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#61
09-13-2007, 08:49 PM
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Re: Game Theory Problem Of The Week
[ QUOTE ]
[ QUOTE ] [ QUOTE ] My answer was like 0.1122222222 and his answer is 0.111111 ? I also get 0.111111 as my answer when we're not working in the discrete case ;ie , we select a number from [0,1]but I get 0.1122222 when we're working in this particular example . Either way , I'm certain I did everything right . [/ QUOTE ] What is the strategy for P1 and P2 that is giving you 0.112222222? [/ QUOTE ] Mykey , my strategy is the same as yours . My answer is nothing but a round-off error . I gave a very close approximation that when player one calls player two , that player two will win 1/3 of the time . This is not exactly true but it's very close which is why i'm off by a slight margin . This is why I prefer not working in the discrete case . [/ QUOTE ] I call BS on the rounding errors. It's formula errors. 
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#62
09-13-2007, 08:53 PM
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Re: Game Theory Problem Of The Week
[ QUOTE ]
Both of these should = 1 if you are calculating for the whole strategy [/ QUOTE ] The other 4/9th of the possibilities should be the times when each player wins 50% of the time, giving a multiplier of zero. I've drawn one very useful conclusion out of this thread: I *really* should pick up Mathematics of Poker. Reading it might save me from looking like a moron when "obvious" things turn out to be false. Oh well. At least I'm willing to publicly admit when I have, in fact, been a moron. I have that going for me, I guess. lol **EDIT** I just went back through my equity calculation, and after changing my 100s to 99s in the appropriate spots, I end up with .5023 of the total probability in spots where there is no EV contribution because both players win half the time. So yeah, something still isn't adding up, because adding that would put the total probability *over* 1. Did I mention I've seen jay_shark calculate EV incorrectly (drastically so in at least one case) in other threads? [img]/images/graemlins/smile.gif[/img]
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#63
09-13-2007, 09:23 PM
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Re: Game Theory Problem Of The Week
[ QUOTE ]
1*33/100*78/99 + 3*33/100*22/99*1/3 -3*33/100*22/99*2/3 +1*66/100*11/99 - 1*66/100*22/99 = 0.11333333 (This is the exact EV without replacement. [/ QUOTE ] Found some problems... With the 78: When P2 has [1,11] and bets, there are only 77 numbers P1 will fold with. When P2 has [79,100] and bets, there are 78 numbers P1 will fold with. With the 66: There are 67 cards that P2 will check with It should be: +1*11/100*77/99 +1*22/100*78/99 + 3*33/100*22/99*1/3 -3*33/100*22/99*2/3 +1*67/100*11/99 - 1*67/100*22/99 = 0.11111111
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#64
09-13-2007, 11:09 PM
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Re: Game Theory Problem Of The Week
[ QUOTE ]
Mykey , you got to be kidding me . Why on earth would you restrict player one to check call with only 57-100 ? [/ QUOTE ] Jay, Do you accept both: P1 [1,56] Fold [57,100] Call and P2 [1,11] Bet [12,78] Check [79,100] Bet are optimal for this game?
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#65
09-13-2007, 11:43 PM
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Re: Game Theory Problem Of The Week
[ QUOTE ]
[ QUOTE ] Both of these should = 1 if you are calculating for the whole strategy [/ QUOTE ] The other 4/9th of the possibilities should be the times when each player wins 50% of the time, giving a multiplier of zero. I've drawn one very useful conclusion out of this thread: I *really* should pick up Mathematics of Poker. Reading it might save me from looking like a moron when "obvious" things turn out to be false. Oh well. At least I'm willing to publicly admit when I have, in fact, been a moron. I have that going for me, I guess. lol **EDIT** I just went back through my equity calculation, and after changing my 100s to 99s in the appropriate spots, I end up with .5023 of the total probability in spots where there is no EV contribution because both players win half the time. So yeah, something still isn't adding up, because adding that would put the total probability *over* 1. Did I mention I've seen jay_shark calculate EV incorrectly (drastically so in at least one case) in other threads? [img]/images/graemlins/smile.gif[/img] [/ QUOTE ] Tnixon , please point to the thread where I made an EV error . Remember that I've been using the definition that EV(fold) =0 which is why I was producing different results . I even stipulated this in one of the problems .
