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

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

[ 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?

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

[ 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.

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

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.

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

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.

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

[ 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.

[/ QUOTE ]

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.

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

Ok, How about this logic:

As the caller demaning at least 3-1 odds on your call, you hace between a .25 and 1 chance of winning evenly distributed accross all calls. That is, to use the calling strategy of calling with .25 or higher against a someone who will raise with any number, when you call with exactly a .25 you have a .25 chance of winning, and when you call with a 1 you have a lock. The fact that the range you are in is from 0-1 does not matter. If the raiser raised with any .5 or over you would be in the range of .625 to 1, but should still win 62.5% of all your calls.

Re computing the above table I get:
Raise Number Expectation
0 -0.03125
0.1 0.08747
0.2 0.17600
0.3 0.23603
0.4 0.26925
0.5 0.27734
0.6 0.27200
0.7 0.26491
0.8 0.25775
0.9 0.25222
1 0.25000


And exploring around .5, .5 seems to be the exact raise point (if you are going to come into this game, raising adds about a 2.7% edge, I think that is what some of the earlier posters posted). Oh well, it was an interesting journey. I may give some thought to the bluffing part later, but I feel like a fool for taking so long to figure the most simplified version out.

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

I dont think this is correct, although its late and my math could be off.

A can beat this strategy simply by calling his best half and folding his worst.

Half the time B bluffs, A will call and B will lose $2. Half the time B bluffs, A will fold and B will win $1.
This bluffing strategy yields a negative EV of -.08 for the bluff itself. This is calculated as follows .167*.5*1 - .167*.5*-2

This yields a net EV of .16

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

The answer to the second problem is:
(There are co-optimal solutions, but where possible I present undominated ones)

B:
folds on [0,19/36]
raise-bluffs on [19/36,7/12] *
limp-folds on [7/12,2/3]
limp-calls on [2/3,5/6]
value-raises on [5/6,1]

A:
calls a raise on [2/3,1]
folds to a raise on [0,2/3]
raises after B limps on [3/4,1]
checks after B limps on [1/12,3/4]
raises after B limps on [0,1/12]

The value of the game is 1/4.

To answer David's question, B should just raise the same amount of hands when the A's blind is made live. B, however, doesn't limp with his thinnest value calls, because he'll face a raise sometimes. He just folds them.

Jerrod Ankenman

My solution for the first game.
 

Without having read any of the other answers I think that player B should always raise when he has a vale greater than .5. He should bluff, randomly, 50% of the time with hands that have a value < .5. If it's exactly .5 it doesn't matter what he does. I don't think that B should ever just call.

A should then call with any hand that has a value > .25.

A) 25% of the time both hands will be > .5 B will win 50% of these for $2 and lose 50% for 2$.
B) 25% of the time B will be > .5 and A < .5 B will win 50% of these for $2 and 50% for $1
C) 25% of the time both hands will be < .5. B will bet 50% of these and will win 50% of the ones he bets.
D) 25% of the time B < .5 and A > .5. B will bet 50% of these and lose them all.

Out of 100 hands B will bet 75% of them for a total investment of $150
A and C are stand offs, so all the profit/loss comes from B and D.
B) 12.5 times B wins 2 and 12.5 times B wins 1. Total = +37.5
D) 12.5 times B loses 2 Total = -25

Net = +12.5 on a $150 investment. EV = +.125/hand

If A were to call more often with numbers < .5 B would win more, and if A were to call less often B would also win more. If it were correct for B to just call I suspect that it would be some random element as well, but I just don't see it adding any EV. If B realizes that A is always calling > .25 he could change his strategy to compensate, then B adjusts etc, and we end up coming back around to what I already have (eventually, with enough adjustments)


So how'd I do? I'll read the rest of the responses and see. I don't feel like tackling the 2nd problem at the moment. And the answer to the more general question requires me to at least take a preliminary stab at the 2nd problem. So as a pure guess I'll say that B should raise more often in the first one and that calling some of the time now would be a valid option in the 2nd.

Re: My solution for the first game.
 

Your answer contains very sub optimal strategy if it only yields 12.5 cents per game.

[ QUOTE ]
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.


[/ QUOTE ]

Re: #3 non-bluffing solution
 

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

Re: #3 bluffing solution
 

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.

Re: #3 bluffing solution
 

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

Re: #3 bluffing solution
 

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).

Re: #3 bluffing solution
 

I've just made it through lessons 1-3, but that helped immensely--although I think I'll need to get further into it to come up with an algebraic solution even on the simple scenario (still unclear to me how to deal with bluff or even finite pot).

But that brings up my question: Once you actually know what you're doing, is the algebraic approach usually easier than analysis (differentiating for maxima and minima)? I still feel irresistably attracted to analytic solutions and am actually, after looking at the way you set it up, thinking I may be able to get a clean bluffing solution (for the "easy" problem) by differentiating for EV (I'll post it if it works out). I'm actually guessing that the indifference equations turn out to be equivalents of the various equations you get when you differentiate... (??)

