Re: Odds question.
What are the odds of finishing with quads in hold'em where you also have to use both your pocket cards to meet the conditions of a site's bonus, please?
There is a very fine line between "hobby" and "mental illness."
Assuming you start with a pair: (using hypergeometric distribution)
50 cards you have not seen before flop; two cards required on board make you quads; &
of course 5 cards on board after river….
Chances of making quads = Combin(2,2)*Combin(48,3)/Combin(50,5) = 1/122.5
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Testing Wilson Holdem Software:
Monte Carlo (10^6) trials Wilson Holdem software (pair 8’s seed=45): Probability = 8509/1000000 = 117.5
Monte Carlo (10^6) trials Wilson Holdem software (pair 8’s seed=57): Probability = 8476/1000000=117.98
Monte Carlo (10^6) trials Wilson Holdem software (pair 8’s seed=649): Probability = 8597/1000000 =1/116.3
Monte Carlo (10^6) trials Wilson Holdem software (pair Q’s seed=349): Probability = 8434/1000000=1/118.57
Monte Carlo (10^6) trials Wilson Holdem software (pair 8’s seed=911): Probability = 8471/1000000=118.05
For the Wilson Monte Carlo trials I listed: you can see that the trials predict Quads a little more frequently than the Hypergeometric distribution probability.
Assuming you get a pair in holdem about 1 in 17 hands. Then the probability of quads (before hand) on the river is (1/17)*(1/122.5) = 1/ 2082.5 = 2 in 4165.
Like Buzz says: just my opinion & subject to correction if in my haste I goofed….
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