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