There's a toy game of 1-2 NL Hold'em with 3 players, the button bets and the small blind and big blind must call and that ends the game. The bet is considered to be separate from the blinds. 'x' is

the probability of the button winning. I compute the EV of the button:


EV = x*(3*b+6) - (1-x)(b+2)

Then I set the EV to zero to find the value between positive and negative EV, between betting and folding. I solve for b and get this:

b = (2-8*x)/(4*x-1) = -(8x-2)/(4*x-1) = -2*(4*x-1)/(4*x-1) = -2 So less than -2 and the button is negative EV and anything greater than -2 and the button can bet. It doesn't depend on the value of x.

What did I do wrong?

You did not do anything wrong, at least with your algebra. You are asking an impossible question and the result you got is really just the way that the math is telling you that what you ask is impossible.

You are essentially asking: given some win probability x, what amount should I bet to give zero EV? To see why this question is not valid simplify the calculation a bit by defining a new variable t = b+2 (t is now the total amount wagered) and write the EV formula

EV = 2tx - (1-x)t = 3xt -t

Now consider an arbitrary value for x (obviously between 0 and 1 since it is a probability) If x is 1/3 the EV will be zero since 3xt = t in that case. For x<1/3, 3x<1 and multiplying by t (valid since t>0) yields 3xt<t which implies 3xt-t<0, I.e. the EV is negative. For x<1/3 the only value for t such that EV =0 is therefore t=0 - if you donÂ’t bet anything the EV is zero.

In like manner, just with the inequalities reversed the EV is positive regardless of bet size for x>1/3, and again the only way to get zero EV is to bet nothing (t=0).

Remembering that we defined t=b+2 then makes it clear why you got b=-2 in your calculation - that would be the bet amount needed to make the total wager zero.

What this is also telling us is that button should only have two bet sizes. If he has less than 1/3 winning chance, he should bet the smallest allowable amount. If he has greater than 1/3 win probability he should bet the maximum allowable amount (or the maximum amount consistent with something like Kelly’s Criterion if that is a lesser amount). The EV will always have the form kt where k is a constant that is negative when the win probability is less than 1/3, zero for 1/3 and positive if greater than 1/3. Maximizing EV then implies betting as small as possible for negative k and as large as possible for positive k.

You nailed it. Thanks for the help!