Probabilities Involving Bridge

[e(ho0n3]

Homework Statement
(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.

 

[haruspex]

assume that it asks for the probability that one player has 5 spades and another has the remaining 8

No, it is specifically one pair, you and your partner.

The book seems to be ignoring who gets the spades and I don't know why.

It further specifies that you have the 5 and your partner the 8, so it is just m/n.

hands in which the cards are all from two suits

There's a much easier way. There are C(39,13) hands which omit a particular suit, (26,13) that omit a particular two suits.

For (3), apply the principle of inclusion and exclusion: https://en.wikipedia.org/wiki/Inclusion–exclusion_principle

 

[Delta2]

@haruspex do you know bridge? I think in order one to solve this problem must know the basics of bridge right?

 

[haruspex]

@haruspex do you know bridge? I think in order one to solve this problem must know the basics of bridge right?

I do, but all you need to know in addition to the info in post #1 is that the cards in a deck have 13 denominations; that there are four suits each containing one card of each denomination; that 'void in a suit' means the hand has none of that suit; that two of the denominations are Ace and King.

 

[Delta2]

So basically the deck is isomorphic to an ordinary deck except that you call the 13 numbers denominations and the hands suits?

 

[pbuk]

No, the deck is an ordinary deck with only 9 numbers [2, 10]. The other denominations of a deck (or "pack" in the UK) are J, Q, K and A, which are not numbers.

There are 4 suits, C, D, H and S. The cards each player is dealt is called their hand: this is the meaning of "hand" in q2.

A "hand" can also mean a whole round in which all 4 players are dealt 13 cards and play proceeds until the round is finished: this is the meaning of "hand" in q1.

 

[pbuk]

In other words the cards and the terminology is exactly the same as most other games played with a 52 card deck, with specific additions such as "partner".

 

[pbuk]

There's a much easier way. There are C(39,13) hands which omit a particular suit, (26,13) that omit a particular two suits.

We don't care about voids in two suits, so the answer is even simpler.

 

[haruspex]

We don't care about voids in two suits, so the answer is even simpler.

Not quite that simple. There are C(39,13) hands which omit a particular suit, but the question says "at least one".

 

[pbuk]

Not quite that simple. There are C(39,13) hands which omit a particular suit, but the question says "at least one".

Of those $\binom{39}{13}$ hands, some of those omit a second suit. Why does this matter?

 

[haruspex]

Of those $\binom{39}{13}$ hands, some of those omit a second suit. Why does this matter?

At least one means void in S, H, D, C or some combination. $\binom{39}{13}$ is the number of ways of being void in S. So to allow other voids we can multiply by four, but then we have overcounted, so need to subtract a bit, etc.

 

[pbuk]

So to allow other voids we can multiply by four, but then we have overcounted, so need to subtract a bit, etc.

Ah yes, good catch.