I was wondering if somebody could help me figure out the following.

In this spot by the river I'm facing a 70% bet.

If I consider that villian has 12 flush combos on his range, how many other combos will the villian need to have in order to break even?



Thank you very much for your help

Note that the exact bet size is 23/33.

If there are no QJ or Jx in his range:

Let the total # of combos in his range = n. We can express the EV in terms of n:

(56/33)(n-12)/n - (23/33)12/n = 0
(56/33)(n-12) - (23/33)12 = 0
n = 237/14, a little less than 17, so you need to be beating at least 5 combos to make the call.

If there are QJ in his range but no Jx: (56/33)(n-12-QJ) - (23/33)(12+QJ) = 0
If QsJs is already one of the 12 flush combos then don't count it again.

If there are Jx in his range then there are chop combos.
Let x = # of combos we're beating.
Let c = # of chop combos.

We need: (56/33)x + .5c - (23/33)*12 = 0

x = 69/14 - 33c/112

Eg if we're losing to 12 combos and chopping with 5, then we need to be beating at least 4 combos (rounding up from 3.455).

While we're at it, let's generalize this to any bet size b (in units of pot):
(1+b)x + .5c - 12b = 0
x = (24b-c)/(2b+2)

Finally, generalizing to y # of combos we're losing to:
(1+b)x + .5c - by = 0
x = (2by-c)/(2b+2)