Re: #3 bluffing solution
 

I have been thinking about the bluffing part. If B's strategy is known then you compute at what point A's call has a 25% chance of winning (same as with no bluffing). I'm going to work on the algebra later today if I have time. If B always raised A calls with any hand over .25, and has a slight edge in the game (it is worth .03 to him I think).

Re: My solution
 

[ QUOTE ]
[ 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.

[/ QUOTE ]

If you select just the next 10% of hands, player A can adjust his strategy knowing he needs a little less to call. Using the bottom 10% to bluff is much more effective. When you raise, player A must decide if you are bluffing or not. If you always raise with the top 40% in your example, then 40% of the time you have a strong hand and player A must be even stronger to call. But player A also knows that 10% of the time you raise, he has you beat. Then to call becomes a guessing game for him.

Where did I go wrong?
 

I am doing something wrong, but I can't find out what.
Suppose we're looking at the bluff/no-fold strategy.

Player A will call a raise whenever he has something greater than z.
Player B will bluff with something smaller than x, and honestly raise with something over y and will call with the rest.

The bluff:
B bluffs with probability x, and when he does, he
wins 1 with probability z, and loses 2 with probability 1-z.
So we have x(3z-2)

The call:
B calls with probability y-z, and when he does, he
wins 1 with probability x+(y-x)/2, and loses 1 with probability (y-x)/2+(1-y).
So we have (y-x)(x+y-1)

The honest raise:
B does this with probability 1-y, and when he does, he
wins 1 with probability z when A folds, and when called (which happens with probability (1-z)) he will win 2 with probability
(y-z)/(1-z)+(1-y)/(1-z)/2 and lose 2 with probability (1-y)/(1-z)/2.
So here we have (1-y)(2y-z)

Summing these three will result in the EV-function for B which now is
EV(x,y,z)=-x^2-y^2-x-y-z+3xz+yz

[ 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. [...] so the value of the game for B is 5/18.

[/ QUOTE ]

For this solution (x,y,z)=(1/18,5/6,2/3), which results in EV=13/162, or about .0802.

This is not even close to .278

Where did it all go wrong?

Thanks for helping.

Question 3 for Dummies
 

You don't need fancy math. A must call with the top one third of his hands to prevent B from profiting from bluffing with bad hands. (B is laying 2-1 on those bluffs)

B should bet the top one sixth (3/18) of his hands for value since those hands are favored to win even if called.

Since A is getting 3-1 when bet into, B should bluff 1/18 of his total hands so that his bet to bluff ratio is 3-1.

Since we know that B makes 25 cents in the non raising game, we need only see how much the existence of a possible raise helps. It doesn't help on his bluffs since they break even. It doesn't help when both players have the top sixth of their hands. It helps only when B is in the top sixth and A is in the SECOND sixth. That gains a bet for B 1/36 of the time. His EV moves from 9/36 to 10/36. .278

Re: Where did I go wrong?
 

"I am doing something wrong, but I can't find out what.
Suppose we're looking at the bluff/no-fold strategy.

Player A will call a raise whenever he has something greater than z.
Player B will bluff with something smaller than x, and honestly raise with something over y and will call with the rest."

Because the bluff/no-fold strategy is not optimal: you need to add (the point I neglected to mention in my post, because it was so obvious) that B will fold hands below .5 that he doesn't bluff with.

This makes for the possibility of many cooptimal strategies: any strategy where B bluffs 1/18 of the time (all below .5), calls with .5-5/6 and bets 5/6-1. Thus the best strategy, if A is playing very poorly by calling with too many hands (more than 1/2), would be to bluff with 8/18-9/18.

I don't see how all this solving of the bluffless case is of any use: ANY PLAYER THAT NEVER BLUFFS IS GIVING AWAY EV (unless he is playing an absolute calling station). DO you solvers see anything wrong with my bluffless analysis that shows that B should use the same strategy for the (B) raise game as for the call or fold game if he is not allowed to bluff? And fnord, you are working too hard on the hard way to solve this, check out the game theory primer. Aisthesis: the math is much easier algebraically, though with programs to differentiate for you the differential analysis is not too bad, and getting results both ways provides a check.
Craig

Thanks, sometimes...
 

[ QUOTE ]
Because the bluff/no-fold strategy is not optimal: you need to add (the point I neglected to mention in my post, because it was so obvious) that B will fold hands below .5 that he doesn't bluff with.

[/ QUOTE ]

Thanks a lot. Obviously he should fold some hands.
Sometimes you just won't see you missed something in the very beginning...

I'll go on now.

But math does make it easy... (5/18 is correct)
 

You're right, you don't need fancy math.
But problems are easily solved that way, that is: if you don't mess up...

If A's strategy is:
[0,x] : raise(bluff)
[x,y] : fold
[y,z] : call
[z,1] : raise
And B's strategy is:
[0,a] : fold if raised
[a,1] : call if raised,
where 0<x<y<a<z<1, the EV of B would be z-2*x-y^2+3*x*a+a*z-z^2+y-a.

