Monday, August 12, 2013

What are the chances of different "splits" in bridge?

If you know how to play bridge, skip to the fourth paragraph!
In bridge, 13 cards are dealt to each of 4 players (so all 52 cards are dealt).  Players sitting across from each other are partners, so we could think of the two teams positions as North and South and East and West on a compass.  A process of "bidding" ensues, in which the team with the highest bid has selected a "trump" suit and a number of rounds, or "tricks" that they have contracted to take.

Suppose North-South had the highest bid and North is playing the hand.  Then East "leads" a card, meaning East places a card (any card he/she wants) face up on the table.  The play goes clockwise, East-> South-> West -> North.  South, West and North must play a card of the same suit that East played.  When four cards are down, the highest one wins the "trick" and that winner puts any card of his/hers down, in order to begin a new trick.  Play continues until 13 rounds of 4 cards each have been played.

Suppose that West wins a trick and thus gets to play a card.  He plays the Ace of Hearts.  North, who is next and otherwise required to play hearts, is out of hearts.  North can play any other suit, but if he chooses to play the "trump" suit (say Spades are trump), then he automatically wins the trick unless East or South is also out of hearts and play a higher card in Spades (the trump suit).  In other words, trumps are very valuable.  In the bidding process, the teams try to bid in such a way that the trump suit is one in which they have a lot of cards.  Generally, the team with the winning bid (the "contract") will have at least 7 of the 13 trumps between the two of them, meaning the other team will have 6 or fewer.  Whatever the number the opponents have, it is generally advantageous to the contract winners if they have the same number each rather than them being skewed to one or the other opponent.

Bridge players begin here:
So here is the probability piece.  Suppose you and your partner hold 7 trumps between you, what are the chances the opponents each have 3?  have 4 and 2?  have 5 and 1?  have 6 and 0?  To solve this sort of problem, we use combinations.  See my earlier post for some detail (and more odds of bridge hands).

The opponents have 26 cards altogether and we want to know the number of different groups of six among those 26 cards.  Think of this process as a process of picking six cards from the 26.  You have 26 choices for the first card, 25 for the second, and so on, and thus there are 26*25*24*23*22*21 total 'permutations' of size 6.  However, we do not care what order they are in so for each first card, there are 6 possible positions, for each second card, 5, etc., and thus we need to divide these permutations by 6*5*4*3*2*1, in order to get the number of unique sets when order does not matter. Again, see my earlier post for a more detailed explanation of this concept.

The R language allows for calculation of this combination of 6 out of 26 with the command "choose(26,6)." This is the denominator when we calculate probabilities, because it gives the total number of equally likely combinations of 6 cards.  The numerator is split into the two bridge hands of 13 cards each.   The number of combinations with an even 3-3 split are "13 choose 3" for both hands.
To calculate that probability in R, we write:   choose(13,3)*choose(13,3)/choose(26,6) and get 35.5%

How about hands with a 4-2 split?  That is the chance that Opponent 1's hand has 4 trumps multiplied by the chance that Opponent 2's hand has 2 trumps PLUS the chances that Opponent 2's hand has 4 trumps multiplied by the chance that Opponent 1's hand has 2 trumps.  Since the chance that either Opponent has 4 are the same, we can just double the probability of Opponent 1 having 4 and Opponent 2 having 2.  We get: choose(13,4)*choose(13,2)*2/choose(26,6) = 48.4% of one opponent having 4 and the other having 2 trumps.

Continuing this calculation, we get the following chances for hands with 6 trumps in the opponents hands( 6 trumps "out"):
3-3 split : 35.5%
4-2 split: 48.4%
5-1 split: 14.5%
6-0 split:  1.5%

For hands with 5 trumps out, we get:
3-2 split: 67.8%
4-1 split: 28.3%
5-0 split: 3.9%

For hands with 4 trumps out:
2-2 split: 40.7%
3-1 split: 49.7%
4-0 split: 9.5%

For hands with 3 trumps out:
2-1 split: 78%
3-0 split: 22%

For hands with 2 trumps out:
1-1 split: 52%
2-0 split: 48%

I find it interesting that the even split (for 2, 4, or 6 trumps out) is only the most likely scenario when 2 trumps are out.  When 4 trumps are out, a 3-1 split is more likely.  When 6 are out, a 4-2 split is more likely.