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#41
10-20-2005, 03:42 AM
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Re: Is this better ?!?
OK it's all coming together now, thanks. Having to break it up at 0.586 seemed rather forced and somehow I wanted to think the answer was a single function all the way through. But going over various cases makes me realize you guys are right. With hands of .57 or .58 and with A standing (and playing optimally), B is in deep enough kaka to have to take a shot.
As for the EV calculation, it was quite a mess and I'd be curious if anyone can do it without the help of a computer. Integrating fractions like (b^2)/sqrt(2b-1) is not something for which I know any techniques. 
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#42
10-20-2005, 02:58 PM
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Re: Classic Type Game Theory Problem
[ QUOTE ]
[ QUOTE ] [ QUOTE ] How? [/ QUOTE ] I think it's pretty obvious that he means from a uniform distribution, and if generating real numbers (because of a countable computable/uncountable uncomputable issue) bothers you, then use a uniform distribution over some large (a trillion will probably do nicely) number of evenly spaced rationals over that range. [/ QUOTE ] It has nothing to do with being bothered by uncountability or computability. There is no uniform distribution on [0,1]. [/ QUOTE ] This is what I was wondering the whole time as well.
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#43
10-20-2005, 03:32 PM
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solution
[ QUOTE ]
[ QUOTE ] That is part of the optimal strategy in some cases, when b > [sqrt(2)-1], which is around .4142. There, B always draws when A does. But the interesting cases are when b < [sqrt(2)-1], where it turns out A stands pat when his hand exceeds [1 - sqrt(1-2b)]. This value is above b, so it does contain an element of bluffing. Furthermore, B randomizes between standing and drawing. Use the usual indifference conditions for these cases. alThor [/ QUOTE ] I posted the solution in the Poker Theory forum, but I've now seen TWO references to this [1 - sqrt(1-2b)] thing, and I'm trying to understand why on earth this would come up. [/ QUOTE ] Well, because it's right [img]/images/graemlins/smile.gif[/img] I notice I was the first to post reference to this number, and the solution, so I think anyone who writes up a solution should call this the "alThor is Awesome" solution. In fact, 1-sqrt(1-2b) shall henceforth be called the alThor constant. (Despite my eternal greatness, notice I made a typo. Clearly A should stand pat when his hand is BELOW the alThor constant, not "above." I think you already understood this was a typo though.) But seriously, sorry I couldn't reply earlier. I am a bit busy these days. Have you since figured out why my solution is right? When computing an equilibrium for a fixed b, look for one where B randomizes between drawing and standing. This requires him to be indifferent between the two. This means that A must be standing pat with a range of hands [0,a] where a > b, and in fact this indifference determines the value of a. But for A to use this strategy requires that, if his hand exceeds a, he would rather draw than stand (and vice versa). So this indifference at card "a" in turn determines the probability with which B randomizes. Finally, show that when b equals the alThor constant [img]/images/graemlins/blush.gif[/img] , this implies a=1. That is, when b exceeds that constant, we can't have an equilibrium like I just described. That is the point at which we switch over to an equilibrium of the other kind, and B always redraws when A does. The mistake that people are making occurs when looking at A's draw/stand decision. They are not comparing the right probabilities. [ QUOTE ] Suppose there existed a strategy where if b < .5, A could make money by standing pat with a hand less than .5. [/ QUOTE ] The issue is not whether EV is positive or negative, it's "which EV is higher than the other." Let me know if there is still confusion. But first, try to work out the indifference equations I alluded in the first post, and which I just alluded to above in determining "a" and P(B redraws | A stood). alThor P.S. David, what's up with triple-cross-posting? Next time please just post pointers to one centralized thread. P.P.S. I hope everyone takes my jabs in good fun.
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#44
10-20-2005, 03:49 PM
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Re: solution
"P.S. David, what's up with triple-cross-posting? Next time please just post pointers to one centralized thread.
P.P.S. I hope everyone takes my jabs in good fun." I did it as sort of a competion. And if you are right it SHOULD be called the alThor is Awesome solution. Meanwhile if this type of problem has not been previously addressed in game theory books or articles, I believe the simplicity in the game and the non simplicity of the solution, does indeed make it a problem that should henceforth be included in texts. With both our names attached of course. PS What's your background?
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#45
10-20-2005, 04:53 PM
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Re: Classic Type Game Theory Problem
[ QUOTE ]
It has nothing to do with being bothered by uncountability or computability. There is no uniform distribution on [0,1]. [/ QUOTE ] Well, that's news to me. Point me to a reference on that one? I congratulate you on what is still a pretty irrelevant nitpick in terms of answering the problem posed.
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#46
10-20-2005, 06:36 PM
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Re: solution
[ QUOTE ]
I believe the simplicity in the game and the non simplicity of the solution, does indeed make it a problem that should henceforth be included in texts. With both our names attached of course. PS What's your background? [/ QUOTE ] I particularly like the discontinuity of strategies that occurs when b crosses the critical value (which I may soon stop calling the alThor constant [img]/images/graemlins/smile.gif[/img] ). If I get some time, I may try to write up a cleaner derivation. If I get to it, I could mail a pdf file or something. It may take a couple of days. As for background, I'm a game theorist in academia. alThor
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#47
10-20-2005, 07:22 PM
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Re: Classic Type Game Theory Problem
[ QUOTE ]
[ QUOTE ] It has nothing to do with being bothered by uncountability or computability. There is no uniform distribution on [0,1]. [/ QUOTE ] [ QUOTE ] Well, that's news to me. Point me to a reference on that one? [/ QUOTE ] [/ QUOTE ] Suppose m is a measure on [0,1] such that m([0,1]) = 1 and m({x}) = m({y}) for any x, y in [0,1]. That's what we mean by a uniform distribution. If m({x}) > 0 for some x in [0,1] choose n so large that n * m({x}) > 1. Then choose n distinct x_1, ... x_n in [0,1]. We'd have 1 = m([0,1]) >= m({x_1, ..., x_n}) = m({x_1}) + ... + m({x_n}) = n * m({x_1}) > 1 so that 1 > 1. This is a contradiction. Hence m({x}) = 0 for any x in [0,1]. Oops. Uniform distributions happen on finite sets; not infinite ones. [ QUOTE ] I congratulate you on what is still a pretty irrelevant nitpick in terms of answering the problem posed. [/ QUOTE ] It's not a nitpick. I'm a mathematician. The statement of the problem doesn't even make sense to me.
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#48
10-20-2005, 07:44 PM
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Re: Classic Type Game Theory Problem
I am pretty sure that generalizing this assumption
[ QUOTE ] If m({x}) > 0 [/ QUOTE ] is the first step towards defining a continuous probability measure, such as this one.
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#49
10-20-2005, 07:49 PM
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Re: Classic Type Game Theory Problem
[ QUOTE ]
I am pretty sure that generalizing this assumption [ QUOTE ] If m({x}) > 0 [/ QUOTE ] is the first step towards defining a continuous probability measure, such as this one. [/ QUOTE ] Yes, but when talking about "deal[ing] two real numbers" we need a measure that assigns the same nonzero mass to each point.
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Linear Mode
