Hey guys.

Let's start with the simple example.
We're HU.
Hero is SB.
Effective stack is S BB.
Initial pot after the blinds've been posted is IP BB.

So we wanna calculate EV(shove).
Let's say the probabilty of villain fold is F and the probabilty of villain calls is c. (yeah I now that in this particular case c=(1-f) but w/e).
And we know hero's equity Eq against his calling range.

So

Ev(shove) = IP*F + Ev(being called)
Ev(being called) = (2S+IP)Eq - S(Eq-1).

Hope I didn't missepeled, but w/e.
I know how to calculate Ev(move) in such simple examples.

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But here's my question.
How to consider more than one viallin?


For instance how to calculate Ev(shove) if there are 4 players in the game and we're on the CO?

My initial thought was to try to determine average (arithmetical mean ) calling /folding range and calcualte just like we're HU again.
But I doubt that this is gonna work.

Not quite right.

EV(shove) = F*1.5 + (1-F)*( -(s-0.5) + eq*(2s))

Well let's start with 3-handed first. Let's assume (very close to correctly unless stacks are very short or one of the villains is very loose) that you're never getting called by both. If SB and BB's call ranges are C1 and C2 respectively, and your equity against those ranges are E1 and E2, then

EV(shove) = (1-C1-C2)*(1.5) + (C1)*(-S+E1*(2S+1)) + (C2)*(-S+E2*(2S+0.5))

It's pretty easy to extend that to more than 3 handed.

Allowing more than one player to call makes the calculation quite a bit more complicated since presumably they will have different ranges depending on how many people are all-in in front of them. But it's just a matter of calculating your equity against each combination of calling opponents, and adding them up weighted by the probability of each combination.

I'm not gonna write it out in detail but it'll look like this for 3 handed:

EV(Shove) = P(both fold) * 1.5 + P(SB calls, BB folds) * EV(SB calls, BB folds) + P(SB folds, BB calls) * EV(SB folds, BB calls) + P(both call) * EV(both call)

After that it's just a matter of filling in the blanks. For more villains it will be a lot longer, there would be 2^N terms for N villains.

missepeled in the original )))

DarkMagus, thank you very much.