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.

[daryn]

all i know so far is B is losing

interesting how bluffing factors into this game. you can force your opponent into making a -EV decision by making him throw away a winner and losing most of the time with his new draw

of course he will then catch on to that and start "calling" bluffs. are we talking about only ONE game here or do they keep playing and remembering past actions?

[Alex/Mugaaz]

I've been trying to solve this a while now, this is a lot harder than I thought.

[Alex/Mugaaz]

[ QUOTE ]
I've been trying to solve this a while now, this is a lot harder than I thought.

[/ QUOTE ]

I give up, my math knowledge is too limited.

I came to the conclusion that the ideal strategy for A is to stand pat on all hands where 1.0 - B < A. If this is true A is a favorite as long as B is less than somwhere between 0.63 to 0.66.

[DougShrapnel]

[ QUOTE ]
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.

[/ QUOTE ]Can i guess? A wins someplace between $66 and $75 a hand. I'll take the midpoint as a in the dark guess. 70.83cents a hand.

[hmkpoker]

The very basic strategy should be for Player A should hit if his hand is less than Player B's, and stand if his hand is greater. Every hit would then give his hand an EV of .5, and Player B should then hit if his hand is less than .5, and stand if greater.

If player A stands, however, this signifies to player B that player A's hand is greater, and that player B should now hit. To have an edge for this move as a bluff, player A must have a hand greater than .50, because it is very likely to induce Player B to hit. Player B is also aware of the potential of this bluff, and should only hit on Player A's stands when Player B's are slightly above .50...let's say .50 to .75 or so. Anything greater than .75 yeilds a small expected return, especially when considering how likely it is that A is bluffing.

The EV of both strategies then becomes more difficult to calculate because it depends on psychology and the players.

[BluffTHIS!]

Not to be a nitpicker David, but is this a repetitive game or one time/occasional only? It matters.

[BB King's]

You have to find the optimal solution.

It doesn't matter wheter it is a repetitive game or a one time/occasional only.

BB King’s ... formerly known as the pokerplayer formerly known as Jack

[BB King's]

If (b<=0.5 and a<=0.5) then A=Draw and B=Draw
If (b<=0.5 and 0.5<=a) then A=Stand and B=Draw

If (0.5<=b and x<=a) then A=Stand and ‘B=Y*Draw+(1-Y)*Stand’
If (0.5<=b and a<=x) then A=Draw and B=Stand


x=sqrt(2*b-1) Y=(1-b)/x


ex.
Given: b=0.82 and a=0.81
Calculating: x=0.8 Y=0.225
A Stands Pat and B is Drawing 22.5% of the time.


EV-Calculation is left as an exercise to the reader.


BB King’s ... Formerly known as the pokerplayer formerly known as Jack

[BluffTHIS!]

[ QUOTE ]
You have to find the optimal solution.

It doesn't matter wheter it is a repetitive game or a one time/occasional only.

BB King’s ... formerly known as the pokerplayer formerly known as Jack

[/ QUOTE ]

It does matter as to whether you should always just use the mathematically optimal solution if there is one or a mixed strategy with opponents who have previously seen how you play. This is similar to knowing in triple draw lowball the frequency with which your opponent will attempt to freeze you on a worse hand or stand pat on a bluff on the 2nd draw and bet out on the end.