Why Position Matters Using Game Theory

[jay_shark]

Marv , each player posts 1$ in antes so there are 2 $'s in total before action begins . Either player is allowed to bet 2$ which is the size of the pot after they've chosen their numbers .

I'm fairly certain that the EV for player 1 is -0.25 which is totally off from yours so maybe you misinterpreted the question .

To Mikey :

Your total EV using your strategy is 4/7*2*13/21 + Equity when you check .

Your equity when you check from [0,3/7] and that player 2 checks with [0,1/7] is 3/7*1/7*2/3 .

Total EV ~0.74829 . If we subtract the $1 ante we get ~ -0.2517.

[mykey1961]

Analysis for (a = 3/7)

From P1's Perspective:

P1(0..1/7) Check P2(0..1/7) Check -> Tie ($0) * 9/441 = 0/441
P1(1/7..3/7) Check P2(0..1/7) Check -> P1 Win ($1) * 18/441 = 18/441
P1(0..3/7) Check,Fold P2(1/7..1) Bet -> P2 Win (-$1) * 162/441 = -162/441
P1(3/7..1) Bet P2(0..13/21) Fold -> P1 Win ($1) * 156/441 = 156/441
P1(3/7..13/21) Bet P2(13/21..1) Call -> P2 Win (-$3) * 32/441 = -96/441
P1(13/21..1) Bet P2(13/21..1) Call -> Tie ($0) * 64/441 = 0/441

EV = -84/441 = -4/21

[jay_shark]

I have to change my formula set up to include the equity you get when both hands get checked .

Here is player 1's equity using a=3/7 with corrections .


The only profit player 1 makes is from his fold equity and when both numbers get checked . He breaks even when both players contest for the pot .

If you shove with 3/7+ your fold equity is

So 4/7 times you will be betting . In order to win the pot , you need to assure yourself that player two will fold which will happen 13/21 times .

FE= 2*4/7*13/21

Equity when both hands get checked is :

To figure out our equity for player 1 we have to compute the probability that player 1 wins given that he checks [0,3/7] and player 2 checks [0,1/7]. This is simply 2/3 .

3/7*1/7*2/3*2 .

If we add the two we get that player 2's EV = 116/147 . If we subtract $1 we get -31/147 . Which is better than my original -0.25 .

[mykey1961]

[ QUOTE ]
Marv , each player posts 1$ in antes so there are 2 $'s in total before action begins . Either player is allowed to bet 2$ which is the size of the pot after they've chosen their numbers .

I'm fairly certain that the EV for player 1 is -0.25 which is totally off from yours so maybe you misinterpreted the question .

To Mikey :

Your total EV using your strategy is 4/7*2*13/21 + Equity when you check .

Your equity when you check from [0,3/7] and that player 2 checks with [0,1/7] is 3/7*1/7*2/3 .

Total EV ~0.74829 . If we subtract the $1 ante we get ~ -0.2517.

[/ QUOTE ]

Gotta love it when you throw together a formula, get a number you like, and think it's the answer to the question.


Lets take "4/7*2*13/21"

What does the 4/7, 2, and 13/21 represent to you?

[jay_shark]

Final EV set up .

EV in fold equity = 3x*(1-x)

EV when both hands get checked = 2*(3x-1)/2*(3x-1)/6*2/3.

After you set the derivative =0 you should get x=7/12 .
So a=3/8

Final answer .

[mykey1961]

[ QUOTE ]
I have to change my formula set up to include the equity you get when both hands get checked .

Here is player 1's equity using a=3/7 with corrections .


The only profit player 1 makes is from his fold equity and when both numbers get checked . He breaks even when both players contest for the pot .

If you shove with 3/7+ your fold equity is

So 4/7 times you will be betting . In order to win the pot , you need to assure yourself that player two will fold which will happen 13/21 times .

FE= 2*4/7*13/21

Equity when both hands get checked is :

To figure out our equity for player 1 we have to compute the probability that player 1 wins given that he checks [0,3/7] and player 2 checks [0,1/7]. This is simply 2/3 .

3/7*1/7*2/3*2 .


