by Aisthesis » Sat Apr 19, 2008 4:14 pm
Examples just for fun to see how to set things up and solve the game.
There's a pot of 1. We have to define that because if there's no money in there, whoever acts has no motivation to bet without the nuts.
Now we have to define the allowed betting sequences, and it's a really good idea to start off very simple at first. When you allow complex betting sequences, it gets really complicated really fast.
I'll do one that's kind of weird just to get the idea. Player A as first to act is allowed only to check or to bet 8. If A bets, then B can call or fold. If A checks, hands are simply shown down (I'm trying to keep it simple).
Here's the way I go about solving this:
First question: How often does B have to call so that it is unprofitable (or break-even) for A to bluff with his worst possible hand, namely 1?
It's costing A 8 for a chance to win 1. So, B has to call on [0,1/9] for bluffing at 1 to be break-even. If we run it 9 times on that call range, A loses 8 one time and wins 1 eight times, so that's break-even.
So, we have solved the first part: On optimal play, B will call on [0,1/9]. That's going to make A indifferent to bluffing on his bad hands.
Now, finding A's optimal VB range is always just the top half of B's optimal call range, namely [0,1/18]. Again, at 1/18, A is indifferent to value-betting or checking, because if B calls, he wins exactly half the time.
One can also set this up in the form of indifference equations. Given B's call range, when is A indifferent to VBing or checking at the point x?
Assume A has x. What we do here is compare the EV of a bet vs. a check at x given B's call range.
Suppose A bets at x (obviously, x < 1/9).
If B has [0,x], then A loses 8. EV = -8*x
If B has [x,1/9], then A wins 9. EV = 9*(1/9-x) = 1 - 9*x
If B has [1/9,1], then A wins 1. EV = 8/9.
Total EV is -8*x + 1 - 9*x + 8/9 = 17/9 - 17*x
Now suppose A checks at x.
If B has [0,x], then A wins nothing (EV 0).
If b has [x,1], then A wins 1 for a toal of 1-x.
A is indifferent to betting or checking when these are equal.
So, we have 1 - x = 17/9 - 17*x
16*x = 8/9
x = 1/18.
Clearly, against this range, if x <1/18, A isn't indifferent to betting but prefers to bet rather than check.
Now let's look at bluffs. Clearly, if A were never bluffing but only made these value bets, B would be better off calling only at some values < 1/18.
It's the bluffs that are going to make B indifferent to calling or folding at 1/9, where he always loses to the VBs but always wins against the bluffs.
B is calling 8 for a chance at 9, so to be indifferent to calling, there must be 8 bluffs for every 9 VBs. (8/9)*(1/18) = 4/81. So, A bluffs [77/81,1].
So, we have:
A VB: [0,1/18]
A bluff: [77/81,1]
B call: [0,1/9]
Now to finish it off, let's see how the pot of 1 is really divided up in this game on optimal play:
I'm going to use Excel to multiply it out because we get into very small fractions. I'll also figure it for both players to check my math (the sum should equal 1).
EV for A:
A[0,1/18], B[0,1/18]. Split the initial pot for .50. Total is (1/18)*(1/18)*.50 = .001543
A[0,1/18], B[1/18,1/9]. A wins 9. Total is 9*(1/18)*(1/18) = .027778
A[0,1/18], B[1/9,1]. A wins 1. Total is (1/18)*(8/9) = .049383
A[1/18,77/81], B[1,1/18]. 0
A[1/18,77/81], B[1/18,77/81]. Split. EV is .400568
A[1/18,77/81], B[77/81,1]. A wins 1. EV .044201.
We set up the bluffs so that they break even, so EV there is 0.
A's EV is: .523472.
Not surprizingly, since A has more options, A wins a bit more than half the pot on this game.
Now let's check by doing EV for B.
B[0,1/18], A[0,1/18]. Split for .50. Total is .001543
B[0,1/18], A[1/18,77/81]. B wins 1. EV .049726
B[0,1/18]. A[77/81,1]. B wins 9. EV .024691
B[1/18,1/9], A[0,1/18]. B loses 8. EV -.02469
B[1/18,1/9], A[1/18,1/9]. Split. EV .001543
B[1/18,1/9], A[1/9,77/81]. B wins 1. EV .046639
B[1/18,1/9], A[77/81,1]. B wins 9. EV .024691
B[1/9,77/81], A[0,1/9]. 0
B[1/9,77/81], A[1/9,77/81]. Split. EV .352385
B[1/9,77/81], A[77/81,1]. 0
B[77/81,1]. 0 Regarldess of A's hand.
Total: .476528
And they add up to 1, so we must have done it correctly.
Anyhow, that's the way it works. The EV calculation is rather tedious, so I usually leave that out, but I thought I'd throw it in for completeness on the initial example.