View Full Version : Analysis of one-card poker

03-04-2003, 04:27 AM
I created a simple one-card poker game and was surprised at the optimal strategy. Here is the game: Two players, they each ante $1. Player A (the dealer) deals a single card to each, and then he is allowed to CHECK, or BET exactly $2. If he checks, they compare cards and the higher card wins the antes. If he BETs $2, Player B is allowed to CALL $2 or FOLD. (If player B calls, the higher card wins the pot; if he folds player A wins the antes).
The deal then shifts to the other player, who now becomes "Player A". [Note that Player A is always the only aggessor, Player B can only play passively.]

Ok, being a game theory nut, when I first created this game I assumed the following: That since Player A is making a bet of exactly the pot size, player B will have to call at least 50% of the time, or player A will show an automatic profit by bluffing. Doesn't that sound correct? But it is in fact wrong, because the optimal strategy for player B (as far as game theory is concerned) is to call with his best 4/9, or roughly 44%, of hands.

The optimal strategy for player A was also unexpected. I expected he should bet at least his best 25% of hands. But actually his optimal strategy is to bet his best 2/9 (approx 22%) of hands, and his worst 1/9 (approx 11%) of hands (as a bluff).

Game theory fans, any comments? I derived these optimal solutions mathematically and tested them with a C++ simulation.

03-05-2003, 03:50 PM
Very Interesting. Before I comment on the weirdness of the solutions, let me first verify your solutions. I have to do some traveling, so I might not get to work on the one-card poker game until next week. In the mean time, let me ask some clarifying questions:

1) Are your players drawing 1-card each from a 52 card deck or are you just randomly assigning them a number between 0 and 1.

2) Have you read Jerrod Ankenman's recent posts on rec.gambling.poker?

Cheers, Irchans

03-06-2003, 12:00 PM
1) You're right, the solution I gave assumes they are assigned a random real number between 0 and 1, and the lower number wins.

Like I said, I came up with the solution mathematically and then tested it with a high speed simulation. The math got kind of annoying but it was stuff like this:

Given x is the percentage of legitimate hands player A should bet, and y is the percentage of hands player B should call with. Assuming y>x and y<(1-x/2), A's profit is

P = x[(1-y)($1) + y[(x/2y)(-$1)+(1-(x/2y))($2)]] + x/2[(1-y)($1)+y($-1)] + (1-(3x/2))[(x/2)($1)+(1-(3x/2))($.50)]

x is the legitimate hands A bets
(1-y)($1) is the times B folds those and A wins $1
y[(x/2y)(-$1)+(1-(x/2y))($2)] is the times B calls, and they show down
x/2 are the times A bluffs
(1-y)($1) are the times B folds when A bluffs
y($-1) are the times B calls As bluff
(1-(3x/2)) are the hands A checks
(x/2)($1) are the times B has a hand inferior to any hand A would check
(1-(3x/2))($.50) are the times A and B have hands both in the range [x, (1-(x/2))]

simplifies to:

P = x/2 - (9/8)x^2 + .50

Which reaches a maximum at x=2/9, P = 5/9

2) I haven't read Jerrod Ankenman's recent posts but I will go check that out... thanks.

03-07-2003, 02:01 PM
oops, I forgot to say that the above computations make a strange assumption -- that the antes are not paid by the players, they are in fact paid by some outside "benefactor". So, if player A justs checks every time, they both will earn 50 cents per play.

03-08-2003, 12:54 PM
I got the same strategy: Player 2 calls 4/9, player 1 bets with the best 2/9 and bluffs 1/9. I think that Player 2 will call 50% if player 2 is allowed to bet after player 1 checks. (Assuming no raising allowed by either side.)


03-08-2003, 01:11 PM
Steve, I notice that the game you mention has the exact same solution as Jerry's Game 7a.

[0,1] game part 7 (http://groups.google.com/groups?q=)

03-08-2003, 08:58 PM
thanks for verifying my work and for the tip about Jerrods posts on rgp. Jerrod and I seem to have worked on the exact same idea. Guess it goes to shows, "And problem you think is unsolved, probably isn't -- the key is finding the person who has solved it."

03-09-2003, 12:48 PM
Those readers that are unfamiliar with the history of game theory might be interested to know that many of these "[0,1] type" poker games were solved in the 1940's and 1950's. For Example, John Von Neuman and Oskar Morgenstern solve a [0,1] game with raising in Chapter 19 of "Theory of Games and Economic Behavior" (1944). /forums/images/icons/club.gif