Classic Type Game Theory Problem

[David Sklansky]

I just thought of this problem recently when a player mistakingly exposed his pat lowball hand too soon in a triple draw game against an all in opponent.

I realized that the situation rephrased in rigorous terms is a simply stated classic type game theory problem, perhaps never addressed before. Just in case that's so, we'll call it the Sklansky Exposed Pat Hand Problem. It goes like this:

Player A and Player B are both dealt a real number from zero to one. Higher number wins. No betting except for antes. Player A looks at his downcard and decides whether to keep it or replace it. If he replaces it he gets that second card face down. After Player A acts, Player B has the same option. And of course his decison will be based partially on what A did. But the thing about this game is that Player B's first card is face up. So B knows that A's decision to replace was based on what A saw.

If there is $100 in the pot and both players play perfectly what is the EV for both players? What is the optimum strategy?

I'm going to put this question in the Poker Theory, Probability, and Science Math and Philosophy Forums at the same time.

[Shandrax]

Let's replace a number > 0.50 with 1 and a number < 0.50 with 0. The reason to do this is that if you redraw with a number higher than 0.50, it is more likely that you will make it worse.

Under this assumption and with "optimal" play for both sides I got an EV of 0!

If player A has 1 he stands (EV: +12.5%) and if he has 0 he redraws (EV: -12.5%). That play has a better expectation than trying to stand/bluff with 0 (overall EV: -25%).

Player B always stands with 1 and always redraws with 0.

[daryn]

what if A gets .457890435087582038946078?

[Shandrax]

I have edited it in the hope that it makes sense now.

Hmmm... what is Player B's move if Player B has 0.85 showing and Player A decides to keep his card?

[Shandrax]

Oh lol, I overlooked that B's number is showing. Geez...especially since it's called "exposed" hand problem.

Let's say that A's inital holding is a and B's initial holding is b.
Then A redraws if:
1. b<.5 and a<.5
2. b>.5 and a<1-sqrt(2*(1-b))

B redraws iff b<.5.

Now assuming the initial pot is 1, the value of the game should be

Player A has the best information in this game, and therefore will have an overall advantage. B only has the information of his own card, and what A does , but A could be bluffing on any given act. a1 = player A's initial card, a2 = replacement if taken, same for b1 and b2.

Based on B's only reliable information (his own card), he will replace with b1 < .5 (since he is most likely to improve), and stand with b1 > .5 (since a replacement will most likely result with b2 < b1)

If a1 < .5, A must replace.
If .5 < a1 < b1, A must replace since he has no chance of winning otherwise (since B will not replace).

Even though A will have an uphill battle drawing against a b1 > .5, at least he has the fortune of knowing whether or not he MUST draw. B in turn will draw sometimes unecessarily and give up a win. I wil have to think harder for winning percentages.

[Shandrax]

Well, of course player A has the advantage. Player B has to decide below which number he will always draw and above which number he will always stand. Let's say just as an example, he always draws < 0.25 and always stands > 0.75 and inbetween it depends on what A does.

A on the other side has to bluff a certain amount of times. Overall it's this all about the Nash equilibrum and very complicated. No idea how you calculate this.

[Gabe]

If I was forced to play this game right now without thinking about it:

If I was A, and B was <.50 I would draw if I was <.50. If B was <.75 I would stand on anything >.50.

If I was B, I would stand on anything >.75.

However, after I give it some thought the over/under maybe somewhere between .66 and .75.

This strategy might be totally out to lunch. Hopefully, I'll have time to give this some thought.