# Probability in a game of Contract Bridge

in a game of contract bridge, partner and me together have 11 cards in a suite. The remaining 2 cards in the same suite can be distributed amongst our opponents as either 1-1 or 2-0.

What is the probability that it will be distributed as 1-1 and what is the probability it will be distributed as 2-0?

Once the above is solved, how can you extend it if more cards are missing. That is, let us assume that partner and me have 8 cards between us. How do you calculate the probability for the distributions (5-0, 4-1, 3-2)?

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A rough rule of thumb for uneven breaks is the chance that cards will break n-m is somewhat less than 100/(n-m)% for n>m+1. This overstates the chances somewhat and gets worse for bad breaks. So 4-1 is less than 33% (actually 28.26%) and 4-2 is less than 50% (actually 48.45) –  Ross Millikan Nov 16 '10 at 14:19

You can use the technique of Vacant Spaces to calculate probabilities.

During play, if you know that there is space for $\displaystyle x$ unknown cards with West and $\displaystyle y$ unknown cards with East, the probability that a specific card is with west is $\displaystyle x/(x+y)$. This assumes you ignore the carding of the opponents so far. This gives accurate results when computing the apriori probabilities and gives reasonable values during the play of the hand.

So, say you want to calculate the apriori probability of a 3-2 split.

First consider a specific holding say (Q32 with West and 54 with East) and try to calculate the chances.

Chances that West gets the Q = 13/26.
Chances that West gets the 3 = 12/25.
Chances that West gets the 2 = 11/24.

Chances that East gets the 5 = 13/23.
Chances that East gets the 4 = 12/22.

Thus the chances of the specific split Q32-54 is

$13/26 \times 12/25 \times 11/24 \times 13/23 \times 12/22$

There are 5 choose 3 + 5 choose 2 = 20 such splits.

Thus the probability of 3-2 split is 20 times

$13/26 \times 12/25 \times 11/24 \times 13/23 \times 12/22$

$= 0.678\dots$

Thus there is a 67.8% chance of a 3-2 break.

A good rule of thumb to remember is that if there are an odd number of cards outstanding, then apriori chances are that they will split as evenly as possible. So with 5 cards outstanding, 3-2 is split is more likely than 4-1 or 5-0.

If there are an even number of cards outstanding, then they tend to break unevenly. i.e with 2 cards outstanding, 2-0 split is more likely than 1-1.

See:

An article on Vacant Spaces: http://www.clairebridge.com/textes/vacantplaces.pdf

Pavlicek's Suit Break Calculator: http://www.rpbridge.net/xsb2.htm

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Actually, as Yuval shows, two cards break 1-1 with 52% chance. For larger even numbers they tend to break one off even, so six cards will most likely break 4-2 than 3-3. In a sense that is because there are two 4-2 breaks (4-2 and 2-4), each less likely than 3-3 but totaling more. –  Ross Millikan Nov 16 '10 at 14:13
@Ross: Yes, that is correct. By "3-2" I meant 3-2 and 2-3 etc. It is common among bridge players :-) –  Aryabhata Nov 16 '10 at 15:44
Yes, your 67.8% is correct for that. –  Ross Millikan Nov 16 '10 at 18:43
@Moron: I have in the past, not recently. But thinking about getting back in –  Ross Millikan Nov 16 '10 at 18:48

The number of cards in suit a given opponent gets is distributed hypergeomtrically. In your particular case, $1-1$ is slightly more likely, with probability $13/25 = 1/2 + 1/50$. This is not surprising, since once you got one of the cards, the probability you'll get another drops, since then there's only one of them.

EDIT: The hypergeometric distribution with parameters $(N,m,n)$ concerns a situation in which the universe consists of $N$ cards, $m$ of which are special, and you are drawing $n$ of them. The probability that the number of special cards you draw is $k$ is exactly $$\frac{\binom{m}{k}\binom{N-m}{n-k}}{\binom{N}{n}},$$ since there are $\binom{m}{k}\binom{N-m}{n-k}$ choices for a hand with exactly $k$ special cards, and $\binom{N}{n}$ total hands.

In order to find the distribution of $1-1$, you plug in $k=1$. In order to find the distribution of $2-0$, you plug in $k=0$ and $k=2$, and add, since you don't care which player gets none and which gets both.

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Perfect! Yes, it is indeed 48% for 2-0 and 52% for 1-1. But, then how do you get the probability for the other distributions? I quite did not understand the hypergeometric distribution. –  Kanini Nov 16 '10 at 6:43