# Probability in a game of Bridge

• Mar 17th 2009, 08:11 PM
cunoodle2
Probability in a game of Bridge
I've researched this quite a bit. First off I'll post the problem then I'll show you some of what I think the work is to prove that I've attempted it.

The question:
In a bridge game each player has 13 cards so you and your partner have a total of 26 cards. What is the probability that you and your partner have exactly x Aces when x equals 0,1,2,3,4.

MY WORK (so far):
4/52 * 3/52 * 2/52 * 1/52 (this would be the equation IF you were only given 4 cards). Right??? Or is it something like (4!/52!). Totally confused.

We now have to add on the odds considering that you are getting 22 more cards which would be something like this...
48/52 * 47/52 * 46/52 * 45/52 * 44/52 * 43/52 * ......... * 28/52 * 27/52. Can you help me finalize this please? I've researched all over gooooogle and everywhere else I can think of.

I tried Probabilities Miscellaneous: Bridge Odds to get some insight but that didn't help me either.
• Mar 17th 2009, 09:08 PM
Soroban
Hello, cunoodle2!

You can't keep the denominator (52) constant.

Quote:

In a bridge game each player has 13 cards, so you and your partner have a total of 26 cards.
What is the probability that you and your partner have exactly $\displaystyle x$ Aces for $\displaystyle x = 0,1,2,3,4.$

There are: .$\displaystyle {52\choose26}$ possible outcomes.

There are 4 Aces in the deck, and 48 Others.

To get exactly $\displaystyle x$ Aces, there are: .$\displaystyle {4\choose x}$ ways.
We want $\displaystyle 26-x$ cards from the 48 Others: .$\displaystyle {48\choose 26-x}$ ways.
. . Hence, there are: .$\displaystyle {4\choose x}{48\choose 26-x}$ ways to get exactly $\displaystyle x$ Aces.

Therefore: .$\displaystyle P(x\text{ Aces}) \;=\;\frac{{4\choose x}{48\choose 26-x}}{{52\choose26}}$