Game Theory: Unusual Question #3 and #4

[fnord_too]

Reducing the equation:

C = R + .25*(1-R) = .75R+.25

((1+R)/2)/((1+C)/2) = (1+R)/(1+C) = (1+R)/(1.25+.75R)


What we want is the derivative of this whole nasty function. I appologize in advance for the verbosity and pedestrian nature of my calculus, I went into discrete math because I was no great fan of the continuous variety.

Lets call 1+R U and 1.25+.75R V

so we want d/dR 0 + .75R - .375 + (1-R)*E(raise)
= 0 + .75 + d/dR E(Raise) - d/dR R*E(raise) = .75 + some other stuff to be computed presently

d/dR E(Raise) = d/dR .75R+.25 +(.75 - .75R)*2*((U/V)-(V/U))
= .75 + d/dR 1.5*(U/V - V/U) - .75R * (U/V)-(V/U)

= .75 + d/dR 1.5 U/V - 1.5 V/U - .75RU/V + .75RV/U (mommy make the hurting stop)

d/dR U/V = (V dU - U dV) / (V^2)

dU = 1, dV = .75, V^2 = .5625*R^2+.9375R + 1.5625

d/dR U/V = (.75R + 1.25 - .75(1+R))/(.5625*R^2+.9375R + 1.5625)
= .5/(.5625*R^2+.9375R + 1.5625)

d/dR V/U = (if I got this right):
(.75*(1+R) - .75R-1.25)/r^2 +2R +1
= -.5/(R+1)^2

d/dR RU/V = d/dR R+R^2/V = ((V*(2R+1)) + (U*(1.5R + 1.25))/(.5625*R^2+.9375R + 1.5625)
The top simplifies to .5 R

for d/dR RV/U we get -.5R/(R+1)^2 (and yes in retrospect I see the easy way to have done that!)

now where the hell am I? I can't finish grinding this out right now. If someone has mathmatica or the likes handy, please maximize the E(x) function and let me know where R is. Alternately if someone really likes Calc, do this by hand. Alternately, I will do it later, but after about an hour or so, mostly doing my work in a text editor (yuk) I am ready to leave this for a while.

[ZeeJustin]

[ QUOTE ]
So, all in all, he wins 4.5 + 2*(.8125*2 + .1875*(-.14)) =~ 7.6975 units over 16 hands or about .48 per hand. Nice improvement.

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I'm not gonna lie, I can't follow the math in these other posts, but fnord_too seems to have gotten a solution very close to mine. Can you dissprove our solutions?

Intuitively, a .27 solution seems too small to be correct.

Anyway, I see where you guys solved for B's EV, but what did you say B and A's optimal strategies were? What hands do you have A calling raises with. Did you adjust A's calling standards based on the fact that B is now bluffing occassionally?

[fnord_too]

[ QUOTE ]
[ QUOTE ]
So, all in all, he wins 4.5 + 2*(.8125*2 + .1875*(-.14)) =~ 7.6975 units over 16 hands or about .48 per hand. Nice improvement.

[/ QUOTE ]

I'm not gonna lie, I can't follow the math in these other posts, but fnord_too seems to have gotten a solution very close to mine. Can you dissprove our solutions?

Intuitively, a .27 solution seems too small to be correct.

Anyway, I see where you guys solved for B's EV, but what did you say B and A's optimal strategies were? What hands do you have A calling raises with. Did you adjust A's calling standards based on the fact that B is now bluffing occassionally?

[/ QUOTE ]

That was not the solution per se, but trying a number for the raise. Let me explain my approach.

First, I am starting with the case of no bluffing where player A knows what formula I use to raise with.

I am also using a strategy that always raises if the value is above a set amount (.75 in the case I worked through, but in general I call it R), and always calls if his number is between .5 and R.

We know that on the calls player A expects to win 75% of the time (actually, we don't know that. There is a flaw in my logic I just realized. The 75% is assuming the call is anywhere between .5 and 1, the win rate should be lower if we are calling between say .5 and .75. That term of the expectation needs to be redone. I'll come back to that later.)

If player B raises anything above R, and player A knows what R is, he should call if he has a 25% chance of winning or better, since the pot is laying him 3 to 1. If B raises with say anything over .75, 25% of the time he will have between .75 and .75+((1-.75)*.25). That is, break up the range of raise values for B and if A has anything above the first quartile, he should call.

From here, you can compute B's expectation E(x) as follows:
E(x) = probability of B folding * Expectation if B folds (call this P(fold)*E(fold) +
Probability of B calling * expectation if B calls (P(call)*E(call)) +
Probability of B raising * expectation if B raises (P(Raise * E(Raise))

E(fold) = 0 so the first component is 0

Now for the second (which I screwed up earlier). Let R be the point at which B raises. Then his expectation is going to be (.5 + R) /2. Here is why: His average calling hand will be half way between .5 and R. Say R = .7, the average calling hand will be .6. This will beat 60% of the hands out there. The chance of calling is R - .5. Again if R = .7, then B calls if he has between .5 and .7, which is 20% of the time.
So the second component of expectation is (R - .5)*(R + .5) / 2, which reduces nicely to (R^2 - .25) / 2. (note: Read R^2 as R to the second power, i.e. R squared).

