# Homework Help: Probabilities Involving Bridge

1. Dec 10, 2007

### e(ho0n3

The problem statement, all variables and given/known data
(1) In a hand of bridge, find the probability that you have 5 spades and your partner has the remaining 8.

(2) Compute the probability that a bridge hand is void in at least one suit.

(3) Compute the probability that a hand of 13 cards contains:
a. the ace and the king of at least on suit;
b. all 4 of at least 1 of the 13 denominations.​

The attempt at a solution
Some quick info. on bridge: a hand has 13 cards and there are four players: N, S, E, W.

(1) The wording of the problem is confusing. I will assume that it asks for the probability that one player has 5 spades and another has the remaining 8.

There are n = C(52,13) * C(39,13) * C(26,13) * C(13,13) ways of dealing the cards.

How many ways are there of dealing the cards such that N gets 5 spades and S gets the remaining 8? Answer: C(13,5) * C(39, 8) * C(8,8) * C(31,5) * C(26,13) * C(13,13). Now if N gets 8 spades and S 5, the answer is C(13,8) * C(39,5) * C(5,5) * C(34,8) * C(26,13) * C(13,13). Both the former and the latter products are equal. Call it m. As a matter of fact, for any of the 12 permutations of two players, there will be m ways of dealing the cards such that one gets 5 spades and the other 8. That's a total of 12m.

The probability sought is 12m/n. The book states that the probability is just m/n. The book seems to be ignoring who gets the spades and I don't know why.

(2)
If a hand is void in at least one suit, then all cards are from one, two or three suits. There are 4 hands in which all the cards are from one suit, there are

$$\binom{4}{2} \sum_{i=1}^{12}\binom{13}{i}\binom{13}{13 - i}$$

hands in which the cards are all from two suits and there are

$$\binom{4}{3} \sum_{i=1}^{12} \binom{13}{i}\sum_{j=1}^{13 - i}\binom{13}{j}\binom{13}{13 - i - j}$$

in which all the cards are from three suits. Divide these by $\binom{52}{13}$ and add them to produce the answer. The sums are difficult to evaluate so is there an easier method to derive the answer?

(3) a. I employ a reasoning similar to problem (2): If a hand contains at least an ace and king of one suit, then it contains an ace and king of exactly one, two, three or four suits. The problem I face when counting is that a hand may contain the ace xor king of other suits.

b. Ditto.