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Topic: A good probability puzzle but what is the right wording?
Replies: 10   Last Post: Feb 24, 2013 12:19 PM

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RGVickson@shaw.ca

Posts: 1,657
Registered: 12/1/07
Re: A good probability puzzle but what is the right wording?
Posted: Feb 4, 2013 11:10 PM
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On Monday, February 4, 2013 4:02:36 AM UTC-8, Paul wrote:
> The following puzzle is copied and pasted from the internet.
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> BEGIN QUOTE:
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> Alice secretly picks two different real numbers by an unknown process and puts them in two (abstract) envelopes. Bob chooses one of the two envelopes randomly (with a fair coin toss), and shows you the number in that envelope. You must now guess whether the number in the other, closed envelope is larger or smaller than the one you?ve seen.
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> Is there a strategy which gives you a better than 50% chance of guessing correctly, no matter what procedure Alice used to pick her numbers?
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> END QUOTE
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> Let X and Y be the two real numbers. People who pose such a problem point out that if you select a probability distribution (it doesn't really matter which one so long as it has the required domain and is continuous), then the probabiilty of making the correct guess is 50% + probability that a random number selected under the solver's distribution is between X and Y.
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> However, it's difficult to claim that the probability that the random number in the required region is > 0. The problem with the claim is that the secret-number-picker is allowed to use intelligence too. Just as the envelope-selector can claim a probability > 1/2 + epsilon, the number-picker can counter that by pointing out that, for any epsilon, there exists an X-and-Y-picking technique to make the selector's probablity < 1/2 + epsilon.
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> But the problem does have a very clever idea -- can it be made to work by the right wording? Or is the commonsense answer of "it's always 50/50" correct and the > 50% arguments only make sense by avoiding rigorous definitions.
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> Thank You,
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> Paul Epstein


A variant I prefer is one that made the rounds of the Operations Research community some years ago.

You are presented with two opaque bags, and are told that one bag contains twice as much money as the other. You are allowed to open one bag before choosing your prize bag. If you open one of the bags and see $10, should you switch to the other bag?

One popular argument at the time was that the other bag contains either $5 or $20, with equal probabilities, and since the expected value is $25/2 = $12.50 > $10, then yes, you should switch.

However, this seems fishy, since the same type of argument would apply to any opened bag.

RGV



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