Another One Card Game Theory Problem

[David Sklansky]

Both players ante the same amount and get a card from zero to one. Player B's card is again face up but this time he cannot draw. Player A can, if he chooses, draw once. After which he can choose to bet the pot.

Obviously if he stands pat, he will always bet. So the questions are.

1. How often should he bet when he draws? (Is it simply those times he wins plus a percentage equal to half of the chance he will outdraw?)

2. How often should he stand pat when he doesn't have B beaten?

3. How large a card would we have to give B to give him an edge. (If there was no draw it would be .67. See why?)

PS. I want this answer for the game I play in as well as for its theoretical interest.

[AaronBrown]

Let a be A's card and b be B's card. B needs at least 1/3 chance of winning to call A's bet of the pot.

If b < 1/3, then A should always bet and B should always fold.

If b > 1/3, the A should bet if a > (3b-1)/2. Actually, it doesn't matter what numbers A bets on if a < b, only how often he bets. So we could just as well say A should bet if a > b or a < (1-b)/2. If A doesn't bet, he should draw, then follow the same betting rule again.

A's expected value for given b is 3(1-b)(1+3b)/4, or 1 if b < 1/3, times the size of the pot. A's unconditional expected value (before she looks at B's card) is 29/36. B's breakeven card is 0.804738.

My original understanding is that the pot bet wasn't optional to call.

Clearly, A's strategy depends on B's holding.

If A has redrawn, he'll want to raise with any holding stronger than B's and with 1/2 that many holdings which are weaker, and will lose the rest.

That means that the value of of a redrawn hand against B holding b and a pot of p will be:
(1-b)(3/2)(p)
Which is obviously better than zero.
Ergo, for b larger than 1/3, A will want to raise with any holding larger than b's and with any holding less than (1-b)/2.

If B's holding is equal to, or worse than 1/3, B will fold to any raise, so it doesn't matter whether A redraws or not.

So, for B<1/3, the value of the game to A is the pot.
For B>1/3, the value of the game to A is (2-((1-B)(3/2)))(1-B)(3/2)(pot)

[bobbyi]

I don't understand all the rules. If player A bets, player B's options are fold or call? If player A checks, player B's only option is to check? Is that right?

[ QUOTE ]
I don't understand all the rules. If player A bets, player B's options are fold or call? If player A checks, player B's only option is to check? Is that right?

[/ QUOTE ]

That's what my impression was.

[David Sklansky]

The bet doesn't come until after A chooses whether to replace. So how could B ever bet?

[Justin A]

I'm confused. Does player A choose to bet before or after he sees the card he's drawn?

[BBB]

[ QUOTE ]
Let a be A's card and b be B's card. B needs at least 1/3 chance of winning to call A's bet of the pot.

If b < 1/3, then A should always bet and B should always fold.

If b > 1/3, the A should bet if a > (3b-1)/2. Actually, it doesn't matter what numbers A bets on if a < b, only how often he bets. So we could just as well say A should bet if a > b or a < (1-b)/2. If A doesn't bet, he should draw, then follow the same betting rule again.

A's expected value for given b is 3(1-b)(1+3b)/4, or 1 if b < 1/3, times the size of the pot. A's unconditional expected value (before she looks at B's card) is 29/36. B's breakeven card is 0.804738.

[/ QUOTE ]

I solved this analytically, and got all of the same answers, except that I got A's unconditional EV to be 7/9 (in units of initial pot size). It should be noted that this value does not include player A's initial contribution to the ante, His total EV for the game, in units of antes, would be 2*[3/4*(-3b^2+2b+1)]-1, for a net unconditional EV of 5/9 (if my 7/9 number is correct)).

From B's perspective, he should fold to A's bet whenever his card b is less than 1/3. Otherwise, if A bets without drawing, he should call with frequency 1/2 - 3(1-b)/4. If A draws, then bets, B should call with frequency 1/2.

[BBB]

A summary of the solution:

Let a = A's final card
Let b = B's card
Let x = A's bluffing frequency after drawing
Let z = B's calling frequency after A draws and bets
Let y = A's pat-bluffing frequency
Let K = B's calling frequency if A stands pat and bets

Obviously, A will always bet without drawing if a > b, and will always bet after drawing if a > b. And of course A will never check without drawing if a < b. What remains to be determined is how often A should bluff, both before and after drawing, when a < b.

First, assume A draws.

B should call with a frequency such that if A misses, his EV are equal for bluffing and for checking.

A's bluffing EV: (1-z)-z
A's checking EV: 0 (since he missed)
Setting these equal gives z = 1/2

A must bluff with frequency such that B is indifferent between calling and folding.

B's EV of calling, given that A bets:
2*bx/((1-b)+bx) - (1-b)/((1-b)+bx)
B's EV of folding: 0
Setting these to be equal gives: x = (1-b)/(2b)

(Note that x > 1 for b < 1/3. So for b < 1/3, A should always bet, and B should always fold.)

Given that A draws, A's EV is:
2(1-b)z + (1-b)(1-z) + bx(1-z) - bxz
Substituting in for x and z, I get:
A's EV if he chooses to draw = 3(1-b)/2

Now consider the situation before A decides whether to draw or pat-bluff:

If A bets without drawing, B should call with a frequency such that A's EV for bluffing and drawing are equal.
A's bluff EV: (1-k)-k
A's draw EV (calculated above): 3(1-b)/2
Setting these equal gives: k = 1/2 - 3(1-b)/4

A must bluff with frequency such that B is indifferent between calling and folding.

B's EV of calling, given that A bets:
2*by/((1-b)+by) - (1-b)/((1-b)+by)
B's EV of folding: 0
Setting these to be equal gives: y = (1-b)/(2b)

(As was the case for after the draw, here y > 1 for b < 1/3. So for b < 1/3, A should always bet, and B should always fold. Technically, I suppose A should always draw if a < b < 1/3, in case B decides to call, and then always bet the river. But if B is playing optimally, he will always fold to any bet if b < 1/3, whether or not A draws.)

A's EV prior to drawing is:
2(1-b)k + (1-b)(1-k) + by(1-2k) + b(1-y)*3(1-b)/2
Subbing in for k and y,
A's initial EV = 3(-3b^2+2b-1)/4

Setting this equal to zero and solving for b gives the break-even value for b as (1+sqrt(2))/3, or 0.804738.

Integrating this expression from 1/3 to 1, and adding 1/3 (from integrating 1 from 0 to 1/3) gives the EV if b is taken from a uniform distribution on [0,1]; I get 7/9 for the answer.