0 to 1 Game -- Problem 2
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August 13th, 2012 at 2:28:54 PM
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Here is your next problem. The same as problem 1, except of X checks then Y can bet. Here are the full rules.
What is the optimal strategy for both players? What is the expected value for X under mutual optimal strategy?
If sufficient interest is show in this exercise, then problem 3 will be the same, but allow betting up to a pot size of $6.
- Player X and Y each ante $1.
- Both are given a random number uniformly distributed from 0 to 1. The higher number wins.
- Player X may bet $1 or check.
- If player X checks then Y may check or bet.
- If player X bets then Y may call or fold.
- If player X checks, then Y bets, then X may call or fold.
What is the optimal strategy for both players? What is the expected value for X under mutual optimal strategy?
If sufficient interest is show in this exercise, then problem 3 will be the same, but allow betting up to a pot size of $6.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 13th, 2012 at 3:12:59 PM
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my initial guess (probably wrong)
X has the same initial strategy, raises with 0.1 or less, raises with 0.7 or more.
Y calls a raise with anywhere from 0.1 to 0.7 or higher (doesn't matter EV) wise.
If X checks, Y raises with 0.16 or less, raises with 0.52 or more.
X calls with anything from 0.16 to 0.52 or higher.
Expected value for X = 0.068
X has the same initial strategy, raises with 0.1 or less, raises with 0.7 or more.
Y calls a raise with anywhere from 0.1 to 0.7 or higher (doesn't matter EV) wise.
If X checks, Y raises with 0.16 or less, raises with 0.52 or more.
X calls with anything from 0.16 to 0.52 or higher.
Expected value for X = 0.068
August 13th, 2012 at 3:23:31 PM
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Wouldn't it make sense that X should call less often than problem 1? In problem 2 Y has more power if X checks.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 13th, 2012 at 3:29:02 PM
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Yes, of course. But this is the only answer I could do in 2 minutes :)
August 13th, 2012 at 3:35:00 PM
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Left side: Player X. Right side: Player Y.
dots are probability to bet, circles are probabiltiy to call.
The scattered data is (as before) almost undecidable - i.e. it oszillates in the simulation. Probably doesn't matter what you pick there.
[Edit^2: As X never bets between ~0.18 and 0.5, Y' call results between ~0.18 and 0.5 does scatter heavily. Possibly Wizards "doesn't matter" decision point. Also valid for the other side (X's call results between ~0.18 and 0.5)
I leave the analytical analysis to the experts around here.
Edit: Summary of the results (in case the graphic won't load):
X should never bet between ~0.18 and 0.5, and never call below ~0.18.
X should call above 0.5.
Y should never bet between ~0.18 and 0.5, and never call below ~0.18
Y should bluff below 0.18, and bet/call anything above 0.5
Edit^2 (see spoiler)
August 13th, 2012 at 3:36:06 PM
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Mango, I couldn't see the graphic.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 13th, 2012 at 3:41:34 PM
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Could you try again ? On my machine it works: http://img401.imageshack.us/img401/1475/strategies2.png
I hope this is not a cookie-related thing, I'm not very familiar with imageshack.
I hope this is not a cookie-related thing, I'm not very familiar with imageshack.
August 14th, 2012 at 1:08:30 PM
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I get that x should bluff on 19/112 or less and bet on 55/112 or higher. This is very close to MangoJ.
If x bets, then y should call on 40/112 or higher.
If x checks, then y should bluff on 15/112 or less and bet on 19/112 or higher.
If x checks, then y bets, x should fold on 31/112 or less, otherwise call.
If x bets, then y should call on 40/112 or higher.
If x checks, then y should bluff on 15/112 or less and bet on 19/112 or higher.
If x checks, then y bets, x should fold on 31/112 or less, otherwise call.
I heart Crystal Math.
August 14th, 2012 at 1:52:58 PM
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I still can't see Mango's graphic. It may be a similar situation to problem 1 where multiple answers result in the same expected values. May I ask what you guys get? Here is my EV...
EV(player X) = -1/18
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 14th, 2012 at 2:30:34 PM
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Quote: WizardI still can't see Mango's graphic.
