# Probability problem with cards

Problem 6. A bridge hand is dealt, so each of 4 players has 13 cards from the 52 card deck. You have 8 clubs in your hand. What is the probability that at least one of the other three hands is void in clubs?

I don't know how to set up the problem at all. What would be the most simplest way to do it?

## 2 Answers

There's several different ways to do this. No guarantee that this is the simplest way to do this. In the following paragraphs, a hand is defined as a collection of 13-cards from the pool of available cards.

One way is to name the three other players, say Alice, Bob, and Charlie. Next, calculate the chance $p_{1}$ that Alice has a void in clubs. This is $$p_{1} = \frac{\texttt{# hands without clubs from 39 cards}}{\texttt{total # hands from 39 cards}}.$$ Multiply by 3, and you get the chance that Alice or Bob or Charlie has a void...

EXCEPT we double-counted the cases where two of the three have a void. Let $p_{2}$ be the probability that Alice and Bob both have a void. This happens if and only if Charlie has all 5 remaining clubs, so $$p_{2} = \frac{\texttt{# hands with all 5 clubs from 39 cards}}{\texttt{total # hands from 39 cards}}.$$

Your final answer would be $3p_{1}-3p_{2}$ (Normally, you would add $p_{3}$, the chance that all three have a void, but this is obviously equal to $0$. This is a case of the inclusion-exclusion principle, which is particularly useful when thinking through probability problems.)

Hopefully, this gives you a good launching point. Sometimes, the hardest part of probability questions is not double-counting and making sure you counted everything.

There are 39 cards left that you don't have, 5 clubs and 34 non-clubs. You need to group the cards into 3 groups of 13 cards for players P1,P2,P3.

The total number of combinations of possible hands is C(39,13)*C(26,13)*C(13,13).

The number of combinations of hands in which P1 has no clubs is C(34,13)*C(26,13)*C(13,13). Multiply this by 3 to cover # of hands where P1,P2, or P3 have no clubs. But we need to adjust for double counting.

We need to subtract for the hands in which more than one hand have no clubs. Obviously they can't all not have clubs, so we look at possibilities where 2 out of the 3 of P1,P2,P3 have no clubs.

The number of combinations where players P1 and P2 have no clubs is C(34,13)*C(21,13)*C(13,13). Multiply that by 3 to cover all combinations of 2 out of 3 players P1,P2,P3 that can have no clubs together. This result needs to be subtracted from our result from two paragraphs above to get the number of unique combinations of hands where at least one player has no clubs.

Therefore the answer is found by diving the number of combinations of hands where at least one player has no clubs by the number of combinations of possible hands.

Cancelling and combining some terms, we get...

P = (C(34,13)*C(26,13) - C(34,13)*C(21,13))*3 / (C(39,13)*C(26,13)) = 4961/14763 = .336