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Poker, Probability, Combinatorics
Date: 11/04/97 at 14:40:51
From: Hein Hundal
Subject: Poker, Probability, Combinatorics
Recently on rec.gambling.poker, we were discussing a question that was
simple to pose, but fairly hard to answer.
Question: If we deal n hands consisting of 2 cards each, what is the
probability that there will be no pairs amoung the hands?
For example:
If hand1 = { Jack of clubs, Queen of clubs }, and
hand2 = { Jack of spades, 3 of diamonds}, then
two hands were dealt and no pairs appeared.
One poster figured out an algorithm the gives the answers, but I was
hoping there might be a power theorem that gives the probability in
terms of the number of values and suits in the card deck.
For a standard card deck with values 2,3,...,9,10, Jack, Queen, King,
Ace and 4 suits, the probabilities were
16 18448 48008 88348248 715247168
{--, -----, -----, ---------, ---------}
17 20825 57575 112559125 968008475
for 1,2,3,4, and 5 hands respectively.
Hein Hundal
Date: 11/04/97 at 18:11:45
From: Doctor Anthony
Subject: Re: Poker, Probability, Combinatorics
With just one pair, we can choose the first card in 52 ways, and the
second card in 48 ways. So the probability of no pair is
52 x 48 48 16
------- = ---- = ---
52 x 51 51 17
For 2 hands, the probability that the first hand is not a pair is
16/17.
For the second pair there are several possibilities depending on
whether the face values are completely different from those in the
first hand, or whether they repeat some of those values.
If completely different, the number of ways of selecting the hand =
44 x 40.
If one card is a repeat, the number of ways = 6 x 44 + 44 x 6
If two cards are repeats, the number of ways = 6 x 3
16[44 x 40 + 12 x 44 + 6 x 3] 36896
Total probability = ----------------------------- = ---------
17 x 50 x 49 41650
18448
= ------
20825
As you can see, the calculation is fairly laborious, as the various
possibilities have to be considered. There is no simple formula for n
hands.
-Doctor Anthony, The Math Forum
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