Right. Nice formula too.
It's a little counter-intuitive to me that p includes villain's last bet, but it does yield a really nice formula. Of course you can do it either way.
So, to illustrate the case of a pot-sized open raise, small blind's already bet .5, big blind raised to 1,
so p = 1.5, B = 1, and we want to make a full pot-sized raise, so f = 1.
Plugging in to R = B + f(p+B), we get R = 1+1*(1.5 + 1) = 3.5, correct.
If you don't want to think of the pot as containing the last uncalled bet or raise, then daviddem's formula becomes
R = B + f(pot + 2B)
ie, match the bet, then add another f times all that, and that's an f-times pot-sized raise, laying odds of 1+f:f to the original bettor.
edit:
We should note that this formula works whether
*hero is the first to bet that street, (in which case, B = 0 in the formula) and then the first caller behind is getting pot odds of 1+f:f, or
*villain is the first to bet and it folds to us, or
*there has been previous action that street and villain is either calling or raising (previous action before villain is just part of the pot), and this will lay odds of 1+f:f to villain, assuming everyone else in between folds.
had to re-edit the edit lol
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