## Perfect Bridge Hand

**Question:** Bridge is a game where you are dealt 13 cards. Assuming the deck is a standard 52 card deck, and it is well shuffled, and no one else is in the deal, what are the chances that you would be dealt 13 cards of the same suit ?

**Answer:**We have 13 events of interest - that being drawing 13 individual cards. Each card draw is an independent event so we can simply multiply the probabilities of each of the 13 events happening.

More concretely, the first card we draw from the 52-card deck has no real constraints, thus the probability of this is 1. Any card will do. The second event is a card drawn from the remaining deck which has 51 cards, and it must be one of the other 12 cards that have the same suit as that in the first drawn card. Thus the probability for the second event is 12/51. Similarly, the likelihood of drawing the 3rd card from the remaining deck and it having the same suit as the first two cards drawn is 11/50.

Continuing this for all 13 cards we get the answer:

(1).(^{12}⁄_{51})(^{11}⁄_{50})(^{10}⁄_{49})(^{9}⁄_{48})(^{8}⁄_{47})(^{7}⁄_{46})(^{6}⁄_{45})(^{5}⁄_{44})(^{4}⁄_{43})(^{3}⁄_{42})(^{2}⁄_{41})(^{1}⁄_{40}) = ^{(12!)(39!)}⁄_{51!}

Alternately, we can consider the problem in terms of the total number of combinations. The number of different ways we can get 13 cards from a deck of 52 is: ^{52}C_{13} = 52! / (39!)(13!). since there are 4 suits there are only 4 ways to get 13 cards of the same suit, thus the probability of getting one of these 4 perfect bridge hands is:

4 / (52! / (39!)(13!)) = 4 (39!)(13!) / 52! = (39!)(13!)/ (13. 51!) = (39!)(12!)/51!

So about 1 in 158,753,389,900.

*02 Apr 2016*
*Damien Wintour*