Game Theory: Unusual Question #3 and #4
 

By now some of you must realize that my "unusual questions" (that I don't profess to have an exact answer for) are meant to help you see why I believe excellent players should limp fairly often in very early position. Even when your opponents are not fish (but do play worse than you).

To complete this process I now ask two very precise game theory questions (again I have not calculated the solutions but I believe I will know the right answer when I see it.)

Two players, A and B receive one card with a real number between zero and one. High card wins. Player A bets one dollar in the blind. Player B can fold, call, or raise one dollar. If he raises, A can call or fold only. End of game.
What is the optimal strategy for each player and the value of the game to B?

To get you started notice that if there were no raises allowed, B would simply call whenever he had above .5. When he called he would win three quarters of the time (half of the time they were both above .5 and all of the time A was below.) Thus in eight tries he would fold four, win a dollar three times and lose a dollar once. The game is worth 25 cents to him. Since the blind is not live, the existence of an option to raise must help player B because if it didn't he could revert to the never raise strategy.

The second version of this question makes A's blind live. So if B just calls, A can raise if he chooses. Game over. Logic says that the value of this second game to B, if both players play optimally, is less than it would be if the blind wasn't live. Solving this second game for optimal strategy is probaly a lot harder than the first one. But I know we have some two plus twoers who can do it.

For the majority of you who know they are not familiar enough with game theory to tackle these proplems precisely, I ask a more general question. Should B raise more often in the first game or in the second game where's A's blind is live?

My solution
 

[ QUOTE ]
Two players, A and B receive one card with a real number between zero and one. High card wins. Player A bets one dollar in the blind. Player B can fold, call, or raise one dollar. If he raises, A can call or fold only. End of game.
What is the optimal strategy for each player and the value of the game to B?

[/ QUOTE ]

B clearly folds anything less than .5, and calls with a .5. The tough question is when is raising profitable? B will raise with #'s > X, where X > .5. A will then call with Y, where Y >X. If Y calls with # < X, he is gaurenteed to lose.

Let's say X = .75 (i like the ol' guess and check method). What would the optimal value for Y be?

Y is getting 3:1 pot odds on the call, so he has to win 1/4th of the time. Calling with .76 would obviously not win 1/4th of the time, and calling with .875 would win half the time. I'm assuming the average of .75 and .875 is what he calls with -> .8125.

So what's the EV of this strategy for B?

1/2 the time, he folds. EV = 0
1/4th of the time, he calls. When he calls, he wins 3/4 of the time. EV = $.75
1/4th of the time he raises, we have to split this up.

When he raises, A calls 18.75% of the time.

25x.1875 = 4.6875% of the time, B raises, and A calls.
25-4.6875=20.3125% of the time, B raises, and A folds. EV = $1

When A calls the raise, he wins 1/4th of the time. So in 4 tries B wins $2, wins $2, wins $2, and loses $2. He wins $1.50 on avg when he raises, and A calls.

50% of the time ev = 0.
25% of the time ev = $.75
20.3125% of the time ev = $1
4.6875% of the time, ev = $1.50

[(.5)(0)+(25)(.75)+(20.3125)(1)+(4.6875)(1.50)] / 100 = .4609375

I think this is player B's ev with this strategy, + $.4609375 per game.

So did I calculate the EV correctly? If I did, is this the highest EV attainable? I prolly screwed up somewhere. I'm still in highschool, so I haven't taken any game theory classes or anything.

Re: My solution
 

Justin, B's optimal strategy will be to raise with some of his best hand but also to bluff raise with some bad hands.

Re: My solution
 

[ QUOTE ]
Justin, B's optimal strategy will be to raise with some of his best hand but also to bluff raise with some bad hands.

[/ QUOTE ]

This is incorrect. Let's say he raises all hands .6 (the number is arbitrary) and above, and then bluffs with 10% of his hands. Wouldn't it make more sense to just change that to raise .5 and above instead? Instead of raising 10% of bad hands, just raise the next 10% of hands after the cutoff point.

Re: Game Theory: Unusual Question #3 and #4
 

I don't think there is any calling in version 1.
Bet top 1/2 and bluff bottom 1/6.
Fold remaining 1/3.

THe blind calls with his top 4/9 of hands

B wins .27 units per hand.

Re: Game Theory: Unusual Question #3 and #4
 

It's been a while since I've played around with this type of game theory, but I think there is some subtlety here.

