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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
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
College Probability
High School Permutations and Combinations
High School Probability

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