Why Position Matters Using Game Theory

[mykey1961]

[ QUOTE ]
If player one bets with [0,0.1] and [0.8,1] then player two should/could call with [0.8,1] .

If player one checks with [0.1,0.8] then player two should/could bet with [1/3,0.8] and check behind everything else .



[/ QUOTE ]

why wouldn't Player 2 Bet with [1/3,1] when Player 1 checks?

[marv]

[ QUOTE ]
If player one bets with [0,0.1] and [0.8,1] then player two should/could call with [0.8,1] .

If player one checks with [0.1,0.8] then player two should/could bet with [1/3,0.8] and check behind everything else .

Under this strategy for player two , player one's EV is :

FE= 2*3/10*8/10 =0.48
Check equity= 2*7/10*1/3*2/3 = 0.31111111

Total EV = 0.79111111 -1 = -20888888

Under my strategy for x=7/12 and a=3/8

Total EV = 0.79166666-1 =-20.833333333 (I plugged x=7/12 into my equation )

Again , it would be nice if one would develop a viable strategy for both players so it's easier to compare .

[/ QUOTE ]

I wildly disagree with this EV. My simulation of the exact strategies mentioned here gives close to +0.065 for player1.

Marv


#include <stdlib.h>
#include <stdio.h>

int main()
{
int n = 0;
double sum = 0.0;

srand48(123);

for (n=0; ; n++) {
double h1 = drand48();
double h2 = drand48();

if (h1 < 0.1 || h1 > 0.8)
if (h2 < 0.8)
sum += 1;
else
sum += (h1 > h2 ? 3 : -3);
else
if (h2 > 1/3.0 && h2 < 0.8)
if (h1 < 0.3 || (h1 > 0.4 && h1 < 0.6))
sum += -1;
else
sum += (h1 > h2 ? 3 : -3);
else
sum += (h1 > h2 ? 1 : -1);


if (!(n % 1000000))
printf("%i %g\n", n, sum/n);
}
}

[mykey1961]

Post deleted by mykey1961

[jay_shark]

Am I the only person trying to find a mathematical solution to this ?

It seems like you guys are spewing random numbers with disregard to the solution .

[marv]

[ QUOTE ]
Marv,

try

SB
[0,0.1] bet
[0.1,0.3] check-fold
[0.3,0.4] check-call
[0.4,0.6] check-fold
[0.6,0.8] check-call
[0.8,1.0] bet

vs

BB
if SB Bets
[0,1/10] fold
[1/10,1] Call
if SB Checks
[0,17/30] Check
[17/30,1] Bet

I'm thinking the EV for SB will be -0.49

[/ QUOTE ]

I get about -0.044 for this pair.

To Jay. If we can't agree on the EV of one strategy vs another we have no hope of solving these problems.

Read the Mathematics of Poker for one way to get exact solutions to [0,1] games. This is old, old stuff, and I'm sure someone will use the method described in that book to produce the correct answer at some point.

My techniques don't apply to the continuous [0,1] game but can handle more general discrete games, so I'm not trying to solve your problem, I'm just trying to show you that your own approach is giving completely rubbish answers. If this is not helpful I'll just stop and let you be.

Marv

[jay_shark]

EV = (3x-1)/2*(3x-1)/6*5/3 + 3x*(1-x)
EV' = -10x/4 -1/6 +3 -6x

Set this equal to 0 and we get x=17*2/(3*14) ~ 0.8095

So opponent 2's optimal calling range should be about [0.8095,1]

Player one should shove with a= 5/7 .

Can one do better than this ?

If so , then I must have my equations wrong .

I may have to start from scratch is this is not working out .

[jay_shark]

[ QUOTE ]


I wildly disagree with this EV. My simulation of the exact strategies mentioned here gives close to +0.065 for player1.

Marv


#include <stdlib.h>
#include <stdio.h>

int main()
{
int n = 0;
double sum = 0.0;

srand48(123);

for (n=0; ; n++) {
double h1 = drand48();
double h2 = drand48();

if (h1 < 0.1 || h1 > 0.8)
if (h2 < 0.8)
sum += 1;
else
sum += (h1 > h2 ? 3 : -3);
else
if (h2 > 1/3.0 && h2 < 0.8)
if (h1 < 0.3 || (h1 > 0.4 && h1 < 0.6))
sum += -1;
else
sum += (h1 > h2 ? 3 : -3);
else
sum += (h1 > h2 ? 1 : -1);


if (!(n % 1000000))
printf("%i %g\n", n, sum/n);
}
}

[/ QUOTE ]

Marv , we can work together on this .

My intuition tells me that this game should be negative EV for player 1 . Are you saying that this is not the case ?

[mykey1961]

[ QUOTE ]
Am I the only person trying to find a mathematical solution to this ?

It seems like you guys are spewing random numbers with disregard to the solution .

[/ QUOTE ]

Is it better to spew calculated numbers that aren't the solution because your formula isn't accurate?

[jay_shark]

Actually my latest answer is very close to Marv's . The only difference is that he's working in the discrete case where as I'm not .

[marv]

With optimal play on both sides, player 2 will have a positional advantage, but under optimal play each player will be using a best response to the other players strategy.

In my example which wasn't quite optimal for player 1, if P1 checks with [0.1,0.8] and folds to a bet too often (as he does in my example), player 2's best response to a check must be to bet everything in [0,1]. Your proposal for player 2 wasn't betting enough so player 1 was getting more EV than he should.

Note that player 1's distribution after check is not that much weaker than after he bets. This is typical when player 1 may have further decisions after checking, and is why we wouldn't expect a strategy of the form 'always fold after checking' to be optimal - it means P1's initial action gives away too much information.

Marv