Here is the EV of dealing it twice, that I think everyone wants to know but is too lazy to work out. When you deal twice, you do not change the expected value! Let's say that two players are all-in on the turn, and only the river card remains. Let's say that player A wins with 13 of the remaining 43 cards, and player B wins with 30 of them. So the EV of player A dealing it once is (+13-30)/43 = -17/43*1. The EV with dealing it twice is (the chance of winning it twice)*payoff + (chance of winning once, losing once)*payoff, and (chance of losing twice)*payoff. winning twice: 13/43*12/42 * 1 + one lose, one win (13/43*30/42 + 30/43*13/42)*0 + two losses 30/43*29/42 * -1 = (156 - 870)/(43*42) = -714/(43*42) = -17/43 So the "scoop the pot" theory is BS, since although you decrease your odds of scooping, you also decrease your odds of losing both by the same amount. So definitively, only the variance is changed, not the EV. There's the math to prove it! But then again maybe I'm stoned and made a mistake :) Jojo Mcbean www.wikipollo.com
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Re: "Maybe I'm just stoned and unable to multiply correctly."
Date: 2008-04-01 23:54 (UTC)winning twice:
13/43*12/42 * 1 +
one lose, one win
(13/43*30/42 + 30/43*13/42)*0 +
two losses
30/43*29/42 * -1
= (156 - 870)/(43*42) = -714/(43*42) = -17/43
So the "scoop the pot" theory is BS, since although you decrease your odds of scooping, you also decrease your odds of losing both by the same amount. So definitively, only the variance is changed, not the EV. There's the math to prove it! But then again maybe I'm stoned and made a mistake :)
Jojo Mcbean
www.wikipollo.com