The third component of expectation gets messy. A should call R+ 0.25*(1-R), or .75*R + 0.25. So, B has a (1-R) probability of raising. Of the times he raises, .75R + .25 is the liklihood that A folds because he is not getting the right odds to call. When he calls, his AVERAGE hand will be slightly better than B's average raising hand, since his calling criteria is slightly higher than B's raising criteria. His average hand is (.75R+.25+1)/2, or .375R + .625. B's average hand is simply (1+R)/2 or .5R + .5. B will win at a ratio of his average hand over A's average hand if I am not mistaken. So B will win
(.5R + .5) / ((.375R + .625) + (.5R + .5)) while A will win (.375R + .625) / ((.375R + .625) + (.5R + .5)). (I just caught another logic error from earlier).

So, if B raises, he expects to win 1 unit .75*R + 0.25 times and the remaining 1-(.75R+0.25) times he expects to win 2 units (.5R + .5) / ((.375R + .625) + (.5R + .5)) and lose 2 units (.375R + .625) / ((.375R + .625) + (.5R + .5)) times.

If you take the overall expectaion which is a function of R and maximize that function, you should get the optimal raise point for the constraints.

[fnord_too]

There is an error in the E(Calls) portion of the above post, I forgot to subtract out the losses. So E(Calls) is
Average Calling hand - (1-Average calling hand) or 2*Average calling hand + 1.
Here's an interesting thing, the Expectation of calling is exactly the probability of calling (assuming the Prob. is <= .5). So if the raise point was .6, p(Call) would be .1 (all the values from .5 to .6), the average calling hand would be .55 which would win .55 and loose .45 for an E of .1. Neat!

I am putting this in a spreadsheet and will find a close aproximation of the Ideal R since I am have no real desire to solve it in closed form. I whould have the results in 10 min or so.

[fnord_too]

Wow, I am getting some strange results. Here is a table of expectations for various raise numbers.

Raise Number Expectation
0 0.1250000
0.1 0.2164059
0.2 0.2756923
0.3 0.3081655
0.39 0.317645741
0.4 0.3176441
0.41 0.317440559
0.5 0.3068750
0.6 0.2878182
0.7 0.2718444
0.8 0.2598767
0.9 0.2524926
1 0.2500000


The maximum seems to fall right around 1/e (which is not too surprising, that number shows up all over the place in probability and statistics).

What does surprise me though is, apparently, raising all the time is a winning strategy, if A calls with every .25 or higher, he wins about 55% of the time and looses about 45% of the time, meaning that he is actually giving up too much by folding his < .25 numbers!!

If anyone wants the spreadsheet I whipped up to check it for errors I'll be happy to send it to you, I am having a hard time believing these numbers. The only thing I can think of is that how I am calculating the winning chances of calling the raise are wrong. I have assumed that if the average raising hand is X and the average calling hand is Y, then the raiser will win X /(X+Y) percent of the time and the caller will win Y /(X+Y) percent of the time.

Any thoughts on this? If I am getting a number between 0 and 1 and you are getting a number between .25 and 1, how often do you win? For reference, computing it my way the split is .44 repeating to .55 repeating (5 to four in favor of the .25 to 1 number. Think I will go monte carlo that to see what what develops.

[fnord_too]

[ QUOTE ]
I have assumed that if the average raising hand is X and the average calling hand is Y, then the raiser will win X /(X+Y) percent of the time and the caller will win Y /(X+Y) percent of the time.

Any thoughts on this? If I am getting a number between 0 and 1 and you are getting a number between .25 and 1, how often do you win? For reference, computing it my way the split is .44 repeating to .55 repeating (5 to four in favor of the .25 to 1 number. Think I will go monte carlo that to see what what develops.

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This appears to be in error, a quick monte carlo is showing the .25 to 1 hand winning about 62% of the time, not 56%, any know the propper way to calculate this value? I'll give it some thought, but if anyone has already cracked this nut I'd love to hear the answer.

[Aisthesis]

I'm a little shaky on method for the bluffing solution, but your equations got me going for a non-bluffing solution anyway. Actually, I think I'll refrain from even going through all the individual cases of fold, call or raise.

I'm pretty sure my equations agree with yours, although I couldn't follow all the way. But the great thing is that they simplify to an EV (with R as the raise threshold and .5 as the obvious call threshold) of 11/8*R - R~2.

Hope I didn't mess up anywhere along the road in adding the decisive scenarios, but in any case the solution from there is simple and does yield the best non-bluffing EV seen yet.

Differentiating over R we get:
11/8 - 2R = 0 (for local maximum)
Hence R = 11/16 as the raise threshold

That should make the EV for B 121/256 or a hair over .47

I think the best non-bluffing solution we had up to now was .46

[Aisthesis]

If anyone could explain to me (even, or perhaps more accurately, PREFERABLY with the simplest possible version of this type of problem) how to set up the equations for a bluffing solution, I'd be most appreciative!!

Having absolutely no background in game-theory, I'm having to re-invent it just to get to the non-bluffing solution.

Ankenman's results are too pretty to be wrong on the more complex case, but I'd love to be able to understand how he got there.

[Jerrod Ankenman]

[link to primer material] (groups.google.com)

[Aisthesis]

Thanks! Will take me a little bit to make it through this stuff, but actually this [0,1] game is most interesting...

The first thing that strikes me is the ability to use the continuity here to come back to the discreet "values" of poker hands (which definitely have a lot of mathematical strangenesses over and above that).