My solution
[ QUOTE ]
Two players, A and B receive one card with a real number between zero and one. High card wins. Player A bets one dollar in the blind. Player B can fold, call, or raise one dollar. If he raises, A can call or fold only. End of game.
What is the optimal strategy for each player and the value of the game to B?
[/ QUOTE ]
B clearly folds anything less than .5, and calls with a .5. The tough question is when is raising profitable? B will raise with #'s > X, where X > .5. A will then call with Y, where Y >X. If Y calls with # < X, he is gaurenteed to lose.
Let's say X = .75 (i like the ol' guess and check method). What would the optimal value for Y be?
Y is getting 3:1 pot odds on the call, so he has to win 1/4th of the time. Calling with .76 would obviously not win 1/4th of the time, and calling with .875 would win half the time. I'm assuming the average of .75 and .875 is what he calls with -> .8125.
So what's the EV of this strategy for B?
1/2 the time, he folds. EV = 0
1/4th of the time, he calls. When he calls, he wins 3/4 of the time. EV = $.75
1/4th of the time he raises, we have to split this up.
When he raises, A calls 18.75% of the time.
25x.1875 = 4.6875% of the time, B raises, and A calls.
25-4.6875=20.3125% of the time, B raises, and A folds. EV = $1
When A calls the raise, he wins 1/4th of the time. So in 4 tries B wins $2, wins $2, wins $2, and loses $2. He wins $1.50 on avg when he raises, and A calls.
50% of the time ev = 0.
25% of the time ev = $.75
20.3125% of the time ev = $1
4.6875% of the time, ev = $1.50
[(.5)(0)+(25)(.75)+(20.3125)(1)+(4.6875)(1.50)] / 100 = .4609375
I think this is player B's ev with this strategy, + $.4609375 per game.
So did I calculate the EV correctly? If I did, is this the highest EV attainable? I prolly screwed up somewhere. I'm still in highschool, so I haven't taken any game theory classes or anything.
|