Maybe try one of these links. Please let me know if the server / 404 or the PNG format is the problem.
http://s7.postimage.org/5f62aa9gr/strategies2.png
http://s14.postimage.org/c9q3lkmr5/strategies2.gif
http://s9.postimage.org/pz89t184v/strategies2.jpg
-0.055 for Player X
August 14th, 2012 at 4:07:13 PM
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EV for x is -289/6272
I heart Crystal Math.
August 14th, 2012 at 4:11:06 PM
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I get to the wizard's answer but haven't yet "proved" it mathematically.
(1) Firstly if X always checks then Y has the same advantage as problem 1 : E(X) = -0.1
(2) Secondly if X always bets then Y will call with .25+ (as stands to lose $1 vs gain $3) E(X)=-0.125
(2 a) Hence bad strategy for X to "always" bet.
(3) From above it can be seen Y should always call any bet with expected chance of winning is 25% or more
(3 a) Y has an advantage to maximise range that X doesn't bet (since he then has an advantage)
(3 a) From problem 1 Y should remain prepared to call with .4 (otherwise this gives X an advantage)
(3 b) So X should adopt the same strategy (otherwise this gives Y an advantage)
(3 c) So X should ensure the range of betting (e.g. 1 to .7) being 3x range for bluffing (e.g. 0 to .1)
(3 d) However probably worth taking a small loss to prevent Y for gaining advantage.
(4) Now Y should always bet 0-(bluff X) and (Bet X)-1 as X would be indifferent to calling
(4 a) So Y and X now fall between (bluff X) and (bet X) with problem 1
(4 b) So Y bets upto 1/6 and anything over .5 with X calling 1/3
Problem number 1
X Range Y Range Y Pr (event) Win to X 0.10
0.00 0.10 0.00 0.40 0.04 1 0.04 X Bluffs, Y folds
0.00 0.10 0.40 1.00 0.06 -2 -0.12 X Bluffs, Y calls and wins
0.10 0.70 0.00 0.10 0.06 1 0.06 X checks and always beats Y
0.10 0.70 0.10 0.70 0.36 0 0.00 X checks, and 50:50 whether wins or loses
0.10 0.70 0.70 1.00 0.18 -1 -0.18 X checks, and will always lose
0.70 1.00 0.00 0.40 0.12 1 0.12 X Bets, Y folds
0.70 1.00 0.40 0.70 0.09 2 0.18 X Bets, Y calls but loses
0.70 1.00 0.70 1.00 0.09 0 0.00 X Bets, Y calls, and 50:50 whether wins or loses
X Bluffs .111 111
X Bets .666 667
Y Calls .400 000
Y Bluffs .166 667
Y Bets .500 000
X Calls .333 333
Problem Number 2
X Range Y Range Y Pr (event) Win to X -.055 556
0.00 0.11 0.00 0.40 0.044444444 1 .044 444 X Bluffs, Y folds
0.00 0.11 0.40 1.00 0.066666667 -2 -.133 333 X Bluffs, Y calls and wins
0.67 1.00 0.00 0.40 0.133333333 1 .133 333 X Bets, Y folds
0.67 1.00 0.40 0.67 0.088888889 2 .177 778 X Bets, Y calls but loses
0.67 1.00 0.67 1.00 0.111111111 0 .000 000 X Bets, Y calls, and 50:50 whether wins or loses
0.11 0.33 0.00 0.17 0.037037037 -1 -.037 037 X Checks, Y Bluffs, X Folds
0.33 0.67 0.00 0.17 0.055555556 2 .111 111 X Checks, Y Bluffs, X Calls and wins
0.11 0.33 0.50 1.00 0.111111111 -1 -.111 111 X Checks, Y Bets, X Folds
0.33 0.50 0.50 1.00 0.083333333 -2 -.166 667 X Checks, Y Bets, X Calls and loses
0.50 0.67 0.67 1.00 0.055555556 -2 -.111 111 X Checks, Y Bets, X Calls and loses
0.50 0.67 0.50 0.67 0.027777778 0 .000 000 X Checks, Y Bets, X Calls (50:50 who wins)
0.11 0.17 0.17 0.50 0.018518519 -1 -.018 519 X Checks, Y Checks, X always loses
0.50 0.67 0.17 0.50 0.055555556 1 .055 556 X Checks, Y Checks, X always wins
0.17 0.50 0.17 0.50 0.111111111 0 .000 000 X Checks, Y Checks, (50:50 who wins)
(1) Firstly if X always checks then Y has the same advantage as problem 1 : E(X) = -0.1
(2) Secondly if X always bets then Y will call with .25+ (as stands to lose $1 vs gain $3) E(X)=-0.125
(2 a) Hence bad strategy for X to "always" bet.