Looking at the one where B can call or raise, but A can only call or fold:

I am going to make the simplifying assumption that A knows B's strategy, and plays optimally against it.

If B raises every time he has a .5 or better (an no other times), then A's best play is to call if he has .625 or better other wise fold. Since he knows that B has somewhere between .5 and 1, 1/4 of the time he will have between .5 and .625 so he is getting break even odds to call here. So before investigating further let me compute the value of the game. Over 16 plays, B will fold 8 and raise 8. Of the 8 he raises, he will win 5 out right for +5, and will be called on 3. Of the three he is called on, A will have between .625 and 1 or an average of .8125 to B's average of .75. So B will gain (.75/.8125) * 2 - (.8125/.75)*2 = ~ -.32 per hand, so ~-.96 for the three hands or just over +4 units for the 8 hands he plays. So that is JUST barely over calling with .5 or above.

I'm going to run through this again with slightly tighter raising standards because I believe I will see a way to solve this in closed form without bluffing. Adding in bluffing will be somewhat trickier.

Say B calls with .5-.75 and raises with .75-1

Then, A should call a raise with .8125 or greater.

So 16 hands, B folds 8, calls 6, raises 2. On the 6 he calls, he wins .75 * 6 = 4.5 units.

On the two he raises, his average hand will be .875. .8125 he wins outright.
of the .1875 he gets called, A will have an average hand of (1.8125/2) = .90625, so B wins (.875/.90625) * 2 - (.90625/.875) * 2 = ~1.931 - 2.071 =~ -.14 per hand or ~ -.28.

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.

I think I am ready to write this out abstractly:
We at least call with anything .5 and above
R = raising point.
C = calling point for A = R + .25*(1-R)
Note: Probability of Raise = 1-R. Probability of call = .5 - (1-R) = R - .5

E(x) = P(folding)*E(folding) + P(calling)*E(calling) + P(raising)*E(raising)
= (0.5)*(0) + (R-0.5)(0.75) + (1-R)E(raising)

E(raising) = P(A Folding) * 1 + P(A calling)*E(A calls) = C + (1-C) E(A calls)

E(A calls) = (((1+R)/2)/((1+C)/2)-((1+C)/2)/((1+R)/2)) * 2

so, deep breath, substituting in R + .25*(1-R) for C and expanding everything out, we need to maximize:

E(x) = (0.5)*(0) + (R-0.5)(0.75) + (1-R)*((R + .25*(1-R)) + ((1-R + .25*(1-R))* (((1+R)/2)/((1+R + .25*(1-R))/2)-((1+R + .25*(1-R))/2)/((1+R)/2)) * 2))

I may have some parens screwed up, but that is about it. I need to go off line and write this out. I will simplify and maximize and post my result for the published, no bluffing, optimal strategy shortly.

Regards,

Eric

Re: My solution
 

B should pretty clearly call with hand .51. So if he was to add bluffing hands he shouldn't do it with hands that have positive EV by calling; he should do it with hands that have no EV at all; in this case his worse hands.

Re: Game Theory: Unusual Question #3 and #4
 

I get that B should bet his best 1/6 hands, bluff his worst 1/18, and A should call with his best 1/3.

I may have made an arithmetic mistake, but the value of the game for B is 5/18=.278, so this is better than ProfessorJC's sol'n.

Justin, just to show you are wrong, I solved for the best non-bluffing strategy for B.

Here, let B bet his hand if it is better than b, and A call with hands better than a. Here a>b, since A knows he is throwing away money calling with a hand worse than b.

a=3b/4+1/4

The value of the game for B is

1/4-1/8*(1-b)^2

This function has a maximum at b=1: namely b never bets, he calls (with hands above 1/2) or folds.

Craig

Re: Game Theory: Unusual Question #3 and #4
 

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.

That\'s correct
 

[ QUOTE ]
I get that B should bet his best 1/6 hands, bluff his worst 1/18, and A should call with his best 1/3.

I may have made an arithmetic mistake, but the value of the game for B is 5/18=.278

[/ QUOTE ]

Damn, you just beat me to it. I verify your algebra.

B raises with [5/6,1], calls with [.5,2/3], (bluff) raises with 1/18 of his hands below .5 (you used [0,1/18], though you could use others), and fold the rest. Value to B is 5/18.

The second question looks a little more tedious!

al'Thor