How do you value bet the river?

[Dan Bitel]

[ QUOTE ]

I've been thinking about bet sizing and "The Price Is Right". Say you know you have the best hand on the river and he
has something mediocre. You want to bet as much as possible such that he'll call. The more you bet (that he calls)
the more profit you make. If you bet too much, he'll just fold and you make $0. This is like the "Price is Right" in that
you want to get as close as possible to his maximum call amount, but without going over.
Your profit as a function of bet
size is :
<P>
P(B) = (B < B_max) ? B : 0
<P>
The problem is you don't know B_max, but you can estimate it with some error. Let's say you know it's between L and H,
with an even probability of being anywhere in there. What should you bet to maximize profit? Clearly it's somewhere in
the range [L,H] , but where exactly?. Maybe I'll do the math and figure out where exactly.

<P>

Well, Sean beat me to the math and sent me this :

<P>
<PRE>
Is it clearly in the middle? Clearly we have the boundary cases,
let N be the bet amount:

N = L -> E = $N
N = H -> E = $0

So you might hope that it's a curve that grows as N increases,
and then starts decreasing again. But that might be overwon by
the person's tendency to fold. Now, if L is 0, and H is non-zero,
clearly there must be some > 0 value, though. So that sounds
good. But I think in practice it's not. (Also, this totally changes
for a gaussian instead of a uniform distributution).

For a bet of N, probability of folding is (N-L)/(H-L), so probabiliity
of calling is 1-(N-L)/(H-L).

E = N - N*(N-L)/(H-L)
E = N - (N*N-N*L)/(H-L)
H-L is constant, so computing the derivative ignores it:

dE/dN = 1 - (2N - L)/(H-L)

set that to 0:

1 = (2N-L)/(H-L)
(H-L) = (2N-L)
H = 2N
N = H/2

Weirdly, L canceled out, so I probably screwed up. But it could
be true, that that's always the peak of the quadratic, and therefore
the optimal result is:

L <= H/2: N = H/2
L >= H/2: N = L

This makes some intuitive sense. Let's ask whether it's a good
idea to bet (H+L)/2, versus betting L.

In other words, we have a sure thing of making $L, so let's factor
that out (I think this makes it more intuitive). Now we want to know
if we should increase that to $(H+L)/2.

If we increase, our increase versus $L is:
$(H-L)/2
and we have a 50/50 chance at it. If we increase and lose,
we lose the $L
-$L

so E = 0.5 * (H-L)/2 + 0.5 * -L
= 0.25*H - 0.25*L - 0.5*L
= 0.25*H - 0.75*L
So this says if H is less than _3 times_ L, it's a losing bet
to take the 50/50 gamble of raising to (H+L)/2. So it makes
sense; when L gets close to H, the rate of extra-money-beyond
L doesn't sufficiently offset the increased chance of losing L.
</PRE>

<P>

That all looks right to me, and is kind of interesting. If L is
significant at all, you want to bet the maximum that you're *sure*
they'll call, and trying to edge it a little higher is -EV because the
risk of losing them is too great. Say for example L is $20 and H is $40, you
must bet just $20 !! Even $21 is worse because it makes them fold 1/20th of
the time so your EV is just (19/20)*21 = $19.95. However if L is zero, eg. they have
something so weak they might not call any bet, the best bet is H/2.

<P>

In reality, they have more like a Gaussian distribution of call values,
though I doubt that changes the answer too much. Also you have to look
at multiple models, say there's a 75% they call based on a model like
this, and a 25% chance they call any bet (but your stacks are not huge
compared to the pot, so this isn't insane). In that case there's a bit
more reward for edging your bet up, the risk of losing them if they have
something weak is hedged a bit by the EV gain when they have something
they call any bet with.

[/ QUOTE ]

[ QUOTE ]
Work out what they have, then work out what they'll call...

[/ QUOTE ]

LOLOLOLOLOLOLOLOLOL

[cbloom]

Pokey,

[ QUOTE ]

cbloom: Your post really screwed with me. I worked out the math six times and every time I got a max at N = H/2.


[/ QUOTE ]

Yeah, H/2 is indeed the local max as long as it's inside the [L,H] bound. Of course H/2 always < H so the question is just if H/2 > L

[ QUOTE ]

Oh, and as to the distributional problem: not to make the problem even more complicated than it already is, but I'd say you might consider something more like a Poisson distribution: (...)


[/ QUOTE ]

Yeah, this distribution is clearly B.S. and better distributions make the math harder, but it was kind of interesting to me that even this simple distribution leads to some conclusions which are a bit un-intuitive and (I believe) would remain correct with more correct distributions.

I think a lot of people probably think "he'll call something between $10 and $20 - okay, I'll bet $15" , which sounds reasonable, but it turns out is very wrong.