Lurker > DrPez

So, there is a flaw in my solution. Conceptually it's on the right track. But it doesn't count the cards left in the deck with pairs and without pairs correctly. Which means that some the terms will get more complicated. Complicated in such a way that I don't particularly care to try to fix at this moment. I'll get back to you guys if I do find a good fix. But don't hold your breath.

DrPez posted...

How to extend this solution.

Note that I had to use P( X_1 ) to find the probability that 2 specific players do not have a pair. Similarly, I'd need to use P( X_2 | X_1 ) to find the probability that 3 specific players do not have a pair. So let me define some more terms.

Let Y_1 = X_1, and then Y_i = X_i | Y_i-1 for i > 1. Then the probability of n specific players not having a pair would be P( Y_1 ) * P( Y_2) * ... * P( Y_n )

Let d_i = # of cards in the deck for a given player i, i.e. d_1 = 48, d_2 = 44, and so on. d_i = 48 - 4*(i-1)
Let a_i = # of cards with no pairs remaining for a given player i (assuming all previous players also have no pairs), i.e. a_1 = 0, a_2 = 4, a_3 = 8, and so on. a_i = 4*(i-1)
Let b_1 = # of cards whose pair remains in the deck under the same conditions, i.e. b_i + a_i = d_i

Then for 1 < n < 12
P( Y_n ) = [ (4 C 0 * b_n C 4) / (d_n C 4) ] * [ (b_n / b_n) * (b_n - 2 / b_n - 1 ) * ( b_n - 4 / b_n - 2) * ( b_n - 6 / b_n -3 ) ]
+ [ (a_n C 1 * b_n C 3) / (d_n C 4) ] * [ (b_n / b_n) * (b_n - 2 / b_n - 1 ) * ( b_n - 4 / b_n - 2) ]
+ [ (a_n C 2 * b_n C 2) / (d_n C 4) ] * [ (b_n / b_n) * (b_n - 2 / b_n - 1 ) ]
+ [ (a_n C 3 * b_n C 1) / (d_n C 4) ] * [ (b_n / b_n) ]
+ [ (a_n C 4 * b_n C 0) / (d_n C 4) ] * 1

and P( Y_12 ) = 1 because no cards with pairs will remain in the deck after the 11th player doesn't have a pair.

In particular if one wishes to find the probability that no player has a pair, you would need to calculate
P(Y) = P( Y_1 ) * P( Y_2 ) * P( Y_3 ) * P( Y_4 ) * P( Y_5 ) * P( Y_6 ) * P( Y_7 ) * P( Y_8 ) * P( Y_9 ) * P( Y_10) * P( Y_11 ) * P( Y_12 )

Then to get the probability that at least 1 player has a pair, you would calculate 1 - P(Y)

Found a small error on a reread. Fixed it.

Moonroof posted...

So what are the final answers to my two questions

The odds of a particular player having at least one pair is 1 - .8738 or 12.62%. No one has a number for the probability that at least one player out of 12 has a pair. But I gave an algorithm that could find that number. One simply has to follow the process repeatedly to find the actual number.

How to extend this solution.

Note that I had to use P( X_1 ) to find the probability that 2 specific players do not have a pair. Similarly, I'd need to use P( X_2 | X_1 ) to find the probability that 3 specific players do not have a pair. So let me define some more terms.

Let Y_1 = X_1, and then Y_i = X_i | Y_i-1 for i > 1. Then the probability of n specific players not having a pair would be P( Y_1 ) * P( Y_2) * ... * P( Y_n )

Let d_i = # of cards in the deck for a given player i, i.e. d_1 = 48, d_2 = 44, and so on. d_i = 48 - 4*(i-1)
Let a_i = # of cards with no pairs remaining for a given player i (assuming all previous players also have no pairs), i.e. a_1 = 0, a_2 = 4, a_3 = 8, and so on. a_i = 4*(i-1)
Let b_1 = # of cards whose pair remains in the deck under the same conditions, i.e. b_i + a_i = d_i

Then,
P( Y_n ) = [ (4 C 0 * b_n C 4) / (d_n C 4) ] * [ (b_n / b_n) * (b_n - 2 / b_n - 1 ) * ( b_n - 4 / b_n - 2) * ( b_n - 6 / b_n -3 ) ]
+ [ (4 C 1 * b_n C 3) / (d_n C 4) ] * [ (b_n / b_n) * (b_n - 2 / b_n - 1 ) * ( b_n - 4 / b_n - 2) ]
+ [ (4 C 2 * b_n C 2) / (d_n C 4) ] * [ (b_n / b_n) * (b_n - 2 / b_n - 1 ) ]
+ [ (4 C 3 * b_n C 1) / (d_n C 4) ] * [ (b_n / b_n) ]
+ [ (4 C 4 * b_n C 0) / (d_n C 4) ] * 1

Prob a particular player does not have a pair.

48/48 * 46/47 * 44/46 * 42/45 = .8738

Each fraction represents a card in the player's hand. For the first card there are no restrictions, the player can receive all 48 cards. For the second card, there is only one card in the deck the player cannot receive. The third card there are 2 cards the player can't receive. And following that pattern, the 4th card there are 3 cards the player cannot receive.

Unfortunately this approach doesn't extend easily to multiple players. You have to keep track of what the first player has.

****

Here's the probability that a specific pair of players each do not have a pair. Let me define some symbols.

Let X1 be the event that player 1 does not have a pair. Let X2 be the event that player 2 does not have a pair. Then, we are looking for P(X1 AND X2) = P(X1) * P(X2 | X1). Luckily we already know P(X1) = .8378. Let's now calculate P(X2|X1). Let Y = X2 | X1.

In this situation there are 20 pairs left and 4 cards without a pair for a total of 44 cards left in the deck. Player 2 can get 0, 1, 2, 3, or 4 of the cards that have no pairs.

Let Z0 be the event that player 2 got 0 of those cards.
Let Z1 be the event that player 2 got 1 of those cards.
Let Z2 be the event that player 2 got 2 of those cards.
Let Z3 be the event that player 2 got 3 of those cards.
Let Z4 be the event that player 2 got 4 of those cards.

Then, P(Y) = P(Y AND Z0) + P(Y AND Z1) + P(Y and Z2) + P(Y and Z3) + P(Y and Z4)
= P(Z0) * P(Y | Z0) + P(Z1) * P(Y | Z1) + P(Z2) * P(Y | Z2) + P(Z3) * P(Y | Z3) + P(Z4) * P(Y | Z4)

P(Y) = [ (4 C 0 * 40 C 4) / (44 C 4) ] * [ (40/40) * (38/39) * (36/38) * (34/37) ]
+ [ (4 C 1 * 40 C 3) / (44 C 4) ] * [ (40/40) * (38/39) * (36/38) ]
+ [ (4 C 2 * 40 C 2) / (44 C 4) ] * [ (40/40) * (38/39) ]
+ [ (4 C 3 * 40 C 1) / (44 C 4) ] * [ (40/40) ]
+ [ (4 C 4 * 40 C 0) / (44 C 4) ] * 1
P(Y) = P (X2 | X1) = .8745

Now we can calculate P( X1 AND X2) = P(X1) * P(X2|X1) = .8738 * .8745 = .7641

You can see that repeating this process would be tedious. But doable.