Probability Analysis in Bridge: Distribution of Aces Among Players

In a typical bridge game, each player is dealt 13 cards from a standard deck of 52 cards. This article will explore the probability that at least 3 players hold an ace, under the condition that at least 2 players hold an ace. We will delve into the calculation steps, providing a detailed breakdown of the mathematical approach used.

Introduction to the Problem

The scenario involves a standard deck of 52 cards, with 4 aces distributed among 4 players. The core question is to find the probability that at least 3 persons hold an ace, given that at least 2 persons hold at least 1 ace. This can be expressed as the conditional probability ( P(A mid B) ), where:

( A ): The event that at least 3 persons hold an ace. ( B ): The event that at least 2 persons hold at least 1 ace.

Step 1: Calculate ( P(B) ) - The Total Distributions of Aces

To calculate ( P(B) ), we need to determine the number of ways to distribute the 4 aces among the 4 players such that at least 2 players have at least 1 ace.

Total distributions of aces:

Each of the 4 players can receive 0 to 4 aces. Using the stars and bars method, the total number of ways to distribute 4 aces among 4 players is:

( binom{n k-1}{k-1} binom{4 4-1}{4-1} binom{7}{3} 35 )

where ( n 4 ) (number of aces) and ( k 4 ) (number of players).

Next, we need to subtract the cases where fewer than 2 players have aces:

Case 1: No player has any aces. This is 1 way. Case 2: Exactly 1 player has all 4 aces. There are 4 ways to choose this player.

The total number of distributions where fewer than 2 players have aces is 1 4 5.

Hence, the number of ways where at least 2 players have at least 1 ace is:

( P(B) frac{35 - 5}{35} frac{30}{35} frac{6}{7} )

Step 2: Calculate ( P(A cap B) ) - The Intersection of Events

To find ( P(A cap B) ), we need to calculate the probability that at least 3 players have an ace, and at least 2 players have at least 1 ace.

Cas 1:

Exactly 3 players have aces. Choose 3 players from 4 to have aces: ( binom{4}{3} 4 ). Distribute 4 aces among these 3 players such that each has at least 1 ace. This can be done in ( binom{13-1}{3-1} binom{3}{2} 3 ).

Therefore, the total for this case is:

( 4 times 3 12 )

Cas 2:

All 4 players have aces. There is only 1 way to distribute 4 aces such that each gets 1 ace.

Therefore, the total for this case is:

( 1 )

Combining both cases:

( P(A cap B) frac{12 1}{35} frac{13}{35} )

Step 3: Calculate ( P(A mid B) ) - The Conditional Probability

Now, we can find the conditional probability:

( P(A mid B) frac{P(A cap B)}{P(B)} frac{frac{13}{35}}{frac{6}{7}} frac{13}{35} times frac{7}{6} frac{91}{210} frac{13}{30} )

Final Answer:

Thus, the probability that at least 3 persons hold an ace given that at least 2 persons hold at least 1 ace is:

( boxed{frac{13}{30}} )