A heads-up player opens 78% of the buttons (let's say he min raises to 2 bbs). He calls shoves with 12% of the hands he opens and folds the rest.

I am interested in calculating the EV of shoving ATC over his raises. Equity of ATC against a 12% range is 0.33 according to Pokerstove. Effective stack size is S.

I calculate it like this and would like to confirm that it is correct:

<shoving ATC> = p(opens)[p(folds)(pot size) + p(calls)p(we win)(pot size)-(eff stack size)] + p(open folds)(blinds)

alternatively I can write the equation as:

<shoving ATC> = p(opens)[p(folds)(pot size) + p(calls)[p(we win)(eff stack size)-p(lose)(eff stack size)] + p(open folds)(blinds)

using the first version:

<shoving ATC> = (0.78)[(0.88)(3) + (0.12)(0.33)(2*S) - S] + (0.22)(1.5)

<shoving ATC> > 0 <=> S < ~3.2

So this is +EV for very small stack sizes (less than 3.2 units).

I think there is some mistake in the calculation because if try it with pocket aces (85% equity against a 12% range) I get that it is still +EV only for very small stack sizes (~3.8).

If the calculation is incorrect I would appreciate if someone points me to the mistakes.

Thanks.

Prior to reading this: I notice your math uses 12% as his call3bet% and 88% as his foldto3bet%. The problem is that you assign him 12% range, when his range should be 9.3% (.78*.12). This slightly changes the equity if you're called.

You're calculating the EV of the decision of shoving ATC given that your opponent minraised.
This means that p(opens) shouldn't show up in your equation (you're examining the game for after he opens, not before it). When dealing with conditional probabilities, you're treating them as if the associated probability is equal to 1 or 0 (depending on if it meets the condition).

So your equation:

<shoving ATC> = p(opens)[p(folds)(pot size) + p(calls)p(we win)(pot size)-(eff stack size)] + p(open folds)(blinds)

Needs to be rewritten as:

[p(folds)(pot size) +p(calls)[p(we win)(eff stack size)-p(lose)(eff stack size)]

Notice that we dropped the p(open folds) term because we're defining this as "the situation where he minraises," which means p(open folds) = 0, so we ignore this term.

Now, when you plugged in these numbers, it appears you forgot to include therm p(lose), which is 1-.33 = .67. Include this, and you end up with:

(0.88)(3) + (0.12)((0.33)*S - .66*S) = 0

Which is equal to 0 at 66.

So, when your opponent minraise / calls with this ridiculous ratio, you can 3bet jam ATC up to 66 BB deep.

So don't minraise 78% and call 12%

I see your point and it makes sense. If he opens 78% and only calls with 12% of those, his range cannot be 12%.

I didn't forget to include the term p(lose) I misplaced the parenthesis! It should be:
<shoving ATC> = p(folds)(pot size) + p(calls)[p(we win)(pot size)-(eff stack size)]

This style of writing EV equations where you multiply the p(win) by the pot size (both effective stacks) and omit the p(lose) term (because it is equal to 1- p(win)) is from the book MoP. The second type of equation that you used (and I also posted in the initial post) is more intuitive.

Thanks a lot for the corrections. I was totally puzzled by my strange results and posted this equations in another sub-forum but no one pointed my mistakes clearly.

i think plancer made a slight mistake.

i think it has to be .88*2 and not *3 since we gain 2 blinds from his minraise. i know that we also get back our own posted blind but i think this is meaningless if we use the method with effective stacks.

.12*.33 we gain a stack and .12*.67 we lose a stack. therefore we gain 2 blinds .88 of the time. wow i talk like a ****** but i hope i got my point across. therefore it is profitable with 44BB or less.

erdnase,
with the MoP way of writing ev equations (i agree that they are smoother) you have to make sure that you distinguish between the two parts of it that are both called "pot size". the first is the initial pot size of 3bb after his minraise. the second is the final pot size of 2*S after both stacks went in. i think it would be better to give them different names to avoid confusion.

You are correct because otherwise I have to discount the big blind I posted from the effective stack. So we are considering that the effective stack size is S after the blinds are posted. Right?

im sorry, i dont understand what you mean.

S are the effective stacks before the hand. (i think)

the formula goes
villain folds * what we get then + he folds * (we win * his stack - we lose * our stack) = 0

the equation equals 0. zero is before the hand (kind of). at no point in the equation do we lose 1bb for posting it. therefore we cant win it back and thats why we only "win" 2bb when villain folds.
the only possible outcomes for us are having one stack more, having one stack less or having 2bb more.

As it happens I recently implemented a shove-equity tool to answer exactly these kinds of questions. Here's a screenshot of this particular situation - as stated above, it becomes neutral EV at around 40 bets.

- bachfan