Probabilities Involving Bridge

In summary: I was thinking in terms of 1 suit or 2 suits. So it's actually ## \binom{4}{1}\binom{39}{13} - \binom{4}{2}\binom{26}{13} + \binom{4}{3}\binom{13}{13} ##.Ah yes, good catch. I was thinking in terms of 1 suit or 2 suits. So it's actually ## \binom{4}{1}\binom{39}{13} - \binom{4}{2}\binom{26}{13} + \binom{4}{3}\binom{13}{13} ##.And we can simplify that further to
  • #1
e(ho0n3
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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

[tex]\binom{4}{2} \sum_{i=1}^{12}\binom{13}{i}\binom{13}{13 - i}[/tex]

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

[tex]\binom{4}{3} \sum_{i=1}^{12} \binom{13}{i}\sum_{j=1}^{13 - i}\binom{13}{j}\binom{13}{13 - i - j}[/tex]

in which all the cards are from three suits. Divide these by [itex]\binom{52}{13}[/itex] 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.
 
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  • #2
e(ho0n3 said:
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.
e(ho0n3 said:
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.
e(ho0n3 said:
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
 
  • #3
@haruspex do you know bridge? I think in order one to solve this problem must know the basics of bridge right?
 
  • #4
Delta2 said:
@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.
 
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  • #5
So basically the deck is isomorphic to an ordinary deck except that you call the 13 numbers denominations and the hands suits?
 
  • #6
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.
 
  • #7
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".
 
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  • #8
haruspex said:
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.
 
  • #9
pbuk said:
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".
 
  • #10
haruspex said:
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?
 
  • #11
pbuk said:
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.
 
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  • #12
haruspex said:
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.
 
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What is the probability of being dealt a specific hand in bridge?

The probability of being dealt a specific hand in bridge depends on the number of cards in the deck and the number of players. With a standard 52-card deck and four players, the probability of being dealt a specific hand is approximately 1 in 635,013,559,600.

How do probabilities affect bidding strategies in bridge?

Probabilities play a crucial role in bidding strategies in bridge. Bidders use their knowledge of probabilities to determine the likelihood of their hand and the potential of their opponents' hands. This information helps them make more informed bids and ultimately improve their chances of winning the hand.

What is the probability of winning a bridge game?

The probability of winning a bridge game depends on various factors such as the skill level of the players, the bidding strategies used, and the cards dealt. Generally, the probability of winning a bridge game is around 50%, assuming all players have equal skill levels.

How do probabilities impact the play of a hand in bridge?

Probabilities heavily influence the play of a hand in bridge. Players use probabilities to make decisions on which cards to play, which suits to focus on, and when to take risks. A thorough understanding of probabilities can greatly improve a player's chances of success in bridge.

What is the role of probability in calculating bridge scores?

Probability is essential in calculating bridge scores. In duplicate bridge, scores are based on the probability of achieving a certain contract, which is determined by the number of tricks bid and made. In rubber bridge, probabilities are used to determine the score for each hand based on the number of tricks won and lost.

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