Another One Card Game Theory Problem

[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...