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Say you divide a properly shuffled deck of poker cards into four equally sized stacks of cards. How do you calculate the chances that one of the four stacks has the most cards with the suit of Hearts?

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    Should "four equally sized cards" be "four equally sized stacks of $13$ cards"? Are you choosing one of the four stacks or just asking whether there exists a stack with more Hearts than all others? – 2017-02-21
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    In other words, the probability that the $13$ hearts distribute as for example $3+5+2+3$ (one of the stacks has the most hearts) rather than $5+1+5+2$ (where neither stack has more hearts than every other stack)? – 2017-02-21
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    That is the idea, Henning. – 2017-02-21
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    I fixed the typo, @michael burr. I am asking for the probability of whether there exists a stack with more hearts than all the others. – 2017-02-21

3 Answers 3

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Hint -

I think its easy to write cases with no stack has more hearts than any other and find probability. Then subtract probability from 1 to get the desired answer.

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Without loss of generality, assume the first group is the first $13$ cards in the deck, the next is the next $13$ cards, and so forth. For any given shuffle of the deck, identify the locations of the hearts by numbers from the set $\{1,2,3,\ldots,52\}$ representing where each heart occurred in the shuffled deck. There are $\binom{52}{13}$ different sets of locations for the hearts, each equally likely. You just need to count how many of those have more hearts in one group than in any other.

If there is one group with more hearts than the others, and there are $n$ hearts in that group (that is, $n$ of the $13$ positions in that group have been occupied by hearts), there are $13 - n$ remaining hearts to put in the remaining $39$ positions. When $n > 13 - n$ is large enough it is easy to see how many ways the rest of the cards can be distributed; just choose one of the $39$ places for each remaining heart, leading to $\binom{39}{13-n}$ arrangements. For smaller values of $n,$ however, you have to exclude the arrangements in which more than $n-1$ of the remaining hearts would be put into the same group. Fortunately, there are not many such values of $n$ where this is possible at all. When any arrangements exist, you should be able to count them on a case-by-case basis, or else count $\binom{39}{13-n}$ arrangements without regard for the constraint and then subtract the number of arrangements that violate the constraint.

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I think the easiest solution is to first find the probability of a draw. See that a four-way draw can never occur. Then see that the only way a three-way draw can occur by having three stacks have 4 hearts each and the other having 1. This can occur in $4\left(\begin{array}{c} 13 \\ 4\end{array}\right)\left(\begin{array}{c} 39 \\ 9\end{array}\right)\left(\begin{array}{c} 9 \\ 4\end{array}\right)\left(\begin{array}{c} 30 \\ 9\end{array}\right)\left(\begin{array}{c} 5 \\4\end{array}\right)\left(\begin{array}{c} 21 \\ 9\end{array}\right)$ ways (the number of ways the first, second and third group can have exactly three hearts, multiplied by 4). Similarly we can have a two-way draw if the leaders have 6 or 5 hearts each, which can happen in $6\left(\begin{array}{c} 13 \\ 6\end{array}\right)\left(\begin{array}{c} 39 \\ 7\end{array}\right)\left(\begin{array}{c} 7 \\ 6\end{array}\right)\left(\begin{array}{c} 32 \\ 7\end{array}\right)+6\left(\begin{array}{c} 13 \\ 5\end{array}\right)\left(\begin{array}{c} 39 \\ 8\end{array}\right)\left(\begin{array}{c} 8 \\ 5\end{array}\right)\left(\begin{array}{c} 31 \\ 8\end{array}\right)$ ways. The total number of permutations is $\left(\begin{array}{c} 52 \\ 13\end{array}\right)\left(\begin{array}{c} 39 \\ 13\end{array}\right)\left(\begin{array}{c} 26 \\ 13\end{array}\right)$. As all permutations are equally likely the probability of a draw is the sum of the first two numbers divided by the third. Your answer is 1 minus this number.