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.

[mike l.]

"I want this answer for the game I play in"

me too. they spread a ripping $300 limit one card single draw ring game at commerce. and the party 6 max is just amazing, ive never seen so many clowns stand pat on a .3127474 and even worse! what a game!

only problem is live when they ask for a set up, those infinite decks seem to take forever to change...

[jdl22]

The ante created pot is a. Player 1's draw is x. Player 2's card is b.

Start post draw:
1 always bets when x is greater than b
1 bets when x<b wp p
2 calls with probability q

to make 2 indifferent need EV(C) = EV(F) = 0
(1-b+bp)EV(C) = 2a(bp) - a(1-b)
setting equal to zero we get p = (1-b)/2b. So player 1 will bet with probability (1-b)/2b if she draws a card lower than b. If b<1/3 than she will always bet no matter the draw. player 2 will always fold.

Now let's find q. 1 must be indifferent between betting and not given a draw below b. In other words we must have
EV(B) = EV(C) = 0
EV(B|q) = a(1-q) - a(q) = a(1-2q) = 0 => q = 1/2.

(ex ante) EV of this round for player 1 is:
(1-b)(1/2)(2a+a) = (3a/2)(1-b)

Look at the first round, given b, 1 chooses to draw or stay pat and bet. 2 must again be made indifferent between calling or folding given 1 stood pat. Again, EV(C) = EV(F) = 0 implies that 1 will stand pat with probability (1-b)/2b.

2 must call with probability q that makes 1 indifferent between standing pat and betting and drawing. We need:
EV(pat) = EV(draw) = (3a/2)(1-b)
EV(pat given card worse than b) = a(1-2q)

Setting these equal you get:
q = (3b-1)/4

So in summary when b>1/3
Player 1 will pat bluff with probability (1-b)/2b if her card is less than b
Player 2 will call with probability (3b-1)/4

if 1 draws:
1 will bet if x > b, if x < b 1 bets with probability (1-b)/2b
2 will call with probability 1/2.

Something to note is that (3b-1)/4 < 1/2 if b < 1 so we need to make sure that 1 could not improve by drawing if her card is better than b. By drawing she gets (3a/2)(1-b), by standing pat she gets 2a(3b-1)/4 + a(5-3b)/4 = 3a(1+b)/4. Setting this equal to (3a/2)(1-b) we find that this is true if and only if b >= 1/3. This means that our equilibrium is verified for the case that b >= 1/3.

If b < 1/3 the most obvious strategy pair is for 1 to always pat and 2 to always fold.

For EV calculations, if b < 1/3, 1 will get a. If b > 1/3 then 1 will get 3a(1+b)/4 when her card is greater than b. If her card is less than b she gets 3a(1-b)/2. So 1's EV given b is:
(1-b)3a(1+b)/4 + b(3a)(1-b)/2 = -3/4a(-1-2b+3b^2

Setting this equal to a/2 and solving for b gives:
b = (1/3)(1+sqrt(2) or about .8047

[jdl22]

[ QUOTE ]

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?)


[/ QUOTE ]
yes it is. He will bet with probability (1-b)/2b if he gets a draw lower than b. Mulitiply that by b (the probability of a draw being below b) and you get (1-b)/2, add that to 1-b and you have exactly the situation you've described.

[ QUOTE ]

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


[/ QUOTE ]

The same probability as post draw, if he draws below b then he will stand pat with probability (1-b)/2b.

[ QUOTE ]

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?)


[/ QUOTE ]

(1/3)(1+2^(1/2)) or about .8047

[ QUOTE ]

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

[/ QUOTE ]

What game?

Just last week on a table next to me I saw these 3 dudes from Jersey drop about $14,000 in an hour trying to play (1-b)3a(1+b)/3 + b(4a)(1-b)/3 = -3/4a(-1-3b+4b^3

Instead of (1-b)3a(1+b)/4 + b(3a)(1-b)/2 = -3/4a(-1-2b+3b^2.

I didn't have anything near the bank roll to get over there but man....

[David Sklansky]

"What game?"

I'll let others elaborate

[AaronBrown]

[ QUOTE ]
A's initial EV = 3(-3b^2+2b-1)/4. . . . 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.

[/ QUOTE ]
If David keeps asking these kinds of questions, he should put a LaTex editor on the boards.

I get +1 instead of -1, 3(-3b^2+2b+1)/4. The integral from 1/3 to 1 of:

1 is 1 - 1/3 = 2/3
b is (1^2 - (1/3)^2)/2 = 4/9
b^2 is (1^3 - (1/3)^3)/3 = 26/81

Substituting those into the unconditional expectation gives:

3[-3(26/81)+2(4/9)+(2/3)]/4
=3[-26+2*4*3+2*9]/(4*27)
=[-26+24+18]/36
=16/36

Adding 12/36 for b < 1/3 gives 28/36 or 7/9 as you say. I don't know where my extra 1/36 came from.

Also, you correctly clarified that by A's expectation, I did not mean to subtract the ante he contributed. It's his expected share of the pot before he bets, not his expectation for playing the game.

[BBB]

[ QUOTE ]

I get +1 instead of -1, 3(-3b^2+2b+1)/4.

[/ QUOTE ]

Oops - I got the same, but I typed it in wrong.

[mike l.]

"I'll let others elaborate"

he means losing wads on the ponies.

[Sadat X]

[ QUOTE ]
"I want this answer for the game I play in"

me too. they spread a ripping $300 limit one card single draw ring game at commerce. and the party 6 max is just amazing, ive never seen so many clowns stand pat on a .3127474 and even worse! what a game!

only problem is live when they ask for a set up, those infinite decks seem to take forever to change...

[/ QUOTE ]

f'ing hilarious...