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CrystalMath
CrystalMath
Joined: May 10, 2011
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August 14th, 2012 at 4:07:13 PM permalink

EV for x is -289/6272
I heart Crystal Math.
charliepatrick
charliepatrick
Joined: Jun 17, 2011
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August 14th, 2012 at 4:11:06 PM permalink
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)
kubikulann
kubikulann
Joined: Jun 28, 2011
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October 12th, 2012 at 8:34:39 AM permalink
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.
Reperiet qui quaesiverit
Wizard
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Wizard 
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October 12th, 2012 at 10:35:25 AM permalink
Quote: kubikulann

Now 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!
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
Administrator
Wizard 
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November 1st, 2012 at 10:11:54 AM permalink
I just added this problem to my Math Problems site, problem number 215.
It's not whether you win or lose; it's whether or not you had a good bet.

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