Solving grad(EV)=(0,0,0,0) gives

x=1/18
y=1/2
z=5/6
a=2/3

And EV becomes 5/18.

Tomorrow I'll look at #4, must sleep now, tomorrow I have an exam...

Next Time.

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

I get the same values for the solution for #4, but I get a game value of 17/72.

Anyone want to check on these?

If this answer is correct, B would rather have the no raising game than the one raise game, even though he acts last when he raises.

Craig

The [0,1] game and poker
 

This [0,1] game idea is all new to me, but since it may also be new to others here, I thought I'd post an idea I had as to the analogy to poker.

My first thought was that these numbers between 0 and 1 turn out to be analogous to some kind of hand-ranking. But I think that's not really the idea.

In fact, what one is in a sense dealt pre-flop is simply a probability of winning. Hence, even at a full table with all the positional complexities, given the cards everyone is dealt, your hand actually is nothing more than a PROBABILITY OF WINNING at showdown--hence, it actually IS a number between 0 and 1.

One odd fact, though, is that you don't even know exactly what your own number "objectively" is, because, while it is determinate, it depends on the cards the other people at the table are holding (similarly, the number each of them has is only partially known to them, since it depends on your hand, etc.).

I guess this still all boils down to your having a subjective probability with all the unknowns out there. It's just that the subjective probability you're dealt will in fact be different from the probabilities that could be assigned by someone who actually knew what everyone was holding but just didn't know what cards were going to hit the board on flop, turn, and river.

An extreme example of this type of discrepancy: You are holding KK, which is going to make your subjective probability of winning pretty high. But if someone else at the table has AA, your subjective probability is very far removed from the actual probability you have of winning the hand. From the standpoint of modelling the game, I have no idea whether that type of distinction would prove useful or not.

Anyhow, at each stage of a hand, you are actually holding a subjective probability, and at the next stage (flop, turn, river), you will hold a different subjective probability (unless, for example, you flop a royal flush with AKs in your hand or other similar cases--as soon as you have the nuts and can't be outdrawn, then you have a "1," and once you have a 1 it's not going down) but one that IS NOT RANDOM given your subjective probability at the previous stage.

Anyhow, the one thing I'm seeing at the moment is that, in contrast to the example here, the highest "player's subjective probability" (even with best calculation of this value) does not necessarily win the hand (or does it?).

But if one does introduce the additional value of a "probability from the standpoint of an outside observer," (who is ignorant only as to what cards will subsequently be dealt in the middle but knows the hands of all players as well as the current board) this value does determine who is ahead at any given stage of the hand and will typically be either 1 or 0 on the river (or a fraction of 1 in the case of split pots).

Anyhow, these are just some thoughts for beginners such as myself on how the relationship between poker and this [0,1] game might work.

Am I making any sense? or just completely off-base here? [img]/images/graemlins/smile.gif[/img]

Re: #3 bluffing solution
 

Yeah, the idea for using differentiation over various "decision point" variables for the bluffing solution just turned into a big mess. I couldn't get the equations to break down into anything that looked like it would go anywhere.

Re: The [0,1] game and poker
 

Basically, the problems tractable to game theory (and also those for which the [0,1] set makes a reasonable approximation) are generally river problems.


Craig

Re: The [0,1] game and poker
 

[ QUOTE ]
Basically, the problems tractable to game theory (and also those for which the [0,1] set makes a reasonable approximation) are generally river problems.

Craig

[/ QUOTE ]

Well, I dunno. The problems that David poses are usually river problems, because those are the ones that he can solve.

Jerrod

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

I'm willing to believe that the game value is wrong; I kinda did it in a hurry.

BTW, you folks that are doing these problems by brute force (ie, coming up with huge EV equations and maximizing them) are going to run into trouble if you try to solve more complicated games.

Jerrod

Re: The [0,1] game and poker
 

lol... If I'm taking this right, it seems like the river, while already pretty complex, is indeed just going to be the most readily solvable case. But it seems like the model would in principle apply to all streets, if one can eventually get the enormous amount of complexity under control.

Once I've digested at least the "primer material" on this, I'll be very interested to see where you guys are at.

Re: The [0,1] game and poker
 

Sure. I mean, all these games can be turned into gigantic matrices of strategy decisions for the two players and Nature, and you could solve it by brute force. It's just that there are too many combinations and not enough computer power.

The point of the analytical solutions, like the ones Bill and I have worked on, is to find ways to solve entirely games that are really hard to do by brute force - like, for example, no-limit games, as well as effectively approximate solutions in different situations and develop intuition about situations that might not arise from usual analysis.

You guys know about our contest, right? The solution to the no-limit AKQ game, which we have and which is fairly startling, is worth a couple hundred bucks and a cite in our book. I dunno if I'll be censured for advertising, since it isn't a 2+2 book, but whatever.

Jerrod