Suppose we deal four 13-card bridge hands from an ordinary 52-card deck. What is the probability that the North and East hands each have exactly the same number of spades?
The solution I got was
$$\frac{\sum_{i=0}^6\binom{39}{13-i,13-i}}{\binom{52}{13,13}}$$
The textbook only answers odd numbered questions, is my solution correct?
Goal:
We need to look at how many ways you could have 0 spades in both, 1 spade in both etc. and we need to add up all these combinations, then divide by the total number of ways you could arrange the 52 cards into the 2 hands.
Explanation of equation:
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For 0 spades in both hands, we need to see how many combinations of cards we can have between both hands but the combinations cannot have any spades. So we subtract 13 spades from the 52 to get 39, then we see how many combinations we can get between the 2 hands.
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If we have 1 spade in both hands, we have 12 slots left to fill in either hand. So we see how many combinations we can fit into these 2 hands with 12 slots, such that no spades are in them (39 non-spade cards).
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As we increase the number of spades in both hands, the number of empty slots for the combinations decreases. This decrease goes on till we have a maximum of 6 spades in each hand and we can only have combinations between a remaining 7 cards in each hand.
Best Answer
$$\frac{\sum\limits_{i=0}^6\binom{13}{i}\binom{39}{13-i} \binom{13-i}{i}\binom{26+i}{13-i}}{\binom{52}{13}\binom{39}{13}}$$
For each possible number of spades: $\sum_{i=0}^6$
Choose the spades in North's hand: $\binom{13}{i}$
Choose the non-spades in North's hand: $\binom{39}{13-i}$
Choose the spades in East's hand from the remaining spades: $\binom{13-i}{i}$
Choose the non-spades in East's hand from the remaining non-spades: $\binom{26+i}{13-i}$
Take the sum of the products: $\sum_{i=0}^6\binom{13}{i}\binom{39}{13-i} \binom{13-i}{i}\binom{26+i}{13-i}$
Divide by the number of unrestricted hands: $\binom{52}{13}\binom{39}{13}$