Re: \"X combinations of being dealt...XX\"
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For example I hold KK and I put the opponent on another High pocket pair... well there are 6 ways of being dealt AA and there are 6 ways of being dealt KK (of which my hand prevents 3 of them)
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Your hand prevents 5 of them, there is only 1 left.
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and 6 ways of being dealt QQ, JJ, 10-10
So...
Part 1) I am ahead of 18 possibilities, tied with 3 possibilities and behind 3 possibilities.
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You are tied with 1, and behind 6 (AA).
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Part 2 (Where I fall down) Of the hands that I am ahead, there are 6 possible cards that can come to beat me (a card to make trips) ... so that is 6:48 or I'm an 8-1 favorite?
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There are only 2 cards that can beat you if you only have 1 opponent. You are against one of JJ, QQ, or TT, not all 3. There are 48 cards remaining, so you are a 46:2 = 23:1.
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Of the hands I am behind, there are two cards that can rescue me (another K) so that is 2:24 or I am a 21-1 underdog?
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46:2 = 23:1 underdog.
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Part 3... When I am ahead, I will win, 78.5% of the time. When I am behind, I will win 5% of the time... so I have a 83.5% chance of winning?
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You have to multiply the 78.5% by the chance that you are ahead, and the 5% by the chance you are behind. Taking the numbers from above, where you are ahead 18 out of 22, behind 3 out of 22, and tied 1/22:
18/22 * 78.5% + 3/22 * 5% = 64.9%.
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