From: ram@zedat.fu-berlin.de
Richard Heathfield writes:
>Did you even notice my reply re the food dish question?
Yes. Sorry. I should have written a reply!
So, what that book by Eijkhout said is in my words:
The last dish could be anything, so that its expected value of
the rating function on the scale from 0 to 1 is equal to 0.5.
Therefore, one would choose the penultimate dish if its
rating is better than 0.5, because then it would be better
than the alternative.
If one assumes that the penultimate dish can be anything, it is
better than 0.5 in half of the cases, in such a case, one would
choose it, and since it then lies somewhere between 0.5 and 1, the
expected value is then 0.75. Otherwise, one takes the last dish with
an expected value of 0.5. Overall, with this strategy and two dishes,
the expected value is then the mean of 0.75 and 0.5, i.e., 0.625.
So, the second-to-last dish would have to be better than 0.625
in order to be chosen.
I wrote a small Python program to compare this with the strategy you
mentioned. The result is that the strategy given above fares better
by a factor of 1.2146 if every dish's quality is randomly between 0
and 1. But the traditional strategy you gave fares better by a factor
between 0.8 and 0.99 when the quality of all the dishes is a priori
restricted to some range [a,b], where a < b and 0 <= a <= 1, at least
for three such cases I looked at.
Which is "the best" solution, then, might depend on details of
the wording of the question, i.e., whether it states that every
dish can have an arbitrary quality between 0 and 1 or whether
a certain cook always produces dishes between a and b (but how
could he know or control the ratings the customer will give?).
--- SoupGate-Win32 v1.05
* Origin: you cannot sedate... all the things you hate (1:229/2)
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