Game Theory Problem Of The Week

[jay_shark]

Tnixon , this is a mute point but I'll say it again .

If one player picks a card , then there are 99 possible choices for the other player .

There is no denying this .

[TNixon]

[ QUOTE ]
If one player picks a card , then there are 99 possible choices for the other player .

[/ QUOTE ]
Yes, I realize this.

That just means I have no clue why a brute-force simulator would come up with a different result than your math, though. They should be identical, since he did disallow combinations where both players had the same card.

[img]/images/graemlins/smile.gif[/img]

[jay_shark]

My answer was like 0.1122222222 and his answer is 0.111111 ?

I also get 0.111111 as my answer when we're not working in the discrete case ;ie , we select a number from [0,1]but I get 0.1122222 when we're working in this particular example .

Either way , I'm certain I did everything right .

[mykey1961]

In a [0,1] game

P2's Optimal Strategy

[0/1,1/9] Bet
[1/9,7/9] Check
[7/9,1/1] Bet

P1's Optimal Strategy
[0/1,5/9] Fold
[5/9,1/1] Call

P1's EV = -1/9
P2's EV = +1/9

[TNixon]

All questions of .111_ vs .113_ aside...

Are we now in agreement that P2's optimal strategy is betting 1-11, checking 12-78, and betting 79+?

[jay_shark]

Mykey , I already proved that this particular case is optimal for the non-discrete case which is the same as in the discrete case . I will supply a proof shortly .

1/9 is the maximum EV that player two can earn .

[mykey1961]

[ QUOTE ]
My answer was like 0.1122222222 and his answer is 0.111111 ?

I also get 0.111111 as my answer when we're not working in the discrete case ;ie , we select a number from [0,1]but I get 0.1122222 when we're working in this particular example .

Either way , I'm certain I did everything right .

[/ QUOTE ]

What is the strategy for P1 and P2 that is giving you 0.112222222?

[TNixon]

[ QUOTE ]
1/9 is the maximum EV that player two can earn .

[/ QUOTE ]

1/9 happens to be .111_, not .112_ or .113_ or any other sort of .112somethingorother that you've posted throughout this thread.

For what it's worth, P1 can pick a calling range anywhere from 12 through 79, and P2's EV will be the same.

Now I'm confused about why there's still any discussion, argument, or futher necessary proffs going on though, since mykey's strategy does appear to be optimal.

[jay_shark]

[ QUOTE ]
[ QUOTE ]
My answer was like 0.1122222222 and his answer is 0.111111 ?

I also get 0.111111 as my answer when we're not working in the discrete case ;ie , we select a number from [0,1]but I get 0.1122222 when we're working in this particular example .

Either way , I'm certain I did everything right .

[/ QUOTE ]

What is the strategy for P1 and P2 that is giving you 0.112222222?

[/ QUOTE ]

Mykey , my strategy is the same as yours . My answer is nothing but a round-off error .

I gave a very close approximation that when player one calls player two , that player two will win 1/3 of the time . This is not exactly true but it's very close which is why i'm off by a slight margin .

This is why I prefer not working in the discrete case .

[mykey1961]

[ QUOTE ]

First of all it's 68+ (not 67+) that you should compare with [1,11] [79,100] .

EV for [1,11] and [79,100] given that player two checks [12,78] since this will not change the outcome .

1*33/100*78/99 + 3*33/100*22/99*1/3 -3*33/100*22/99*2/3
+1*66/100*11/99 - 1*66/100*22/99 = 0.11333333 (This is the exact EV without replacement.

You will win a $ when you bet (33/100) and your opponent folds (78/99). You will win $3 when you bet (33/100) , your opponent calls(22/99) ,one-third of the time . You will lose $3 when you bet(33/100), your opponent calls (22/99) , two-thirds of the time . You will win a $ when you check (66/100)and your opponent checks a worse hand (11/99) . You will lose a $ when you check (66/100) , your opponent has a better hand (22/99).

Here is the calculation for the second part worked out analogously .

EV for [68,100] and player one check calls with [79,100] is :

1*33/100*78/99 + 3*33/100*22/99*1/3 -3*33/100*22/99*2/3
-67/100*33/99 = -0.036666666

[/ QUOTE ]

1*33/100*78/99
+ 3*33/100*22/99*1/3
- 3*33/100*22/99*2/3
+ 1*66/100*11/99
- 1*66/100*22/99

= 0.11333333

1*33/100*78/99
+ 3*33/100*22/99*1/3
-3*33/100*22/99*2/3
-67/100*33/99

= -0.036666666


There is something seriously missing from your calculations.

You are only accounting for about 5/9th of the possibilities.

33/100*78/99+33/100*22/99*1/3+33/100*22/99*2/3+66/100*11/99+66/100*22/99 = 249/450 = 0.553333

33/100*78/99+33/100*22/99*1/3+33/100*22/99*2/3+67/100*33/99 = 167/300 = 0.556666


Both of these should = 1 if you are calculating for the whole strategy