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Topic: ENVELOPE GAMBLE II (Read 3304 times) |
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AlexH
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ENVELOPE GAMBLE II
« on: Jul 30th, 2002, 6:10pm » |
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I'm assuming that the envelope I get is random otherwise I think I'm out of luck. Pick some well behaved distribution which covers the real line (e.g. normal distribution). Look at the number on the envelope and draw one number from your distrtibution. If your random number is larger than the number in the envelope, switch. For any pair X< Y of numbers in the envelopes, the odds of you switching are higher if you get X than if you get Y. So you're going to get the higher number of the pair slightly more than half the time.
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S. Owen
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Re: ENVELOPE GAMBLE II
« Reply #1 on: Jul 30th, 2002, 6:54pm » |
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Brilliant answer, on this one I totally agree. I see how the argument works if we assume X>0 and Y>0, which seems implied by "amounts" in the problem, but does this hold even without that constraint?
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AlexH
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Re: ENVELOPE GAMBLE II
« Reply #2 on: Jul 30th, 2002, 7:53pm » |
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Yes. We never assumed anything about the numbers being positive and I think the question leaves negatives as a possiblity. A more negative number has a bigger chance of being less than my random draw than does a less negative number. The normal distribution goes from -infinity to infinity.
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S. Owen
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Re: ENVELOPE GAMBLE II
« Reply #3 on: Jul 31st, 2002, 9:16am » |
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Yeah, nevermind, I agree. I guess there is no way to say any more about the payoff? Like, I wonder if the expected value of the game is still 0 as the "edge" you gain is vanishingly small. I don't know... this sounds good though.
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AlexH
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Re: ENVELOPE GAMBLE II
« Reply #4 on: Jul 31st, 2002, 12:31pm » |
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If your opponent gets to know your random number distrtibution in advance he can make your payoff arbitrarily small but it won't be zero for any fixed choice he makes.
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| « Last Edit: Jul 31st, 2002, 12:31pm by AlexH » |
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tim
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Re: ENVELOPE GAMBLE II
« Reply #5 on: Jul 31st, 2002, 3:41pm » |
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A better strategy is to pick the mean of the observed envelope values so far as your criterion. (Pick any number for the first one).
The expected value of the game in the long run is the mean deviation of your opponent's distribution.
Mathematical derivation: Let the density function of distribution of envelope values be f(x), with independent random variables A and B for the values of the first and second envelopes, respectively. Let your criterion value be C. C partitions the distribution into two areas; "<C" with probability p and ">=C" with probability q = 1-p.
Let the mean of the whole distribution be M, divided into a weighted sum of S0 below C and S1 above C. By definition, M = S0 + S1. The mean of the lower section is S0/p, that of the upper section is S1/q.
The partial payoff in the A<C case is p.(M - S0/p) = p.M - S0. The payoff in the A>=C case is S1 - q.M. The total value of the game is V = S1 - S0 + (p-q).M. We want to select a value of C that maximises this quantity.
Now; dp/dC = f(C), dq/dC = -f(C), dS0/dC = C.f(C), dS1/dC = -C.f(C). So dV/dC = 2.M.f(C) - 2.C.f(C). This takes its extreme value when C = M. That is, when our criterion value is the mean of the (unknown) distribution. The value of the game in that case is the integral of |x - M|.f(x) over all x, which is the definition of the mean deviation.
Now, we can't know what the true mean is in any finite time. We can however converge upon it by using the mean of observed values to estimate it. In the limit of large numbers of games, this will converge to the true mean and hence the long-run value of the game is the mean deviation.
Now, there are only two holes that I can see in this derivation:
There may still be a better strategy than a simple criterion-based one. I have some preliminary results that indicate otherwise though, so I don't think this is likely to be a problem. Don't stop looking on my account though 
The distribution may not have a mean. This can happen for some pathological distributions such as the Cauchy distribution, which is symmetric about zero but lacks a mean and has infinite variance. In practice this is not a problem though -- any distribution without a mean results in a long-run unbounded payoff per game if you follow this strategy. Even better 
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AlexH
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Re: ENVELOPE GAMBLE II
« Reply #6 on: Aug 1st, 2002, 12:47am » |
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The original problem only included 1 run of the game, but it's certainly true that we can learn our opponents distribution over time if we assume it is static and we get multiple plays.
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tim
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Re: ENVELOPE GAMBLE II
« Reply #7 on: Aug 1st, 2002, 1:50am » |
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It didn't look like it was excluding multiple runs to me, just describing how the game proceeds.
If there is only ever one round, then no, it is not possible to reliably make any money. There is always quite a high chance that you will lose money in any single round. Only on average can you reliably make money.
Certainly the "strategy" in a single-round game is trivial.
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srn437
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Re: ENVELOPE GAMBLE II
« Reply #8 on: Aug 26th, 2007, 9:03pm » |
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Distribution of the real numbers? What is that?
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towr
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Re: ENVELOPE GAMBLE II
« Reply #9 on: Aug 26th, 2007, 11:47pm » |
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on Aug 26th, 2007, 9:03pm, srn347 wrote:| Distribution of the real numbers? What is that? |
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http://mathworld.wolfram.com/DistributionFunction.html
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| « Last Edit: Aug 26th, 2007, 11:48pm by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Hippo
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Re: ENVELOPE GAMBLE II
« Reply #10 on: Aug 27th, 2007, 3:04am » |
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What about to modify AlexH solution in the following way:
Choose 2k+1 random numbers in your distribution and switch if more than k of them are greater than given value.
But does not it mean ... if the given value is under expected average switch?
So let the expected average be 0, oponent choses two numbers at random and if both have the same sign he gives you the smaler one and if one is negative and one positive he gives you random of them. ... Now if you switch, you always gain in same sign case and with 50% chance when the signs differ.
If you don't switch, you gain only with 50% chance when the signs differ.
... so at least averaging is much worse then one experiment, but if the opponent knows your distribution, he can gain in either case. The only strategy against which oponent cannot gain is choosing switching with 50% chance at random.
He also has strategy you cannot beat ... choose two random distinct numbers (and order them at random such that with 50% the public one is higher than the secret one).
Ooops ... the riddle is named envelope gamble ... so there is no public/secret number ... you get it at random ... than this thoughts should be revised:
If both have the same sign, you get the higher with 50% chance, if they have different sign, you always gain ... so the average method works well.
And the opponents strategy is to give you both numbers with the same sign to prevent loss, but not gaining with it.
One try method again wins as for the given pair of numbers (even of the same sign), the probability of switching is bigger when their order is wrong than if their order is correct.
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| « Last Edit: Aug 27th, 2007, 3:27am by Hippo » |
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