(3) From above it can be seen Y should always call any bet with expected chance of winning is 25% or more
(3 a) Y has an advantage to maximise range that X doesn't bet (since he then has an advantage)
(3 a) From problem 1 Y should remain prepared to call with .4 (otherwise this gives X an advantage)
(3 b) So X should adopt the same strategy (otherwise this gives Y an advantage)
(3 c) So X should ensure the range of betting (e.g. 1 to .7) being 3x range for bluffing (e.g. 0 to .1)
(3 d) However probably worth taking a small loss to prevent Y for gaining advantage.
(4) Now Y should always bet 0-(bluff X) and (Bet X)-1 as X would be indifferent to calling
(4 a) So Y and X now fall between (bluff X) and (bet X) with problem 1
(4 b) So Y bets upto 1/6 and anything over .5 with X calling 1/3
Problem number 1
X Range Y Range Y Pr (event) Win to X 0.10
0.00 0.10 0.00 0.40 0.04 1 0.04 X Bluffs, Y folds
0.00 0.10 0.40 1.00 0.06 -2 -0.12 X Bluffs, Y calls and wins
0.10 0.70 0.00 0.10 0.06 1 0.06 X checks and always beats Y
0.10 0.70 0.10 0.70 0.36 0 0.00 X checks, and 50:50 whether wins or loses
0.10 0.70 0.70 1.00 0.18 -1 -0.18 X checks, and will always lose
0.70 1.00 0.00 0.40 0.12 1 0.12 X Bets, Y folds
0.70 1.00 0.40 0.70 0.09 2 0.18 X Bets, Y calls but loses
0.70 1.00 0.70 1.00 0.09 0 0.00 X Bets, Y calls, and 50:50 whether wins or loses
X Bluffs .111 111
X Bets .666 667
Y Calls .400 000
Y Bluffs .166 667
Y Bets .500 000
X Calls .333 333
Problem Number 2
X Range Y Range Y Pr (event) Win to X -.055 556
0.00 0.11 0.00 0.40 0.044444444 1 .044 444 X Bluffs, Y folds
0.00 0.11 0.40 1.00 0.066666667 -2 -.133 333 X Bluffs, Y calls and wins
0.67 1.00 0.00 0.40 0.133333333 1 .133 333 X Bets, Y folds
0.67 1.00 0.40 0.67 0.088888889 2 .177 778 X Bets, Y calls but loses
0.67 1.00 0.67 1.00 0.111111111 0 .000 000 X Bets, Y calls, and 50:50 whether wins or loses
0.11 0.33 0.00 0.17 0.037037037 -1 -.037 037 X Checks, Y Bluffs, X Folds
0.33 0.67 0.00 0.17 0.055555556 2 .111 111 X Checks, Y Bluffs, X Calls and wins
0.11 0.33 0.50 1.00 0.111111111 -1 -.111 111 X Checks, Y Bets, X Folds
0.33 0.50 0.50 1.00 0.083333333 -2 -.166 667 X Checks, Y Bets, X Calls and loses
0.50 0.67 0.67 1.00 0.055555556 -2 -.111 111 X Checks, Y Bets, X Calls and loses
0.50 0.67 0.50 0.67 0.027777778 0 .000 000 X Checks, Y Bets, X Calls (50:50 who wins)
0.11 0.17 0.17 0.50 0.018518519 -1 -.018 519 X Checks, Y Checks, X always loses
0.50 0.67 0.17 0.50 0.055555556 1 .055 556 X Checks, Y Checks, X always wins
0.17 0.50 0.17 0.50 0.111111111 0 .000 000 X Checks, Y Checks, (50:50 who wins)
October 12th, 2012 at 8:34:39 AM
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Hello,
This is a typical example of "Bayesian Nash equilibrium" developed by John Harsanyi.
Write out the expected value as a function of the 'strategies', i.e. the probabilities of moves at each decision step.
