Dale Dough
02-13-2007, 04:35 PM
It's long. If you don't want to read it, don't. Also, apologies if this is too basic, but it took me quite a while to figure all this out on my own. What can I say, I stupidly went for a business major. Anyway, on to the problem. Question is at the end.
The pot is 1. Hero can either bet 1 or check. Villain can either call, fold or check. Both have a uniformly distributed hand ranging from 0 to 100 percent.
Assumptions I made: Villain has to call 50% of his hands, or it becomes profitable for Hero to bet everything. Obviously this will be the top 50%. Hero will bluff 1/2 of the good hands he bets, offering Villain exactly 2 to 1. Hero will bet top x% of his hands. He will bet any hands that are more profitable to bet than to check. The EV of a bluff is exactly zero, which is worse than checking. He must bluff, though, to keep Villain from profitably lowering his calling range. So he bluffs the hands that have the worst value when checked.
I used two different approaches to get x, and came up with two different answers.
Approach A:
EV(bet) >= EV(check)
EV(bet) = 50% Villain folds and Hero wins the pot, 1. (x-50%) Villain calls with a worse hand and Hero wins 2. (1-x) Villain calls and wins.
EV(check) = x
.5 + 2(x - .5) - 1(1 - x) = x
.5 + 2x - 1 - 1 + x = x
-1.5 + 2x = 0
x = .75
So, he bets hands 75% or better.
Let’s test it: I have hand 74%. If I check, I take the pot 74% of the time, for an EV of .74. Sweet. Now I bet it. 50% of the time, I win 1, for an EV of .50. 24% of the time, I win 2, for an EV of .48. 26% of the time, I lose 1, so that’s -.26.
Total EV: .72. Yup, check’s better.
Let’s test it: I have hand 76%. If I check, I take the pot 76% of the time, for an EV of .76. Sweet. Now I bet it. 50% of the time, I win 1, for an EV of .50. 26% of the time, I win 2, for an EV of .52. 24% of the time, I lose 1, so that’s -.24.
Total EV: .78. Yup, bet is better.
That was Approach A. Bet top 25%, bluff with bottom 12.5%. I get max value out of those top hands, and Villain can't profitably tighten up his calling range because of the bluff.
Approach B
Our EV can be expressed as a function of how often we bet. Maximize that, and we have our answer. I chose to divide it in three terms: what happens when we bluff, when we value bet, and when we check. The first term will work out to be zero because bluffs are neutral EV, but together these terms cover the range of outcomes.
Let x be the top % of hands we value bet. We will bluff the bottom x/2 hands.
When we bluff, our EV is always 0, because we never beat Villain. Half the time we take the pot, half the time we lose a pot-sized bet. Not that this is relevant, but this happens x/2 times.
When we value bet, there are three scenarios:
a) He folds. This happens 50%, and we win 1. Hence, .5 * 1
b) He has a hand in the same range that we bet. We both put in a bet and split the final pot, so our EV is 50% of the main pot. This happens x times, hence x * .5
c) He has a calling hand above 50% but below x, and we win 2. Hence, (.5 - x) * 2.
This contains the entire range of possibilities: .5 + x + .5 - x = 1. The sum of these three outcomes is multiplied by the probability that the we value bet scenario occurs, which is x.
So we have x(.5 * 1 + x * .5 + (.5 - x) * 2).
When we check, there are again three scenarios:
a) He has a hand above our checking range. We bet the top x, so x hands in his range are automatic losers for us. So we have x * 0
b) He has a hand in the range that we bluff with. This is the bottom range, will happen x/2 times, and since we can't have that bad a hand here we always win. EV = .5x * 1
c) He has a hand in the same range as us. This happens (1 - x - x/2) times, and our EV is .5
Again, a b and c cover all possible outcomes and add up to 1.
