Re: Odds of at least 1 player of 6 having pocket pair
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Do you really care for an exact number? Just assuming that everyone is 16:1 to get a pair and doing 1-(16/17)^5 should get you close enough for practical purposes.
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Now let's assume that I open/raise with 88 and the five remaining players will call with 99+,AQo+,AJs+.
That's 72 different hands.
(1326-72)/72 = 17.4167:1 odds that at least one of 5 players have one of the above hands.
Using your formula:
1-(17.4167/18.4167)^5 = 24.36% chance.
My formula:
1-[(1225-72)/1225]^5 = 26.13%
Now we have a bit of a difference. My formula is easier in this situation, however, I don't want to use it, if yours is more accurate.
Anyone know which one is a more accurate approximation?
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