Hello! Assuming in a 3-way pot, Player1 folds 40% of the time and Player2 fold 50% of the time. How often will we get a fold from both players if we bet?

Assuming they act independently of each other, 20%. But that may overstate the real chance because the first fold affects the second decision.

thanks, Iguess its just 0.4*0.5 then?

Yes, to calculate the probability of two independent events BOTH occurring, multiply the individual probabilities.

The probabilities here are not independent because usually each player's ranges will intersect. However, this effect is not significant, so you are fine treating the probabilities as independent here.

thanks! what confuses me a little:
If Player1 folds 40% and Player2 folds 50%. That would mean Player1 calls 60% and Player2 calls 50%. That would then mean total calls 30%. If its now 20% total folds and 30% total calls, where did the remaining 50% go?

The 20% is when BOTH players fold, and the 30% is when BOTH players call. The other 50% is when either player folds and the other player calls.

ah right, that makes a lot of sense... sorry, my bad

I don't agree. The second villain to act will have a different fold percentage if the first villain calls or raises, vs if the first villain folds. We don't know which of those scenarios are the stated percentages, if either. More likely those percentages would be against a single opponent. The player who folds 50% if we bet may get priced in if the other villain first calls, and then folds a much smaller percentage. And that's just one example of dependence.

From a pure math standpoint the question is straightforward and has been answered. But it isn't fine to assume independence here.

I think I may have used terminology incorrectly. Let me try to explain better. It depends on how the percentages are derived. If they are derived empirically, then the second fold rate is likely the probability villain folds, given the first villain has already folded, and they can just be multiplied. If they are derived by theoretical continuing ranges, each probability probably represents the percentage of each villain's range that continue. Because ranges intersect it is not perfectly accurate to say that the probability that both villains fold is the product of the percentages of their ranges that continue, but it is close enough under almost all circumstances.

For an example, let's say that the board is 33344, and villain's ranges are both {KK+}, and they only continue against a bet with AA. Each villain continues with 50% of their range. However, because the events are not independent, the actual probability is P(V1 folds)*P(V2 folds | V1 folds), which is 1/2*1/7 = 1/14 = 7.1% chance they both fold, rather than 25%.

I might be making this too complicated.