If we add the two we get that player 2's EV = 116/147 . If we subtract $1 we get -31/147 . Which is better than my original -0.25 .

[/ QUOTE ]

You need to be a little more careful when you combine Expected Value and Equity.

Expected Value = Pot * Equity - Cost

When P1 and P2 checks, P1's Equity = 5/6, Pot = $2, Cost = $1, EV_chk_chk = $2*5/6-$1 = 2/3

When P2 Bets, and P1 Folds, P1's Equity = 0, Pot = $4, Cost = $1, EV_chk_bet_fold = $4*0-$1 = -1

When P1 Bets and P2 Folds, P1's Equity = 1, Pot = $4, Cost = $3, EV_Bet_Fold = $4*1-$3 = 1

When P1 Bets and P2 Calls, P1's Equity = 1/3, Pot = $6, Cost = $3, EV_Bet_Call = $6 * 1/3 - $3 = -1


P1 and P2 Checks: 3/7 * 1/7 * 2/3 = 6/147
P2 Bets and P1 Folds: 6/7 * 3/7 * -1 = -18/49 = -54/147
P1 Bets and P2 Folds: 4/7 * 13/21 * 1 = 52/147
P1 Bets and P2 Calls: 4/7 * 8/21 * -1 = -32/147

(6 -54 +52 -32)/147 = -28/147 = -4/21

[marv]

[ QUOTE ]
I have to change my formula set up to include the equity you get when both hands get checked .

Here is player 1's equity using a=3/7 with corrections .


The only profit player 1 makes is from his fold equity and when both numbers get checked . He breaks even when both players contest for the pot .

If you shove with 3/7+ your fold equity is

So 4/7 times you will be betting . In order to win the pot , you need to assure yourself that player two will fold which will happen 13/21 times .

FE= 2*4/7*13/21

Equity when both hands get checked is :

To figure out our equity for player 1 we have to compute the probability that player 1 wins given that he checks [0,3/7] and player 2 checks [0,1/7]. This is simply 2/3 .

3/7*1/7*2/3*2 .

If we add the two we get that player 2's EV = 116/147 . If we subtract $1 we get -31/147 . Which is better than my original -0.25 .

[/ QUOTE ]

Jay. Consider the following strategy for player1 (which I do not claim is optimal):

[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

This is much better than either of your strategies.

Marv

[jay_shark]

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 .

[mykey1961]

[ QUOTE ]

Equity when both hands get checked is :

To figure out our equity for player 1 we have to compute the probability that player 1 wins given that he checks [0,3/7] and player 2 checks [0,1/7]. This is simply 2/3 .

3/7*1/7*2/3*2 .


[/ QUOTE ]

This is not true.

When P2 is [0,1/7]:

P1 has 1/2 equity for [0,1/7]
P1 has 1 equity for [1/7,2/7]
P1 has 1 equity for [2/7,3/7]

Therefore P1 [0,3/7] vs P2 [0,1/7]: P1 has 5/6 equity.

[jay_shark]

[ QUOTE ]
[ QUOTE ]

Equity when both hands get checked is :

To figure out our equity for player 1 we have to compute the probability that player 1 wins given that he checks [0,3/7] and player 2 checks [0,1/7]. This is simply 2/3 .

3/7*1/7*2/3*2 .


[/ QUOTE ]

This is not true.

When P2 is [0,1/7]:

P1 has 1/2 equity for [0,1/7]
P1 has 1 equity for [1/7,2/7]
P1 has 1 equity for [2/7,3/7]

Therefore P1 [0,3/7] vs P2 [0,1/7]: P1 has 5/6 equity.

[/ QUOTE ]

Given that both players have checked , the probability player one has a number from [0,1/7]is 1/3. When this happens , he will win $2 , one-half of the time .

1/3*2*1/2 = 1/3 .

The probability player two has a number from [1/7,3/7] is 2/3 . When this happens , he will win $2 , 100% of the time .

2/3*2*1 =4/3

His EV = 4/3+1/3 =5/3 =2*5/6 .

I will have to go back and make yet another adjustment .
The problem isn't too difficult it's just very easy to get sidetracked .