How does knowing the second number establish the range ? Is there any work
on this.

Sarbajit

On Thu, Jun 9, 2011 at 1:15 AM, Russ Abbott <[email protected]> wrote:

> Russell Standish has the right idea.  If you knew the range, say the first
> number is higher/lower than the third depending on whether the first numbers
> is greater than or less than the middle of the range. Since you don't know
> the range, the second random number is used instead.  Say higher/lower
> depending on whether the first number is higher/lower than the second.
>
> Also, think about it this way. If the middle number is either greater than
> or less than both the first and the third, you have a 50% chance of being
> right. If it's between the first and the third, the strategy described will
> always be right. Presumably there is a non-zero probability that the second
> number will be between the first and the third. Therefore one has a greater
> than 50% chance of being right.
>
> *-- Russ *
>
>
> On Wed, Jun 8, 2011 at 10:06 AM, Owen Densmore <[email protected]>wrote:
>
>> Do you have a pointer to an explanation?
>>
>>        -- Owen
>>
>> On Jun 8, 2011, at 12:11 AM, Russ Abbott wrote:
>>
>> > Although this isn't new, I just came across it (perhaps again) and was
>> so enchanted that I wanted to share it.
>> >
>> > Generate but don't look at three random numbers. (Have someone ensure
>> that they are distinct. There is no constraint on the range.) Look at the
>> first two. You are now able to guess with a better than 50% chance of being
>> right whether the first number is larger than the unseen third.
>> >
>> > I like this almost as much as the Monte Hall problem.
>> >
>> > -- Russ
>> >
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>>
>>
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>> Meets Fridays 9a-11:30 at cafe at St. John's College
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>>
>
>
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> FRIAM Applied Complexity Group listserv
> Meets Fridays 9a-11:30 at cafe at St. John's College
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>
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