Im trying to work out the probability that a player missed the flop given that he cbets (not sure how useful it will be).
So far I have

A = missed flop B = cbets P(AlB) = probability of missed flop given cbet


P(AlB) = [P(A)xP(BlA)]/P(B) is this correct and or useful?

if P(B) = 1 then P(AlB) = P(A) as it should (as P(BlA) = 1)

P(A) and P(B) i get from players range and hud but I have to guess P(BlA)?

Thanks

P(A) = probability that player missed the flop
P(B) = probability that player cbets
P(x|y) = probability of x given y
P(B|A) = 1 (you said that, I wouldn't bet on it)
P(B|not A) is unknown

Now, probability of him cbetting with air is P(B|A)*P(A) and the probability of him cbetting with something is P(B|not A)*[1-P(A)].

Hence, P(B) = P(B|not A) + P(A) * [P(B|A)-P(B|not A)] =
= P(B|not A) + P(A) * [1-P(B|not A)] =
= P(B|not A) + P(A) - P(A) * P(B|not A) =
= P(A) + P(B|not A) * [1-P(A)]

You know both P(A) and P(B), so P(B|not A) = [P(B) - P(A)]/[1-P(A)].

Now, P(A|B) = P(B|A) * P(A) + P(B|not A) * [1-P(A)] =
= P(A) + P(B) - P(A) =
= P(B)

I invite someone to check my math, since this seems totally non-intuitive to me and I might be very wrong here.

Well, you both need to take his range into account, since he will ALWAYS bet TPTK, meaning flops lik AT9, KQ4, K42 will be bet.

Whereas flops like 973 will have a lower expected c-bet percent from the player, so its kinda hard doing what you do =))

Jspazz,

I meant P(B|A) = 1 ONLY IF P(B) = 1 otherwise its flop/player dependant.

You put probability of him betting with air is P(B|A)*P(A) but isnt that the probaility he misses the flop AND bets with air?

You also have P(B) = P(A) + P(B|not A) * [1-P(A)]

shouldnt that be = P(A)*P(BlA) + P(B|not A) * [1-P(A)] ?

Also P(B) can be taken from his stats. I dont think I can get P(BlA) in terms of given probabilities.

P(AlB) = [P(A)xP(BlA)]/P(B).

Say villains cbet is 0.7 so P(B) = 0.7
We are in the bb with 88 and the button with a 30% steal raises and we call.
Flop is K92. Inputting his range as 30% gives him middle pair or better 27%.

Assume he cbets this flop 90% of the time if hes missed then

P(AlB) = .73x0.9/0.7 = 93% . is this correct?

just realised i cant make up numbers for P(BlA) as it depends on P(B)
so

P(AlB) = P(BlA)*P(A)/[P(BlA)*P(A) + P(BlnotA)*P(notA)]

assuming P(BlnotA) = 0.9

then P(AlB) = 0.9*0.73/[0.9*0.73+0.9*0.27] = 73%

im confused....

Are we talking probability or poker?

Probability

Your original math is correct. It might help you think about this if you write out all four possibilities. Let X be the probability that he misses the flop and c-bets, Y be the probability that he misses the flop and doesn't c-bet, Z be the probability that he hits the flop and c-bets and 1-X-Y-Z be the probability that he hits the flop and doesn't c-bet. Then:

P(A) = X + Y
P(B) = X + Z
P(A|B) = X / (X + Z)
P(B|A) = X /(X + Y)

Now you can see why P(B|A) = P(A) * P(B|A) / P(B)

Poker

Good poker players c-bet when they've hit the flop, or when it's likely no one hit the flop. You can generally look at the flop and figure out which it is. A c-bet is not intended for deception. So probabilities are not very useful. Of course, good poker players also mix it up, so they'll sometimes c-bet when they didn't hit the flop and it's likely someone else did. They'll rarely fail to c-bet if the flop looks dry, but sometimes they'll have hit that flop because they bluffed the preflop raise. But past statistics are not useful for predicting good poker players' actions.

With bad players, statistics might be useful.