Game Theory Resolution

[mykey1961]

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x = 0.625, a = 0.4375


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Ok it took a while but the lightbulb finally flickered.

As far as the BB knows, the SB may or may not play optimal, therefore the BB plays an optimal strategy "x" which locks in a win rate >= 5/64BB per hand.

SB also doesn't know if the BB is playing optimal, so SB plays an optimal strategy "a" which locks in a loss rate <= 5/64BB per hand.

Seems to me that if SB determines that the BB is also playing optimal, then SB should open up the range, and look for the BB trying to exploit the new range. Once that happens, SB should tighen up the range to exploit the BB.. and the race is on.

[jay_shark]

Hey Mykey , let me put you on the spot .

How often should you bluff in this game ?

[jay_shark]

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

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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]

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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 ??


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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.