#11
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Re: Another One Card Game Theory Problem
"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... |
#12
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Re: Another One Card Game Theory Problem
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 |
#13
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Re: Another One Card Game Theory Problem
[ 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? |
#14
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Re: Another One Card Game Theory Problem
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.... |
#15
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Re: Another One Card Game Theory Problem
"What game?"
I'll let others elaborate |
#16
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Re: Another One Card Game Theory Problem
[ 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. |
#17
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Re: Another One Card Game Theory Problem
[ 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. |
#18
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Re: Another One Card Game Theory Problem
"I'll let others elaborate"
he means losing wads on the ponies. |
#19
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Re: Another One Card Game Theory Problem
[ 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... |
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