September 5th, 2012 at 5:16:51 PM
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Same as problem 2, except Y may raise after a bet. X may not check raise. Here are the full rules:

What is the optimal strategy for each player?

This one is about twice as complicated as problem 2, just to warn you.

- 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 Y bets, then X may then call or fold.
- If player X bets then Y may raise, call or fold. If Y raises, then X may call or fold.

What is the optimal strategy for each player?

This one is about twice as complicated as problem 2, just to warn you.

It's not whether you win or lose; it's whether or not you had a good bet.

October 25th, 2012 at 7:45:38 AM
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Method of solution:

Call x and y the random numbers received at beginning.

b(x) = P (X bets when she holds value x).

c(y) = P (Y calls if X bets). r(y) = P (Y raises if X bets). Constraint: c+r<1

k(x) = P (X calls if Y raises).

d(y) = P (Y bets if X checks), g(x) = P(X then calls).

Writing I(E) for the indicator function (0 or 1) of the event E, we write

SH(x,y) = [I(x>y) - I(x<y)] (SH for "showdown")

Then

EV(x,y) = b { r [k 3 SH + (1-k) 2] + c 2 SH + (1-c-r) } + (1-b) { d [g 2 SH + (1-g)(-1)] + (1-d) SH }

X maximizes EEV(x) = int(y=0 to 1) EV

= b { k (6R(x)+2R(1)-3) + 4c - 3R(1) -C(1) - 1 } + (1-b) { (2g-1) [2D(x)-1+D(1)/2] + (1-D(1))/2 }

where an uppercase letter is the integral of the correponding lowercase.

Do the same for Y, integrating over x.

You get a set of six maximizing conditions, fiendishly difficult to solve...

Using information from the previous problems, one can probably ease some by assuming the betting strategy is deterministic (a threshold below which you bluff, another above which you bet). But I see no hint about the optimal form of strategy for raising.

Work in progress.

Call x and y the random numbers received at beginning.

b(x) = P (X bets when she holds value x).

c(y) = P (Y calls if X bets). r(y) = P (Y raises if X bets). Constraint: c+r<1

k(x) = P (X calls if Y raises).

d(y) = P (Y bets if X checks), g(x) = P(X then calls).

Writing I(E) for the indicator function (0 or 1) of the event E, we write

SH(x,y) = [I(x>y) - I(x<y)] (SH for "showdown")

Then

EV(x,y) = b { r [k 3 SH + (1-k) 2] + c 2 SH + (1-c-r) } + (1-b) { d [g 2 SH + (1-g)(-1)] + (1-d) SH }

X maximizes EEV(x) = int(y=0 to 1) EV

= b { k (6R(x)+2R(1)-3) + 4c - 3R(1) -C(1) - 1 } + (1-b) { (2g-1) [2D(x)-1+D(1)/2] + (1-D(1))/2 }

where an uppercase letter is the integral of the correponding lowercase.

Do the same for Y, integrating over x.

You get a set of six maximizing conditions, fiendishly difficult to solve...

Using information from the previous problems, one can probably ease some by assuming the betting strategy is deterministic (a threshold below which you bluff, another above which you bet). But I see no hint about the optimal form of strategy for raising.

Work in progress.

Reperiet qui quaesiverit

October 9th, 2013 at 7:50:40 AM
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For those interested, here is a solution. There may be others.

Solution

(Values of x or y) ACTION

X:

(0) BET/bluff + FOLD if Y raises (.079) CHECK + FOLD (.352) CHECK + CALL (.762) BET + FOLD if raise (.810) BET + CALL if raise (1)

Y if X checks:

(0) BET/bluff (.148) CHECK (.557) BET (1)

Y if X bets:

(0) RAISE/bluff (.019) FOLD (.352) CALL (.905) RAISE (1)

The Expected Value for X is - 0.49/18 = -0.0274 .

Surprisingly, alternative solutions include X calling the raise even though she was bluffing. That is because Y might be bluffing in his raise.

Other alternatives have Y folding for small values and bluffing (raise) for higher values.

(Values of x or y) ACTION

X:

(0) BET/bluff + FOLD if Y raises (.079) CHECK + FOLD (.352) CHECK + CALL (.762) BET + FOLD if raise (.810) BET + CALL if raise (1)

Y if X checks:

(0) BET/bluff (.148) CHECK (.557) BET (1)

Y if X bets:

(0) RAISE/bluff (.019) FOLD (.352) CALL (.905) RAISE (1)

The Expected Value for X is - 0.49/18 = -0.0274 .

Surprisingly, alternative solutions include X calling the raise even though she was bluffing. That is because Y might be bluffing in his raise.

Other alternatives have Y folding for small values and bluffing (raise) for higher values.

Reperiet qui quaesiverit

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