Let me explain. Call x and y the random numbers received at beginning.
Set b(x) as the Prob that X bets when she holds value x.
c(y) = P (Y calls if X bets).
d(y) = P (Y bets if X checks), g(x) = P(X then calls).
Write I(E) for the indicator function (0 or 1) of the event E.
Now EV = b(x) { c(y) 2 [I(x>y) - I(x<y)] + (1-c) (1) } + (1-b) { d(y) [ g(x) 2 [I(x>y) - I(x<y)] + (1-g) (-1) ] + (1-d) [I(x>y) - I(x<y)] }
Integrate along y to get the EV for X, along x to get the EV for Y. (Actually, you perform the expectation, but here the pdf is uniform.)
X maximizes EVX with respect to b and g ; you get two first-order conditions (partial derivative = 0).
Y minimizes EVY w.r.t. c and d.
Those four conditions are then solved simultaneously (that's the most difficult part) to yield the solution b(x), g(x), c(y) d(y) . You also have to care for subgame perfection, which means that threats must be credible. Here, it means that you must ensure the four conditions also work when probabilities are strictly different from 0 or 1.
X bets on a x < 1/9 (bluff) or x > 2/3; Y calls when y > 1/3
If X checks, Y bets whenever y < 1/6 or y > 1/2; then X calls on a x > 0.4
The EV is -1/18 for X (= - 0.05555).
You see how the EV, which was in favour of X in Problem 1, switches towards Y in this game.
Why are the strategies deterministic ("With this number, always act like this")? Because the stochastic part, destined to confuse the opponent, has been embedded in the unknown (x,y) received at beginning. But the players can still add a bit of fuzziness in their threshold for calling. Simply, they must ensure that the average remains the value quoted.
This is a typical example of "Bayesian Nash equilibrium" developed by John Harsanyi.
Write out the expected value as a function of the 'strategies', i.e. the probabilities of moves at each decision step.
Let me explain. Call x and y the random numbers received at beginning.
Set b(x) as the Prob that X bets when she holds value x.
c(y) = P (Y calls if X bets).
d(y) = P (Y bets if X checks), g(x) = P(X then calls).
Write I(E) for the indicator function (0 or 1) of the event E.
Now EV = b(x) { c(y) 2 [I(x>y) - I(x<y)] + (1-c) (1) } + (1-b) { d(y) [ g(x) 2 [I(x>y) - I(x<y)] + (1-g) (-1) ] + (1-d) [I(x>y) - I(x<y)] }
Integrate along y to get the EV for X, along x to get the EV for Y. (Actually, you perform the expectation, but here the pdf is uniform.)
X maximizes EVX with respect to b and g ; you get two first-order conditions (partial derivative = 0).
Y minimizes EVY w.r.t. c and d.
Those four conditions are then solved simultaneously (that's the most difficult part) to yield the solution b(x), g(x), c(y) d(y) . You also have to care for subgame perfection, which means that threats must be credible. Here, it means that you must ensure the four conditions also work when probabilities are strictly different from 0 or 1.
X bets on a x < 1/9 (bluff) or x > 2/3; Y calls when y > 1/3
If X checks, Y bets whenever y < 1/6 or y > 1/2; then X calls on a x > 0.4
The EV is -1/18 for X (= - 0.05555).
You see how the EV, which was in favour of X in Problem 1, switches towards Y in this game.
Why are the strategies deterministic ("With this number, always act like this")? Because the stochastic part, destined to confuse the opponent, has been embedded in the unknown (x,y) received at beginning. But the players can still add a bit of fuzziness in their threshold for calling. Simply, they must ensure that the average remains the value quoted.
Reperiet qui quaesiverit
October 12th, 2012 at 10:35:25 AM
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Quote: kubikulannNow V = b(x) { c(y) 2 [I(x>y) - I(x<y)] + (1-c) (1) } + (1-b) { d(y) [ g(x) 2 [I(x>y) - I(x<y)] + (1-g) (-1) ] + (1-d) [I(x>y) - I(x<y)] }
I have a feeling you'll fit in well around here. Welcome to the forum!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 1st, 2012 at 10:11:54 AM
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I just added this problem to my Math Problems site, problem number 215.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
