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ron8
02-23-2004, 06:52 AM
I am reasonably new to the game of Limit Hold'em and have been reading and learning the various odds.
I read that the odds of flopping a set are 7.5 to 1. when you hold a pocket pair. As you need one of only 2 cards left in the pack of 50 unseen cards I do not understand these odds.
I would be grateful if anyone can help me with the maths.

Regards
ron8

Henke
02-23-2004, 08:26 AM
P(flop set)=1-C(48,3)/C(50,3)=0.1176

Odds against flopping a set: 1/P(flop set)-1=7.5

Lost Wages
02-23-2004, 09:21 AM
To be exact, 7.51:1 is the odds against flopping a set, full house or quads when holding a pocket pair. The odds against flopping exactly a set are 8.28:1.

Lost Wages

BruceZ
02-23-2004, 10:17 AM
To make at least a set, you have 3 chances to catch one of the 2 cards. The probability of catching is 1 minus the probability of NOT catching 3 times in a row, and this is
1 - (48/50 * 47/49 * 46/48) = 11.76% = 7.5-to-1.