We check (1 - 1.5x) of the time, so this term becomes
(1-1.5x)(x * 0 + .5x * 1 + (1 - 1.5x) * .5)
The three scenarios occur either x, x/2 or (1 - 1.5x) of the time. This adds up to 1. Our EV is the sum of the three final terms:
x(.5 * 1 + x * .5 + (.5 - x) * 2) + (1-1.5x)(.5x * 1 + (1 - 1.5x) * .5)
(the bluffing term x*0 is omitted)
Eventually, this works out to be -1.125x^2 + .5x + .5 .
It's been a while since I looked at any calculus text, but I'm 99% sure the value for x that yields maximum EV is 2/9, or 22.2222%
Just to be sure, I put that whole scenario into Excel to double-check (not the final function, I did it step by step like I described here) and using Solver I arrived at the same number (also the quadratic function yielded the same result as my spreadsheet).
According to that formula, EV(x=2/9) is 5/9
Plugging in 25% yields a value close to, but smaller than 5/9: .554688
That was Approach B.
According to Approach B, betting 25% is suboptimal, 22.22% is better. According to Approach A, the cutoff point is 25%: the top 74% hand can better be checked, and the top 76% hand can better be bet.
I'm assuming that Approach B is the better one, and that the difference can be accounted for by the fact that bluffs cost money; EV is 0, so we give up whatever the showdown value of that range would have been.
My Question: Is my assumption about him calling 50%, and us bluffing according to pot odds, correct? And if it is, given what we know, what prevents us to do THIS:
According to 'optimal' strategy, we bet top 2/9. He can't profitably alter his calling range (in fact, I believe he could call either nothing or everything except the absolute worst hands if he felt like it, and our EV wouldn't change).
But we can alter our range. The top 1% (based on original 100% distribution) of our bluffing hands 'costs' us from 10.1111% to 11.1111% in EV. Taking those 1% of hands from there, and betting an extra % at the top instead, is therefore +EV for us given his calling range.
That means somehow, the equilibrium I have arrived is not optimal, or the last assumption is not valid. But I'm WAAAY to tired right now to think straight, I'll sleep on it and hope some of you can enlighten me.
Thanks in advance.
The pot is 1. Hero can either bet 1 or check. Villain can either call, fold or check. Both have a uniformly distributed hand ranging from 0 to 100 percent.
Assumptions I made: Villain has to call 50% of his hands, or it becomes profitable for Hero to bet everything. Obviously this will be the top 50%. Hero will bluff 1/2 of the good hands he bets, offering Villain exactly 2 to 1. Hero will bet top x% of his hands. He will bet any hands that are more profitable to bet than to check. The EV of a bluff is exactly zero, which is worse than checking. He must bluff, though, to keep Villain from profitably lowering his calling range. So he bluffs the hands that have the worst value when checked.
I used two different approaches to get x, and came up with two different answers.
Approach A:
EV(bet) >= EV(check)
EV(bet) = 50% Villain folds and Hero wins the pot, 1. (x-50%) Villain calls with a worse hand and Hero wins 2. (1-x) Villain calls and wins.
EV(check) = x
.5 + 2(x - .5) - 1(1 - x) = x
.5 + 2x - 1 - 1 + x = x
-1.5 + 2x = 0
x = .75
So, he bets hands 75% or better.
Let’s test it: I have hand 74%. If I check, I take the pot 74% of the time, for an EV of .74. Sweet. Now I bet it. 50% of the time, I win 1, for an EV of .50. 24% of the time, I win 2, for an EV of .48. 26% of the time, I lose 1, so that’s -.26.
Total EV: .72. Yup, check’s better.
Let’s test it: I have hand 76%. If I check, I take the pot 76% of the time, for an EV of .76. Sweet. Now I bet it. 50% of the time, I win 1, for an EV of .50. 26% of the time, I win 2, for an EV of .52. 24% of the time, I lose 1, so that’s -.24.
Total EV: .78. Yup, bet is better.
