A bridge hand has been dealt, i.e. thirteen cards are dealt to each player. Given that your partner has at least one ace, what is the probability that he has at least two aces? Given that your partner has the ace of hearts, what is the probability that he has at least 2 aces? Answer these questions for a version of bridge in which there are eight cards, namely four aces and four kings, and each player is dealt two cards. (The reader may wish to solve the problem with a 52-card deck.)
I only solve the 8 card case for simplicity. The solution proceeds by using a counting argument.
I enumerate all possible ways to deal the 8 cards to the 4 players. We can assume all players are arrange in a line from left to right and numbered as players 1, 2,3 and 4. Assume the partner is positioned always as player 1. Assume cards are dealt in all possible permutations to the players from positions \(1 .. N\) where \(N=8\).
Each such permutation maps to a unique deal in which player \(i\) get the cards \(2i-1, 2i\) for \(i=1,2,3,4\).
There are \(8!\) possible outcomes of the permutations.
We need to calculate the number of those permutations with 2 aces divided by the number of those permutations with 1 ace to get the conditional probability.
The number of permutations with 2 aces is:
\[N_2 = 2{4\choose2} 6!\] This is because there are \(2{4\choose2}\) ways to pick 2 aces for positions 1 and 2. For each permutation of positions 1 and 2 with aces, there are \(6!\) ways to order the remaining positions \(3..8\).
The number of deals with 1 ace or more is equal to the complement of deals with 0 aces.
The number of deals with no aces is equal to the number of deals with 2 kings. By symmetry, the number of deals with only kings is the same as the number of deals with 2 aces.
This is \(N_2\).
Therefore, the solution is:
\[\frac{ 2{4\choose2}6! }{ 8! - 2{4\choose 2}6! } = \frac{ 12 \cdot 6!}{ 8 \cdot 7 \cdot 6! - 12 \cdot 6! } = \frac{12 \cdot 6!}{ 56 \cdot 6!- 12 \cdot 6!}= \frac{12}{44}=\frac{3}{11}\]
The answer is the conditional probability he has 2 aces given that he has 1 ace is \(3/11\).
By a counting argument for the numerator and denominator of the conditional probability, we can get the answer.
The denominator is the number of deals where ace of heart is included in hand 1 to player 1. This is equivalent to hands with ace of hearts in positions 1 or 2. Since these scenarios are disjoint, the total is \[ 2 \cdot 7!\] permutations.
The numerator is the number of deals where ace of heart and another ace are together. In these deals, ace of hearts could be in position 1. The other ace can be one of 3 cards. The remaining 6 cards can be distributed in any fashion (i.e. in any permutation). This gives the numerator to be:
\[ 2 \cdot 3 \cdot 6!\] This gives the ratio or conditional probability: \[ \frac{2 \cdot 3 \cdot 6!}{2 \cdot 7!} = \frac{2 \cdot 3}{2 \cdot 7} = \frac{3}{7}\]
Thus, the answer is \(3/7\) for the conditional probability of two aces where one of them is ace of hearts.