Game Theory Problem Of The Week

[Perestroika]

Oh sorry i thought 0-50 was better than 51-100. And I'm way off with my guess anyway. But should he not bet more hands than 33%?

[jay_shark]

Whatever hand player two bets with means that the lowest number in player 1's calling range must be able to beat one third of player 2's range .

ie 2 bets with [0.5,1] then the best strategy for player 1 is to check call with [0.666666,1]; 0.66666 beats one-third of the numbers from [0.5,0.66666]and loses to (0.66666,1]

[mykey1961]

[ QUOTE ]
I finally spotted another error that we were all making .

In the last example , if player 1 bets with any number , then player 2 must call with any number !

ie , Hero bets everything from 0-1 . We mistakenly suggested that villain should call with 1/3-1 . This is blatantly wrong !!

If villain hadn't posted the ante , then this assumption is certainly valid . However , since he posted an ante , calling with any number is better than foregoing the loss of that ante .

To make this question and the previous question more interesting , we should add the stipulation that only Hero posts the ante and that he acts first .

I can't believe , nobody picked up on this .

[/ QUOTE ]

I think the reason you have such problems is you don't look at a fold as -EV. It's EV=0 for that decision, but that choice makes the overall EV negative.

In a hypothetical game where we both ante $1, you are dealt in the range of [1,50] and I'm dealt in the range of [51,100] When I bet $2, you should always fold.

You consider folding EV = $0.

Your strategy has an EV of -$1

[mykey1961]

[ QUOTE ]
For this week's game theory problem we will take a look at another situation .

There are two players who pick numbers from 1-100 without replacement . Each player posts a $1 ante but player one must always check even though he's first to act . Player two has the option of betting the pot or checking behind . Given this knowledge , what strategy must player two employ to maximize his EV ?

We may make the assumption that player one and two are playing optimally aside from the stipulation placed on player one .

[/ QUOTE ]


Optimal is:

P1
[1,56] Fold
[57,100] Call

P2
[1,11] Bet
[12,78] Fold
[79,100] Bet

P1 EV = -1/9

P1 can't improve by making any changes against P2
P2 can't improve by making any changes against P1

[jay_shark]

Here is the EV set up using the cash game definition of EV .

Player two bets with [x,1] and player one check calls with [(2x+1)/3,1] . Notice that the answer to this will be the same as in the discrete case .

EV(P2)= 1*(1-x)*(2x+1)/3 + 3*(1-x)*(2-2x)/3*1/3 -3*(1-x)*(2-2x)/3*2/3 - (1-x)/3

There are 4 different product terms :

The first is your EV|player 1 folds .
The second is your EV|player 1 calls and you win
The third is your EV|player 1 calls and you lose
The fourth is your EV|player 2 checks

After simplifying of the EV formula you should get

EV(P2)=(-4x^2+6x-2)/3

EV' = -8x/3 +2
So x=3/4 .

This means that player two should bet with 75-100 and player one should call with 84-100 . This is better than my previous attempt .

[mykey1961]

Ok here is what I just don't seem to understand.

You apply a formula, get an answer. But do you do anything else to verify it actually answers the question?

Your P1
[1,83] fold
[84,100] call

Your P2
[1,74] Check
[75,100] Bet

My P1
[1,56] Fold
[57,100] Call

My P2
[1,11] Bet
[12,78] Check
[79,100] Bet


My P1 vs Your P2
EV = -0.094545

Your P1 vs My P2
EV = -0.118182


My Strategy wins 0.011815 Ante's per hand from yours while rotating positions.



Maximal against your P1 "P1MO"
[1,65] Bet
[66,92] Check
[93,100] Bet

Maximal against your P2 "P2MO"
[1,83] Fold
[84,100] Call

Your P1 vs Your P2
EV = +0.023636

Your P1 vs P1MO
EV = -0.447879

P2MO vs Your P2
EV = +0.023636

Maximal would win 0.2357575 Ante's per hand from your strategies while rotating positions

[mykey1961]

[ QUOTE ]
Here is the EV set up using the cash game definition of EV .

Player two bets with [x,1] and player one check calls with [(2x+1)/3,1] . Notice that the answer to this will be the same as in the discrete case .


[/ QUOTE ]

Why do you assume P2 bets with [x,1]?

why not [0,x] and [1-2x,1]?

For this problem you if you have P2 bet only with 1, 99, and 100:
Player 1 would optimally fold 1, call with 99, 100 and be indifferent to calling or folding with 2 thru 98.

Since calling and folding would both produce the same EV for P1, folding means losing the ante, then P2's EV when betting with a 1 is (-3*2+1*97)/99 = 0.919191

That is clearly better than if P2 checked with a 1.


[ QUOTE ]

EV(P2)= 1*(1-x)*(2x+1)/3 +
3*(1-x)*(2-2x)/3*1/3
-3*(1-x)*(2-2x)/3*2/3
-(1-x)/3

There are 4 different product terms :

The first is your EV|player 1 folds .
The second is your EV|player 1 calls and you win
The third is your EV|player 1 calls and you lose
The fourth is your EV|player 2 checks

After simplifying of the EV formula you should get

EV(P2)=(-4x^2+6x-2)/3

EV' = -8x/3 +2
So x=3/4 .

This means that player two should bet with 75-100 and player one should call with 84-100 . This is better than my previous attempt .

[/ QUOTE ]

Again, you get an answer, but not the correct answer. It's correct for your formula, but your formula isn't correct for the question.

[Perestroika]

mykey1961 could you explain where you get the numbers (-3 and 2+1 in your equation below. thank you

(-3*2+1*97)/99 = 0.919191

[mykey1961]

[ QUOTE ]
mykey1961 could you explain where you get the numbers (-3 and 2+1 in your equation below. thank you

(-3*2+1*97)/99 = 0.919191

[/ QUOTE ]

Assume P2 has a value of 1:

If P1 has a value of 99, or 100, P1 calls and beats P2 for a 3 ante profit. That's a 3 ante loss for P2. 99 and 100 are 2 numbers

If P1 has a value of 2 thru 98, P1 has the same EV if it calls or folds. If it folds, it's loss is 1 ante, and that's a 1 ante win for P2. 2 thru 98 represents 97 numbers.

When P2 has a 1, there are 99 possible numbers for P1

P2's expected profit is (-3 antes * 2 numbers + 1 ante * 97 numbers) / 99 numbers = 0.919191 ante per number

This calculation is just for the case when P2 has a 1.

[jay_shark]

Why not avoid betting 1-11 and bet with 67+ instead ?

I'm not sure the significance in why you're betting with 1-11 here .