LogFAQs > #938655222

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.

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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.