choose from the non-spades.

Thus the final answer to the problem is ${5\choose3}{21\choose 10}\over{26\choose 13}$.

The probability does not depend on which of the $8$ spades are chosen for N and S and therefore it is not a conditional probability type problem.

The problem can be restated as "We have a deck of card consisting of $5$ spades and $21$ non-spade cards and we have two players, what's the probability first player gets $3$ spades?"

It becomes much easier to understand.

cr001
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