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#1
11-10-2005, 03:34 PM
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Factoring fold equity percentage.
Can someone please explain the numbers in the following equation to me in detail?
let x = % time he folds. so (1-x) = times he calls 7200 in pot 9200 left in your stack Assuming 22% equity when called Then your push is EV neutral when 7200x + (1-x){(.22)(22000) - 9200} = 0 Sorry, but I am not a math guy and need it spelled out for me. In the equation above, I believe he comes up with the 22% equity because he is drawing to the nut flush on the turn. This equation came from the following thread: link Thanks 
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#2
11-10-2005, 11:12 PM
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Re: Factoring fold equity percentage.
OK, how about a different situation...
If there is $5,000 in the pot, and you know you have 22% pot equity, how much fold equity do you need to push in your remaining $7,500 against a single opponent? I figure if the total pot is $20,000, and your pot equity is 22% ($4,400), I estimate you need at least 28% of fold equity to equal the remaining $5600 of your half to hit at least 50% to make it at least neutral EV. Is this correct?
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#3
11-12-2005, 01:44 PM
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Re: Factoring fold equity percentage.
The missing piece of information is that the bet is $3,600 to you. So if you go all-in, you call $3,600 and raise $5,600. If he calls your $5,600 raise, there is $7,200 + $9,200 + $5,600 = $22,000 in the pot.
If he does not call, you win the $7,200 in the pot. If he does call, you win $22,000 - $9,200 with probability 0.22 and lose $9,200 with probability 0.78. Since you lose the $9,200 either way, you can write this as 0.22*$22,000 - $9,200. Therefore, your overall expectation is 7200x + (1-x){(.22)(22000) - 9200} = 0 as you wrote. Note that you might have put in some of the $7,200 in the pot at the beginning (in heads up play, you will have put up $3,600 of it), but this is left out of the calculation. Once the money's in the pot, it doesn't matter who put it there.
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#4
11-12-2005, 01:51 PM
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Re: Factoring fold equity percentage.
Not quite. In this case, you are assuming there is no bet to you. If you bet $7,500 and he calls, you win $20,000 with probability 0.22, which is worth $4,400 as you say. You paid $7,500 to get this, so it's negative $3,100 EV.
So you're betting he folds, you get $5,000 if you win and pay $3,100 if you lose. You need to win with probability $3,100/($5,000+$3,100) = 31/81 or about 38% to break even. If you do this 81 times, you get the $5,000 when he folds 31 times for $155,000. 11 times he calls and you win, then you get $12,500 for a total of $137,500. 39 times he calls and you lose $7,500, for a total of negative $292,500. After all 81 hands, you're even.
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