note
polettix
Hi,
your link to the puzzle has to be updated, it now points to a nonexistent page.
<p>Nevertheless, I found the solution and also [http://groups.google.com/group/rec.puzzles/browse_frm/thread/e6a31d179a093b8d/d81fc34edc96ac8e?tvc=1&q=high+or+low#d81fc34edc96ac8e|this discussion] about the problem, but the solution is flawed IMHO. Apart from the fact that the author is arguing about the phrasing in the original solution, the algebra should be the following (quoted from one of the messages in the thread):<c>
P(correct guess) =
P(we were shown the higher number H)
* P(we guessed "high" given H)
+ P(we were shown the lower number L)
* P(we guessed "low" given L)
=> (plugging in from the original selection method)
P(correct guess) = (1/2) * (1 - F(H)) + (1/2) * F(L)
=>
P(correct guess) = (1/2) (1 - (F(H) - F(L)))
=> Since F(H) - F(L) > 0 (by assumption)
P(correct guess) = (1/2) (1 - (a positive value))
=>
Therefore,
P(correct guess) < 1/2.
</c>
What I find very amusing is that we're using two different values (namely H and L) to feed the F function, but this is not applicable. We have to use the same value - that is the value we figure out in our mind, let's call it y. This leads to:<c>
P(correct guess) = (1/2) (1 - (F(y) - F(y))) = 1/2
</c>as it should clearly be.
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<p align=right>Flavio <br />
perl -ple'$_=reverse' <<<ti.xittelop@oivalf</p>
<i>Don't fool yourself.</i>
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