EV of an all-in and why the books are wrong



Sep 21, 2007
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Assume all effective stacks are $200

I’m in EP.
Dealt to Hero Kd, 2d
6 limpers for $5 including blinds who check.
Pot = $30
*** FLOP *** Ad, Qd, 2h
Blinds check.
Hero bets $15 – kind of a probe bet, see where I am, build the pot.
Button calls, SB calls.
Pot $75
*** TURN *** 5c
Hero checks.
Button bets $25
SB calls $25

I was really torn here between calling and raising. If I call I’m getting 5:1 and I have probably 14 outs so 2.6:1 against – this is where I think a lot of the books are wrong because they would just tell you to call based on pot odds. But if I miss I will lose the whole pot. On the other hand I felt that there was a really good chance I could take down $125 for sure with an all in. The button could be betting with anything but probably a Q or else he would have raised the flop. The SB just called so he can’t be that strong, maybe a flush draw drawing dead.

Look at the EV. If I go all in and he calls I am 2.6:1 but only getting 1.7:1 on my $$. That has an EV of -$44.2. However there is a pretty good chance he’ll fold. In fact I only need there to be a 26% chance that he folds to be breakeven EV.

EV = 26% * $125 [if he folds] + 74% * { 28% * $305 – 72% * $180 } [if he calls]
EV = $0



Cardschat Elite
Sep 2, 2007
Total posts
true, but by flat calling, we allow ourselves to get away from it when we miss, but then also have implied odds after we hit. As you cant put a guess on what implied odds you will be getting, you have to put out an estimate. So lets say after hitting our flush we will get another 50$ from 1 of the villains, seems reasonable.

Therefore, we are getting 5:1 on our money to call the initial bet making it profitable to call that in the long run. But if we include implied odds, we can change that value to about 7:1. Meaning we only have to hit once in every 7 to make a profit. Law of averages say that we will hit our draw at least once.

I hope that makes sense. Also, im not really sure if it completes the arguement about which is most profitable, but if it doesnt, its a pretty good article about why pot odds nad implied odds are profitable :smile:

Edit: Please dont limp with k,2 suited if the above situation is based on a real event
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Jul 4, 2005
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In practice, I wouldn't be so sure about your assessment of 14 outs (especially if your all-in gets called), but let's assume that's true.

EV(calling) = 0.28 * $125 - 0.72 * $25
EV(calling) = +$17

EV(raising) = p(F) * $125 + (1 - p(F)) * (0.28 * $305 - 0.72 * $180)
EV(raising) = ($125 * p(F)) + ($44.2 * p(F)) - $44.2
EV(raising) = ($169.2 * p(F)) - $44.2

Since EV(raising) must be at least EV(calling) to make it worthwhile...

$17 < ($169.2 * p(F)) - $44.2
$61.2 < $169.2 * p(F)
p(F) > 0.36

So if you think there's more than a 36% chance of both of them folding to an all-in, then all-in is better than calling.


Sep 14, 2007
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^^^^^ what viking said, plus you get the ev from implied odds. If you hit with money behind, you can win more when you hit, and lose nothing more if you miss.


Jul 4, 2005
Total posts
I think you can get folds more than 36% of the time, but if you factor in the implied odds of calling, I'm inclined to say it's a wash. And when it's equal EV, I go with the lower risk option, so just calling.