Game Theory Resolution

[jay_shark]

The object of this game is to select a number from the closed interval [0,1] and to bet if you think your number is the highest . You only play one round so if you fold , the game is over .

a) A generous man decides to give you (hero) and your friend (villain) a free roll to enter this game . Hero posts the sb and villain posts the bb and you can raise to 3bb's or fold . Villain on the other hand can only call . What number should you raise with ??

Solution: Let a be your optimal pushing range ; a>=0
Let x be your opponent's optimal calling range . 1/3<=x<=1

We can write a in terms of x . Notice that (1-x)/(x-a) = 2
x=(2a+1)/3

The probability that your opponent wins given that he calls is 2/3 which is verified since 1/3 + 2/3*1/2 = 2/3

So the probability that you win given that he calls is 1/3 .

EV(x) = (3-3x)/2*[[1.5x + 3.5*1/3*(1-x) -2.5*2/3*(1-x)]

We wish to maximize this function using calculus .

After simplifying you get

EV(x) = 3x-3x^2
EV'(x)= 3 -6x
Set this =0 so we get x=1/2 .

Finally we're done !!

[jay_shark]

So we plug x =0.5 into the equation relating x and a .
0.5=(2a+1)/3
a=1/4

Hero's EV for employing this strategy is 0.75 which you get from plugging 0.5 into the EV(x)function .

So our shoving range for hero is to push with any number greater than 1/4 . Villain should call with any number greater than 1/2 which ensures that this is the best result for both players.

[HokieGreg]

LOL YOUAREGOODATMATHEMATICSAMENTS

[teteatot]

can you explain where you get "Notice that (1-x)/(x-a) = 2" ?

[TNixon]

Only one problem.

The optimal play isn't actually anywhere near a=0.25, x=.5.

This is very easy to prove by just calculating a couple EVs.

Figure the EV at a=0.25, and then figure the EV at a=0.5. The EV at a=0.5 will be higher, meaning that it is more +EV to bet 50+ than it is to bet 25+.

Which is a direct contradiction to this statement:

[ QUOTE ]
So our shoving range for hero is to push with any number greater than 1/4 . Villain should call with any number greater than 1/2 which ensures that this is the best result for both players.

[/ QUOTE ]

[TNixon]

Your simplification of the EV equation is incorrect.

Here it is step-by-step, so you can point out specific errors if you disagree:

(3-3x)/2 (1.5x + 3.5 * 1/3 * (1-x) - 2.5 * 2/3 * (1-x))
...
(3-3x)/2 (4.5x/3 + 3.5/3 - 3.5x/3 - 5/3 + 5x/3)
...
(3-3x)/2 * (4.5x + 3.5 - 3.5x - 5 + 5x)/3
...
(1-x)/2 * 6x - 1.5
...
(6x - 6x^2 - 1.5 + 1.5x)/2
...
(-6x^2 + 7.5x - 1.5)/2
...
-3x^2 + 3.75x - .75

The derivative is:
-6x + 3.75
so
6x = 3.75
x = 3.75/6 = 0.625

x = 0.625, a = 0.4375

Which, strangely enough, is pretty much exactly what my exhaustive search with the simulator found. (bet with 44+, call with 63+) Finally, math matches reality.

So I guess the small blind didn't matter after all, except for the fact that it lead to EV calculations that appear to be profitable, but really aren't.

[jay_shark]

Yup , thx for correcting the algebra .

So you understand how the EV equation works , right ?

Specifically the term (3-3x)/2 ?

Now it's easy to solve more elaborate problems using derivatives which is pretty neat .

[xSCWx]

For a second I thought I had been doing something wrong by not calling bets with 6[img]/images/graemlins/diamond.gif[/img]3[img]/images/graemlins/heart.gif[/img].

[jay_shark]

[ QUOTE ]
can you explain where you get "Notice that (1-x)/(x-a) = 2" ?

[/ QUOTE ]

since x is the minimum for your opponent's optimal calling range , this means he will be calling 1-x of the time .

Likewise , a is hero's minimum for his optimal betting range. X-a are the numbers that hero would be losing to when villain calls .

So , since villain is getting 2:1 pot odds , he should be calling with any hand with any equity greater than 1/3 . This is equivalent to (1-x)x-a) = 2:1

[mykey1961]

[ QUOTE ]
The object of this game is to select a number from the closed interval [0,1] and to bet if you think your number is the highest . You only play one round so if you fold , the game is over .

a) A generous man decides to give you (hero) and your friend (villain) a free roll to enter this game . Hero posts the sb and villain posts the bb and you can raise to 3bb's or fold . Villain on the other hand can only call . What number should you raise with ??


[/ QUOTE ]

You're so obsessed with the fold EV = 0 idea, that you resort to this?

Clearly since it's a freeroll, the BB should never fold.
And if the BB isn't going to fold, neither should the SB.