Find the probability of a player being given exactly one ace in a Bridge deal

Rules of Bridge

 * http://www.pagat.com/boston/bridge.html
 * http://en.wikipedia.org/wiki/Contract_bridge

Method

 * 1) Find the probability that player 1 has exactly one ace.
 * 2) Find the probability that player 2 has exactly one ace given that player 1 has one ace.
 * 3) Find the probability that player 3 has exactly one ace given that players 1 and 2 each have one ace.
 * 4) Find the probability that player 4 has one ace given that players 1, 2, and 3 each have one ace.

Breaking down the solution
The probability that player 1 has exactly one ace is \[\frac{52\choose13}=0.4388\] The probability that player 2 has exactly one ace given that player 1 has one ace is \[\frac{39\choose13}=0.4623\] The probability that player 3 has exactly one ace given that players 1 and 2 each have one ace is \[\frac{26\choose13}=0.52\] The probability that player 4 has one ace given that players 1, 2, and 3 each have one ace is \[\frac{13\choose13}=1\] The overall probability is \[(0.4388)(0.4623)(0.52)=0.1055\]
 * \({52-4\choose12}={48\choose12}\) is the number of ways the 1st player can receive 12 cards with no aces.
 * \({4\choose1}\) is the number of ways the 1st player can receive 1 ace.
 * \({52\choose13}\) is the number of ways of drawing 13 cards from a deck of 52.
 * \({52-13-3\choose12}={36\choose12}\) is the number of ways the 2nd player can receive 12 cards with no aces.
 * \({3\choose1}\) is the number of ways the 2nd player can receive 1 ace.
 * \({52-13\choose13}={39\choose13}\) is the number of ways of drawing a 2nd hand.
 * \({52-13-13-2\choose12}={24\choose12}\) is the number of ways the 3rd player can receive 12 cards with no aces.
 * \({2\choose1}\) is the number of ways the 3rd player can receive 1 ace.
 * \({52-13-13\choose13}={26\choose13}\) is the number of ways of drawing a 3rd hand.
 * \({52-13-13-13-1\choose12}={12\choose12}\) is the number of ways the 4th player can receive 12 cards with no aces.
 * \({1\choose1}\) is the number of ways the 4th player can receive 1 ace.
 * \({52-13-13-13\choose13}={13\choose13}\) is the number of ways of drawing a 4th hand.

The mathematical information from the solution above will be used to write the understanding and method sections. The process of finding those mathematical facts will be written out as explicit commands in the method section. Also the rules of the game will be used to deduce why the non-mathematical facts are true in the understanding section (and why the mathematical facts are true).

We'll also write a summary of the main facts about the game of Bridge as given in the above links. Then when that's complete the links can be placed at the bottom in the References section and written in proper referencing format like the existing reference.

The following C program obtains a similar result to the above solution.
 * 1) include 


 * 1) define N 1000000

char deck[52] = { 'A','2','3','4','5','6','7','8','9','0','J','Q','K', 'A','2','3','4','5','6','7','8','9','0','J','Q','K', 'A','2','3','4','5','6','7','8','9','0','J','Q','K', 'A','2','3','4','5','6','7','8','9','0','J','Q','K', };

main {	int i, h[4], j, n = 0; for (i = 0; i < N; i++) { shuffle; h[0] = h[1] = h[2] = h[3] = 0; for (j = 0; j < 52; j++) if (deck[j] == 'A') h[j % 4]++; if (h[0] == 1 && h[1] == 1 && h[2] == 1 && h[3] == 1) n++; }	printf("%g\n", (float) n / N); }

shuffle {	int i, j, t;	for (i = 51; i > 0; i--) { j = rand % (i + 1); t = deck[i]; deck[i] = deck[j]; deck[j] = t;	} } $ gcc sim.c $ ./a.out 0.105337 The following Eigenmath session computes the probability as the exact fraction \(\frac{2197}{20825}\) > p1 = choose(48,12) * choose(4,1) / choose(52,13) > p2 = choose(36,12) * choose(3,1) / choose(39,13) > p3 = choose(24,12) * choose(2,1) / choose(26,13) > p1 * p2 * p3 2197 --- 20825 > float 0.105498

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