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.

[catlover]

Of course the tricky thing about this one is that A can "bluff" by keeping a card when he is beat. Interesting problem.

[catlover]

Let's call the initial cards for A and B "a" and "b", respectively.

A few thought experiments on what player A should do for different values of b:

b=0.25. A will never bluff (i.e, keep a card below b), because that will give a chance of winning below the 50% he can get by drawing. A should keep if a>0.5 and draw otherwise. B will always draw. A's EV on his keeps is 0.75. A's EV on his draws is 0.5. Overall A's EV is 0.625.

As for B, B will always draw with this b.

b=0.5. Same as above, except that if A draws, it now doesn't matter if B draws or not. b=0.5 looks like a boundary value. For higher b, bluffing will come into play.

b=0.75. If a>0.75, A will always keep. Furthermore, A should keep a few cards below 0.75, so that if A keeps, B has the same EV for keeping or drawing. Let's say A keeps if a>x, where x is to be determined.

We have:

EV for B keeping if A keeps = EV for B drawing if A keeps.

The left side is 1-(0.25/(1-x)). The right side is (1-x)/2.

So 1-(0.25/(1-x)) = (1-x)/2.

Multiplying both sides by 2(1-x), we have

2-2x+0.5 = (1-x)^2 = x^2-2x+1
1.5 = x^2 + 1
0.5 = x^2
x = sqrt(0.5), which is roughly 0.707.

I think this problem will turn out to be doable. I am way too tired right now, but I think generalizing the above will lead to a general solution for x in terms of b. We can then use that to find the EV for both players.

[EnderW27]

Regardless of the frequency of A’s bluff, B’s optimal strategy should be to always pick again if he’s below .5 and stand pat above .5 when A picks a card himself.

If A NEVER bluffs, B should repick a card whenever A stands pat regardless of what B is showing.

If A ALWAYS bluffs, B should always stand pat when A stands pat.

So it really comes down to what the optimal frequency of bluffing is.

What I find quite interesting is that the strategy I’ve laid out thus far has very little to do with what you actually hold and quite a bit with how well you know your opponent.

[Greeksquared]

I broke down the game into 6 indpendent outcomes after the first two cards are dealt.

1) A < B < .5
2) B < A < .5
3) A < .5 < B
4) B < .5 < A
5) .5 < B < A
6) .5 < A < B


Now, assuming no third level thinking
for 1) A < B < .5 P[1]= 1/8

A has no reason to bluff (and stay pat) because B will always redraw and will have +EV if A is actually bluffing

EV=0

2) B < A < .5 P[2]=1/8

Again A has no reason to bluff (stay pat) as B will always redraw. So both redraw.

EV= 0
3) A < .5 < B p[3] = 1/4

Here A may be able to bluff successfully to increase EV

If A always redraws then his EV is 100(1-B) because B will never redraw when A redraws.

Now if

A*P[B redraws] > (1-B) then A should bluff. The Probability of B redrawing when ahead could be player dependent.

So for example we have A=.48 B = .6

So if A redraws its EV= (1-.6)*100 = 40
if A bluffs its EV is .48*100*P[B redrawing] . Here P[B redrawing] must be greater than 40/48= .833 for A to bluff.

P[B redrawing] could be estimated with 2*(1-B). For some players possibly sqrt((1-B)*2)

In this example if the B's probability of redrawing is 2*(1-B) then no bluff would occur, but if it was sqrt(2*(1-B)) then a bluff would happen.



4) B < .5 < A p[4] = 1/4

Here A will always stay pat and never redraw, but B will always redraw

5) .5 < B < A p[5]=1/8

Here A will always stay pat and B will sometimes stay and sometimes redraw.

The chance B redraws again might be (1-B)

SO EV for B could be (1-B)*100*(1-A)+ B*0


6) .5 < A < B P[6]=1/8

Again here it could be reasonable for A to bluff for the same reasons stated above in 3).


So if A is reasonably close to a B > .5 then a bluff by can help its expectation.

Basically if,
A*P[B redraws] > (1-B) then A should bluff

[AaronBrown]

I think A has a very small advantage, about 0.5%. It’s true he has more information, knowing B’s card, but he also has to act first.

I think the optimal strategy is if B < 0.433, A should stay if A < 1 – SQRT(1-2B). If B > 0.433, A should stay if A < 0.5.

There are two tricks to this solution. First is to notice this is a different game for every value of B, we do not need to worry about interactions between A’s strategy for one value versus another. Second is to solve for an equilibrium solution, the one where A doesn’t care if B stays or draws, and then prove it’s optimal. This often works for simple game theory problems.

[danq]

I'm pretty sure the equilibrium strategies work out as follows. (Let a and b be player A and B's first cards.)

If b < 1/2, there is no bluffing: player A draws if he's below 1/2, and player B always draws.

If b is between 1/2 and 5/8, then player A still draws if he's below 1/2 and stands when he's above 1/2; but now player B draws if A did not, and does not draw if A did.

Finally, if b is above 5/8, player A draws if a is below sqrt(2b-1), a cutoff which is bigger than 1/2 but smaller than b (so player A sometimes keeps a losing hand); player B does not draw if A did, but if A stood, then B mixes his strategies, drawing with probability (1-b)/sqrt(2b-1).

Calculting each player's EV is some messy double integrals, but a quick simulation suggests player A wins 50.5% of the time, as AaronBrown said.

Dan

Edit: Looks like AB solved for low-card-wins and I solved for high-card-wins, otherwise our strategies are basically identical. I think there's something off about my middle case, but I don't have time to fix it right now.

[AaronBrown]

Whoops, you're right. The problem says "higher number wins." I was misled by the lowball in the introduction.

[wrestler_118]

I think case 3 has sub cases:

if (.5 - a) > (b - .5) then a should never bluff.
If a redraws then a wins 100(1-b). If a bluffs and stands and is sucessful 100% of the time in getting b to redraw then a wins 100a. so bluffing in this situation is always negative EV.

Example: a = .3, b = .6
1) a redraws, wins 100(1-.6) = 40
2) a stands:
if b stands then a wins 0
if b redraws then a win 100(.3) = 30

in this case a must redraw.

If (.5 - a) < (b - .5) then a may bluff with postive EV


Matt

[Vincent Lepore]

David,

After reading the responses to your problem it has become evident, at least to me, that you should retire from doing math. These fellows will outshine you every time. Knowing you I don't think you will be able to live with that. Oh...the younger generation.

Vince