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

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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?

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

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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?

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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
 

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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.

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.

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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.


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