[LOW] Jam OTF with a really marginal hand: Why not or massive spew?

... let's say it's an ambitious move.

Looks cool, A3dd would be better...
What ended up happening?

Going to post the full hand once I have a bit more replies

What's our equity when called, around 20-25%?

Not that easy to work out given all the diff size stacks but if this was the BB alone you'd need him to fold about 64% to break even. vs sb you need him to fold about only 5% of the time since he's so short and CO needs to fold around 27%. How you add all that together I haven't a friggen clue but I'm guessing you need them all to fold roughly 75% of the time or there abouts. imo they won't all fold as much as that so I think it's a -EV shove.

Or maybe I'm working this out wrong and we only account for the FE we need against the biggest stack.

Someone who knows maff chime in pls

Why though?

Ask and you shall receive:

I am approximating a bit here, but pot is 28bb and hero shoves 114. There are three possible outcomes - all fold, in which case hero wins 28, a call and hero wins 142 (28 in pot plus shove amount), or a call and hero loses 114. (IÂ’m neglecting cases where 2 or 3 villains call; this seems unlikely and heroÂ’s equity is probably very low in these cases).

Let x be the probability of all villains folding. LetÂ’s assume hero has 0.25 (25 percent) equity when called. Then we can calculate the EV of heroÂ’s shove as

28x + 142*0.25(1-x) - 114*0.75(1-x). Algebraically (I wonÂ’t bore with the details) this simplifies to 78x-50. Set this equal to zero to get the fold probability hero needs to break even on the shove and we get 50/78 = 0.641. If everyone folds more often than this heroÂ’s shove is profitable. That is the probability that everyone folds though. Hero would need (approximately*) the third root of this value for the probability of each individual villain to fold. This works out to 86.2%. That is because the probability of all villains folding is the product of the fold probability for each individual villain.

If this is heads up, hero probably has a good shove. Against multiple villains itÂ’s questionable IMO. I havenÂ’t worked out what the percentage of PF calling ranges would be in villainÂ’s shove calling range here, but it only needs to be 13.8%, which seems low IMO.

* This result is approximate for two reasons, one math based and the other poker based. The math based reason is that the probability of a given villain having a hand in his calling range is not independent of the probability of another villain having a hand in his calling range. For example if villain A folds, we now know that he did not have JJ. Therefore villain B and villain C are slightly more likely to have JJ, and similarly for other calling hands.

The poker based reason is that my calculation assumes that all villains will have the same calling probability, If two are indeed fish, you might expect them to have a higher calling probability. (Fish might call with hands like KJ, QJ or suited heart combos). The non-fish would then need to have a correspondingly lower call probability.

Really impressive, hats off 😀

The thing I don't get though is the 86% number. Like thinking about it logically why would we need them as a whole to fold more often? When one of them calls we still have the same equity and we can still only lose up to our full stack depending on which one calls. Or does that 86% mean that theoretically it's like needing each individual to fold 86% of the time?

Edit: or is it like having one individual that we need to fold 86% of the time? That seems to make more sense.

We need everyone to fold approximately 64% of the time to break even. So individually, if all three players fold 86% of their range, we get to 64% total (0,86 * 0,86 * 0,86 = 0,64).

I don't like the play. In a vacuum, maybe. Even then I don't know. But if you keep doing this, others will soon catch on to what you're doing.

The chance of having everyone fold when there are two dudes with 30 and 16 big blinds pre is almost non-existent.

Those fish are more likely to think among the lines of "whatever let's play those last 5/10 bucks and go to sleep" than about what they're supposed to fold lol.

This is the correct explanation, subject to the assumption that all villains fold with equal probability. If not, then if the probability that villain A folds is a, and similar for villains B and C, the probability of all folding is abc, the product of the individual probabilities. If all are the same and we need all to fold with probability 0.64, then each individual probability is the third root of 0.64 (I.e. the value that solves the equation x^3=0.64), which is 0.86.

Mathematically, this is the solution that gives the minimum value for the highest individual fold probability. This can be proven, but intuitively it should be clear. If villain A, for example has a lower fold probability than 0.86, then for the product abc to equal 0.86, either b or c must fold with probability higher than 0.86. In real terms this matters if the read that we have two fish and a non-fish as our villains. It means that, assuming fish will call more often and likely have a lower fold probability, we would likely need the non-fish to have an even higher fold probability than 0.86. There will be some hands in non-fish calling ranges here, so it is questionable whether non-fish will fold often enough to overcome the propensity of the fish to call too often.

The chances that everyone folds is miniscule, and when you get called you're basically flipping or almost dead, so I don't think it's good.