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I'm working on a poker problem where I'm trying to calculate how much of the pot I want to bet (as a percentage). I'll give the nuts and bolts of my request first and then an explanation of it's usage in case that helps or matters.

The request:

If you look at this post in the poker stack exchange you will see a comment by paparazzo where he gets the formula s / (1 + 2s) = f. I want to reconfigure that formula to solve for "s". i.e. I know my bluff frequency "f" and I want to know what percentage of the pot "s" I should bet to make my opponent indifferent to calling. So: how do I solve for "s" in this formula; or if more appropriate, write an entirely different formula that accomplishes my goal.

https://poker.stackexchange.com/questions/1045/the-right-bluff-frequency

Detailed explanation Often in poker you come at this problem from the perspective of "f" in the formula above. e.g. There is 1 in the pot and my opponent bet 1. How often do I have to call so that he can't profitably bluff with any two cards. In this scenario you are calling 1 to win 3 (original pot + opponent bet + your call). 1/3 = 33%.

Then you can do something like say: I have 10 value hands here and I want to bet pot. Betting pot gives my opponent odds of 2-to-1 on a call. So I need to bluff with 5 hands to make him indifferent to calling or folding. 5 bluffs making my bluff % the same as his pot-odds. If I only wanted to bet half-pot then I would be giving him 3-1 odds to call so I need to bluff 2.5 hands.

But I want to come at this problem from the other direction. Analyze a hand and when I get to the river I might end up in a situation where I just have a lot of value but not very many bluffs in my range. If I have 18 value hands and only 4 possibly bluffing candidates; then I can't go looking for other bluffs to add because they don't exist. So knowing that I'm going to be bluffing 22% of the time; what percentage of the pot do I need to bet to make my opponent indifferent to calling or folding?

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  • $\begingroup$ It looks like you are trying to solve for $s$ in terms of $f$, given the equation $f=s/(1+2s)$. Is this your question? [This particular question can be stated more simply without the poker application] $\endgroup$
    – Michael
    Oct 6, 2019 at 0:15
  • $\begingroup$ Yes that is the question. I figured so; which is why I gave "the request" separate from the explanation. But I figured the explanation was potentially important in case I am thinking about the math in a fundamentally flawed way which would prevent me from seeing a better/easier solution. $\endgroup$
    – William
    Oct 6, 2019 at 1:39

1 Answer 1

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You can write $$\frac s{1+2s}=f\\ s=f(1+2s)\\ s=f+2fs\\ s-2fs=f\\ s(1-2f)=f\\ s=\frac f{1-2f}$$

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  • $\begingroup$ This is it. Thank you! $\endgroup$
    – William
    Oct 6, 2019 at 1:42
  • $\begingroup$ Ok I found an issue with the formula; maybe. If you say that you have 18 value combos and 22 bluff combos that makes your bluff percentage 55%. If you plug anything over 50% into the formula the answer doesn't make sense. Is this an issue with the formula or is it effectively saying "don't bluff more than you value bet because it's impossible to give your opponent a 0 EV call that way"? $\endgroup$
    – William
    Oct 6, 2019 at 2:13
  • $\begingroup$ If you look at your original formula $f$ can never be greater than $\frac 12$ for $s \gt 0$. It will approach $\frac 12$ when $s$ is very large, but always be below. $\endgroup$ Oct 6, 2019 at 2:28
  • $\begingroup$ Note that if $f \gt \frac 12$ your opponent should always call as they will win more than half the time. $\endgroup$ Oct 6, 2019 at 13:35
  • $\begingroup$ Ok that makes a lot of sense. If you are bluffing more than 50% then they should in theory always call because their call is always <= 50% of the pot. Ok thanks a lot; that clears it up. $\endgroup$
    – William
    Oct 7, 2019 at 1:49

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