8-8-05 [poker]
Another interesting situation in poker. The case where someone is going all-in overbetting the pot frequently. Your counterplay
is to just fold with bad hands and call with good ones to hopefully bust him. However, exactly how often should you call? I'm going
to first consider a very simple toy situation, and just in a cash game. I'm going to assume you both start with the same amount of chips
each hand, and when he steals from you when you fold your big blind, that profit just goes into a bank and your chip count is reset. When
you win a hand, that profit goes into your bank and the chips are reset. In a tournament, you'd have to worry about your stack getting
whittled down because the chips dont reset. I'll try to address that later.
So, here's the toy situation I'll analyze : you are in the big blind for 1 chip every hand. You both have stacks S (really what
matters here is the stack to big blind ratio; I make the BB just be 1 chip here so that the stack to big blind ratio is S). He's going
all-in with the best F fraction of hands and folding the rest. You can either call or fold. How much should you call to maximize your
EV ?
If you fold, your ev is -1. If you call, your chance of winning is P(C) where C is the fraction of hands you're calling with. In
that case your ev is 2*S*P(C) - S . Overall, your EV is :
EV = C * [ 2*S*P(C) - S ] + (1-C) * (-1)
EV = C*2*S*P(C) - S*C - 1 + C
The problem is to choose C to maximize this. The tricky thing is that P(C) is complex and nonlinear. We do know C is less than F,
you'll be calling with better hands than him.
If we were on the river, it would be easy because the better hand just wins. In that case, P(C) is just
P(C) = (C/F) * .5 + ((F-C)/F) * 1 = (F - .5C) / F = 1 - C/2F
EV = C*2*S*(1 - C/2F) - CS - 1 + C
EV = 2CS - C^2*S/F - CS - 1 + C
= C*(S+1) - C^2*S/F - 1
maximize :
S+1 - 2CS/F = 0
C = F*(S+1)/2S
So, you're calling with a fraction proportional to his, but reduced by (S+1)/2S. If S is large (eg. the stacks are much bigger than
the blinds), you're calling with 50% of the hands he calls with. If the stacks are very small, eg. as S goes close to 1, you should
call with roughly the same hands he's going all-in with.
But that's wrong because it's preflop and the winning hands aren't that simple. However, it is almost that simple. All we have to
look at is the region where you have a hand in the best fraction C and he has a hand in the next best fraction (F-C). In this case,
you're almost certainly around 65% to win on average; we'll just call it 2/3 and see what we get. For the cases where you're playing hands in
the same region, you might be an 80% favorite or a 30% dog, but it all averages out to 50/50 since you're playing the same range of
hands.
P(C) = (C/F) * .5 + ((F-C)/F) * (2/3) = (3C + 4F - 4C)/6F = (4F - C)/6F = 2/3 - C/6F
EV = 2CS*(2/3 - C/6F) - CS - 1 + C
EV = 4/3*CS - C^2*S/3F - CS - 1 + C
= CS/3 + C - C^2*S/3F - 1
= C*(S/3+1) - C^2*S/3F - 1
maximize :
(S/3+1) - 2C*S/3F = 0
(S/3+1) = 2C*S/3F
3F*(S/3+1)/2S = C
F*(S+3)/2S = C
Very similar to before, but you have to call with more hands. Note that this is wrong when S is very small because we assumed C <= F,
which would break down as S gets close to 1.
Now, what are the hands like in practice here? Let's consider a typical scenario. Say S is 10, and F is 1/4 , he's going all-in with
the best 1/4 of hands. What should you call with? C = F*13/20 = 16.25% of hands. What are these hands exactly? Well, the best 1/4 of
hands is all the hands like K9s or KTo or QJ or better (all pairs, of course, and any ace).
The best 16.25% of hands is A5s and A8o or better, and KJ or better.
Let's check our approximation; if you're playing a hand in the good region C and he's playing a hand in the region F-C , you're on something
like A5s or better and he's on something like K9 or KT. If you actually had A5, you'd be a 60/40 favorite; if you have a low pair, it's almost 50/50 ,
but you could also be on KJ,KQ,AK, etc. that dominate him 80/20. So, the 2/3 guess looks good.
In a tournament you have to worry about your stack bleeding down each time you fold. In that case you
can't just look at the EV of each hand, because if you wait too long your stack is smaller to double up with.
However, Sklansky has shown that this is a very small factor unless you are just about to pay your blind and
the blinds are very large compared to your stack. In that case we can look at - what if he's going all-in with
50% of hands, and S is a mere 4 big blinds. In that case C = 7/8 * .5 = 43.7 % of hands. What are these hands?
The top 50% of hands starts around J5s, Qxs or Q4o or J8o. The top 43.7% is only a little better, J7s, Q4s,
Q7o, J8o and better. Note that this is different than the famous "computer hand" Q7, which is 50/50 against a
random hand. We're not talking about win percentage, we're talking about the fraction of hands when ranked in
order of best to worst.