That was Approach A. Bet top 25%, bluff with bottom 12.5%. I get max value out of those top hands, and Villain can't profitably tighten up his calling range because of the bluff.
Approach B
Our EV can be expressed as a function of how often we bet. Maximize that, and we have our answer. I chose to divide it in three terms: what happens when we bluff, when we value bet, and when we check. The first term will work out to be zero because bluffs are neutral EV, but together these terms cover the range of outcomes.
Let x be the top % of hands we value bet. We will bluff the bottom x/2 hands.
When we bluff, our EV is always 0, because we never beat Villain. Half the time we take the pot, half the time we lose a pot-sized bet. Not that this is relevant, but this happens x/2 times.
When we value bet, there are three scenarios:
a) He folds. This happens 50%, and we win 1. Hence, .5 * 1
b) He has a hand in the same range that we bet. We both put in a bet and split the final pot, so our EV is 50% of the main pot. This happens x times, hence x * .5
c) He has a calling hand above 50% but below x, and we win 2. Hence, (.5 - x) * 2.
This contains the entire range of possibilities: .5 + x + .5 - x = 1. The sum of these three outcomes is multiplied by the probability that the we value bet scenario occurs, which is x.
So we have x(.5 * 1 + x * .5 + (.5 - x) * 2).
When we check, there are again three scenarios:
a) He has a hand above our checking range. We bet the top x, so x hands in his range are automatic losers for us. So we have x * 0
b) He has a hand in the range that we bluff with. This is the bottom range, will happen x/2 times, and since we can't have that bad a hand here we always win. EV = .5x * 1
c) He has a hand in the same range as us. This happens (1 - x - x/2) times, and our EV is .5
Again, a b and c cover all possible outcomes and add up to 1.
We check (1 - 1.5x) of the time, so this term becomes
(1-1.5x)(x * 0 + .5x * 1 + (1 - 1.5x) * .5)
The three scenarios occur either x, x/2 or (1 - 1.5x) of the time. This adds up to 1. Our EV is the sum of the three final terms:
x(.5 * 1 + x * .5 + (.5 - x) * 2) + (1-1.5x)(.5x * 1 + (1 - 1.5x) * .5)
(the bluffing term x*0 is omitted)
Eventually, this works out to be -1.125x^2 + .5x + .5 .
It's been a while since I looked at any calculus text, but I'm 99% sure the value for x that yields maximum EV is 2/9, or 22.2222%
Just to be sure, I put that whole scenario into Excel to double-check (not the final function, I did it step by step like I described here) and using Solver I arrived at the same number (also the quadratic function yielded the same result as my spreadsheet).
According to that formula, EV(x=2/9) is 5/9
Plugging in 25% yields a value close to, but smaller than 5/9: .554688
That was Approach B.
According to Approach B, betting 25% is suboptimal, 22.22% is better. According to Approach A, the cutoff point is 25%: the top 74% hand can better be checked, and the top 76% hand can better be bet.
I'm assuming that Approach B is the better one, and that the difference can be accounted for by the fact that bluffs cost money; EV is 0, so we give up whatever the showdown value of that range would have been.
My Question: Is my assumption about him calling 50%, and us bluffing according to pot odds, correct? And if it is, given what we know, what prevents us to do THIS:
According to 'optimal' strategy, we bet top 2/9. He can't profitably alter his calling range (in fact, I believe he could call either nothing or everything except the absolute worst hands if he felt like it, and our EV wouldn't change).
But we can alter our range. The top 1% (based on original 100% distribution) of our bluffing hands 'costs' us from 10.1111% to 11.1111% in EV. Taking those 1% of hands from there, and betting an extra % at the top instead, is therefore +EV for us given his calling range.
That means somehow, the equilibrium I have arrived is not optimal, or the last assumption is not valid. But I'm WAAAY to tired right now to think straight, I'll sleep on it and hope some of you can enlighten me.
Thanks in advance.