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#66
09-13-2007, 11:55 PM
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Re: Game Theory Problem Of The Week
Here is the proof that you should bet with [0,1/9] , [7/9,1]where x is your optimal betting range and that your value bets to bluffs should be in the ratio 2:1 .
EV= 1*(1-x)*(2x+1)/3 +3*(1-x)(2-2x)/3*1/3 -3*(1-x)*(2-2x)/3*2/3 -x*(1-x)/3 EV= (-3x^2+4x-1)/3 after simplifying . EV' = -2x+4/3 Set EV'=0 and we get 4/3=2x and x =2/3 . So Mykey , you haven't proved a thing . All you are doing is spewing random numbers because you're incompetent to figure it out yourself . You're relying on simulators as a means to find your answer which is unethical .
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#67
09-14-2007, 12:02 AM
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Re: Game Theory Problem Of The Week
[ QUOTE ]
[ QUOTE ] 1*33/100*78/99 + 3*33/100*22/99*1/3 -3*33/100*22/99*2/3 +1*66/100*11/99 - 1*66/100*22/99 = 0.11333333 (This is the exact EV without replacement. [/ QUOTE ] Found some problems... With the 78: When P2 has [1,11] and bets, there are only 77 numbers P1 will fold with. When P2 has [79,100] and bets, there are 78 numbers P1 will fold with. With the 66: There are 67 cards that P2 will check with It should be: +1*11/100*77/99 +1*22/100*78/99 + 3*33/100*22/99*1/3 -3*33/100*22/99*2/3 +1*67/100*11/99 - 1*67/100*22/99 = 0.11111111 [/ QUOTE ] Mykey , you are such a nit . This is a round off error !! Stop trying to take credit for something I've done .
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#68
09-14-2007, 01:30 AM
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Re: Game Theory Problem Of The Week
Taking credit?
On the 9th message of this thread I had to optimal strategy for P2, and on the 15th I had the optimal for P1 What exactly did you do present here that you think I'm taking credit for?
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#69
09-14-2007, 01:41 AM
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Re: Game Theory Problem Of The Week
[ QUOTE ]
Here is the proof that you should bet with [0,1/9] , [7/9,1]where x is your optimal betting range and that your value bets to bluffs should be in the ratio 2:1 . EV= 1*(1-x)*(2x+1)/3 +3*(1-x)(2-2x)/3*1/3 -3*(1-x)*(2-2x)/3*2/3 -x*(1-x)/3 EV= (-3x^2+4x-1)/3 after simplifying . EV' = -2x+4/3 Set EV'=0 and we get 4/3=2x and x =2/3 . So Mykey , you haven't proved a thing . All you are doing is spewing random numbers because you're incompetent to figure it out yourself . You're relying on simulators as a means to find your answer which is unethical . [/ QUOTE ] How exactly do you go from x = 2/3 to [0,1/9] and [7/9,1]? The amazing thing is that my "random" numbers were correct. My "simulator" didn't tell me the strategy, it only gave me the EV once I selected a pair of strategies. And when did using the tools at one's disposal become unethical? Just accept the fact that your methods never would have arrived at the proper answer for P2 [1,11] and [79,100] bet, and may have never arrived at P1 [57,100] call.
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#70
09-14-2007, 04:36 AM
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Re: Game Theory Problem Of The Week
[ QUOTE ]
Tnixon , please point to the thread where I made an EV error . [/ QUOTE ] If I answer this in the way that I really want to, this thread will turn into a bigger flamefest than it already is. Stupidity hits us all at some point or other. Some of us can admit when it's happened, and some of us can't. So I'll choose to remain silent. Leader, see how much restraint I'm showing? I *like* to flame. [img]/images/graemlins/smile.gif[/img]
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