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Re: A good probability puzzle but what is the right wording?
Posted:
Feb 4, 2013 9:33 AM
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On Mon, 04 Feb 2013 08:15:59 -0500, quasi <quasi@null.set> wrote:
>quasi wrote:
>>Paul wrote:
>>>
>>>The following puzzle is copied and pasted from the internet.
>>>
>>> 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 youve seen. 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?
>>
>>Yes.
>>
>>Let R denote the set of real numbers and let (0,1) denote
>>the open interval from 0 to 1.
>>
>>Let f : R -> (0,1) be a strictly decreasing function.
>>
>>Use the following strategy:
>>
>>If the initially exposed value is t, "switch" with probability
>>f(t) and "stay" with probability 1 - f(t).
>>
>>Suppose Alice chooses the pair x,y with x < y (by whatever
>>process, it doesn't matter). After Alice choose that pair,
>>then, by following the strategy I specified above, the
>>probability of guessing the highest card is exactly
>>
>> (1/2)*f(x) + (1/2)*(1 - f(y))
>>
>>which simplifies to
>>
>> 1/2 + f(x) - f(y)
>
>I meant:
>
>which simplifies to
>
> 1/2 + (1/2)*(f(x) - f(y))
>
>>and that exceeds 1/2 since f is strictly decreasing.
Huh. I thought the answer was obviously no. But
this seems right. Huh.
>>Of course, it's not the case that probability of guessing
>>correctly is more than c for any fixed c > 1/2, but the
>>problem didn't require that.
_If_ we assume in addition that Alice used some fixed
probability distribution on {(x,y) : x < y} to choose
x and y then it does give such a c, namely the
expected value of 1/2 + 1/2(f(x) - f(y)). The expected
value of a strictly positive random variable is
strictly positive.
>
>quasi
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