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re: getting a pair in five cards
Posted:
Mar 28, 2003 6:32 PM


In the simplest scenario, looking at getting one pair or better, and totally disregarding flushes and straights, from a 52 card deck the calculation goes something like this: (Calculation is for not getting a pair and then subtract from 1)
1. Deal a card, any card, say you get an Ace. 2. For the second card you want anything except an Ace, ie 48 out of the fifty one card keeps you pairless. Say you get a King. 3. For the third card you don't want an ace or a king so 44 of the fifty remaing cards keeps you pairless. Say you get a Queen. 4. For the fourth card you do not want an ACE, a King, or a Queen, so 40 of the remaining 49 cards keeps you pairless. Say you get a Deuce. 5 For the fifth card 36 of the 48 remaining cards leaves you pairless. so The probability of getting a pair or better is 1 [(48/51)*(44/50)*(40/49)*(36/48)]
The probability of getting exactly a pair is more complicate and might be appproach using the previous result and computing the probability of getting 4 fof a kind etc. This might tget you started. There is a direct calculation, but I don't have time to recreate it.
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