Why Position Matters Using Game Theory

[jay_shark]

I thought I mentioned that .

There are two antes so you should bet the pot which is 2 antes . So your opponent is getting 2:1 odds to call .

[TNixon]

Oh. Duh. You're right. I missed the "pot" part of "bet the pot" in one sentence, and everywhere else you just say "bet". [img]/images/graemlins/smile.gif[/img]

[shat4brains]

why dont u become a math teacher ?

[mykey1961]

[ QUOTE ]
Here is another interesting problem that shows why position matters .

There are two players who pick random numbers from the closed interval [0,1] . Each player posts an ante and has to make a decision whether or not to play for the pot or fold . The first player to act has three choices to make . He may either bet the pot, check-call , or check and fold . Player two can only call a bet , check or fold . There is no raising in this game .

a) What numbers should player one bet with ?
b) What numbers should player one check and call with ?
c) What numbers should player one check and fold with ?

*Bonus Question*

What is the EV of this game for both players ?

[/ QUOTE ]

I'm not 100% sure but here goes:

Player 1 should bet with 3/7 thru 1
Player 1 should never check and call.
Player 1 should check and fold with 0 thru 3/7.

If Player 1 checks:
Player 2 should check with 0 thru 1/7
Player 2 should bet with 1/7 thru 1

If Player 1 bets:
Player 2 should fold with 0 thru 13/21
Player 2 should call witn 13/21 thru 1

EV is -4/21 for Player 1

[jay_shark]

Here is one possible solution . I will follow with the same set up as I did for my Game Theory Resolution problem .

Solution: Let a be your optimal pushing range ; a>=0
Let x be your opponent's optimal calling range . 1/3<=x<=1

We can write a in terms of x . Notice that (1-x)/(x-a) = 2
x=(2a+1)/3

The probability that your opponent wins given that he calls is 2/3 which is verified since 1/3 + 2/3*1/2 = 2/3

So the probability that you win given that he calls is 1/3 .

Lets compute your optimal EV when you bet .

EV = 3*(1-x)/2*[2x + 4*1/3*(1-x) -2*2/3*(1-x)]
EV= 3x*(1-x) after simplifying .

Take the derivative of the EV function

EV' = -6x+3 which means that x>=1/2 if we set the derivative =0 .

Therefore a=1/4 .

So we should only check and call or possibly check and fold if our number is less than 1/4 and that we will always bluff with hands [1/4,1/2]. Note that there are variations to one's bluffing interval .

If we check , then our opponent knows that our number is less than 1/4 .

I haven't worked out the check and call or check and fold scenario but I'll come up with an answer shortly .

[jay_shark]

So player one should check and fold 100% of the time with numbers less than 1/4 .

Clearly , player two's EV is positive just for being in position !

[mykey1961]

Analysis of (a = 1/4)

P1 bets (1/4..1)
P2 fold (0..1/2), call (1/2..1)

P1 checks (0..1/4)
P2 checks (0..1/12)

P2 bets (1/12..1)

P1 folds (0..7/18)
P1 calls (7/18..1) but if P1 had (7/18..1) he would have bet.


Calculating EV from P1's perspective:

P1 (0..1/12), P2 (0.,1/12) : Check Check Tie 1/144 * ($0) = 0/144
P1 (0..1/12), P2 (1/12..1) : Check Bet Fold P2 Win 11/144 * (-$1) = -11/144
P1 (1/12..1/4), P2 (0..1/12) : Check Check P1 Win 2/144 * ($1) = 2/144
P1 (1/12..1/4), P2 (1/12..1) : Check Bet Fold P2 Win 22/144 * (-$1) = -22/144
P1 (1/4..1/2), P2(0..1/2) : Bet Fold P1 Win 18/144 * ($1) = 18/144
P1 (1/4..1/2), P2(1/2..1) : Bet Call P2 Win 18/144 * (-$3) = -54/144
P1 (1/2..1), P2 (0..1/2) : Bet Fold P1 Win 36/144 * ($1) = 36/144
P1 (1/2..1), P2 (1/2..1) : Bet Call Ties 36/144 * ($0) = 0/144

P1's EV = -31/144

That's worse for P1 than (a = 3/7)

[jay_shark]

Mykey , you have my EV wrong .

If player one shoves with 1/4 + , then his

EV= 3*0.5*0.5 - 1 = -0.25

Just plug the numbers into my EV equation .

EV using your strategy is :

EV = 4/7*2*13/31 ~ 0.7074 . If we subtract the $ fee then we get ~ -0.2925.

Again , my strategy is optimal .

[mykey1961]

Your doing it again.

You plug the numbers into a formula, get a number and assume that's the right answer to the question.

Your formula doesn't consider the times player 1 checks, then player 2 bets.

[marv]

[ QUOTE ]
Mykey , you have my EV wrong .

If player one shoves with 1/4 + , then his

EV= 3*0.5*0.5 - 1 = -0.25

Just plug the numbers into my EV equation .

EV using your strategy is :

EV = 4/7*2*13/31 ~ 0.7074 . If we subtract the $ fee then we get ~ -0.2925.

Again , my strategy is optimal .

[/ QUOTE ]



Hmmm.

This is a jam/call/fold problem with ante=1 and stacksize=3

My program doesn't handle the continuous [0,1] game but can do the game with a finite number of cards (like your previous problem). The optimal EV for various numbers of cards in the deck looks like:

ncards EV
10 -0.0888889
50 -0.0847619
100 -0.0841414
150 -0.0838628
200 -0.0837353
250 -0.0836627
300 -0.0836046
350 -0.0835667

So I'm pretty sure your SB strategy isn't optimal.

Marv