Wrong Answer in the Archive! Re: decision/high.or.low
Siqi Chen
I pick two numbers, randomly, and tell you one of them. You are
supposed to guess whether this is the lower or higher one of the two
numbers I picked. Can you come up with a method of guessing that does
better than picking the response "low" or "high" randomly (i.e.
probability to guess right > .5) ?
==> decision/high.or.low.s <==
Pick any cumulative probability function P(x) such that a > b ==> P(a)
> P(b). Now if the number shown is y, guess "low" with probability
P(y) and "high" with probability 1-P(y). This strategy yields a
probability of > 1/2 of winning since the probability of being correct
is 1/2 * ((1-P(a)) + P(b)) = 1/2 + (P(b)-P(a)), which is > 1/2 by
assumption.
#1. 1/2 * ((1-P(a)) + P(b)) = 1/2 + (P(b)-P(a)) <-- that doesn't
follow.
#2. P(b) - P(a) < 0, by assumption.
#3. 1/2 * ((1 - P(a) + P(b)) is not the right probability equation.
Shouldn't it be 1/2 * ((1-P(a) + P(a))??
Jonathan Dushoff
: ==> decision/high.or.low.p <==
: supposed to guess whether this is the lower or higher one of the two
: numbers I picked. Can you come up with a method of guessing that does
: better than picking the response "low" or "high" randomly (i.e.
: probability to guess right > .5) ?
: ==> decision/high.or.low.s <==
: Pick any cumulative probability function P(x) such that a > b ==> P(a)
: > P(b). Now if the number shown is y, guess "low" with probability
: P(y) and "high" with probability 1-P(y). This strategy yields a
: probability of > 1/2 of winning since the probability of being correct
: is 1/2 * ((1-P(a)) + P(b)) = 1/2 + (P(b)-P(a)), which is > 1/2 by
: assumption.
: #1. 1/2 * ((1-P(a)) + P(b)) = 1/2 + (P(b)-P(a)) <-- that doesn't
: follow.
Sure it does.
You are right that the solution is confused, though. Not only does it
seem to suddenly switch a and b, but it also has the probability of
guessing "low" increase as the number increases, which is clearly wrong.
The following would be better:
==> decision/high.or.low.s <==
Pick any cumulative probability function P(x) such that a > b ==> P(a)
> P(b). Now if the number shown is y, guess "high" with probability
P(y) and "low" with probability 1-P(y). This strategy yields a
probability > 1/2 of winning since the probability of being correct
is 1/2 * (P(a) + (1-P(b))) = 1/2 + (P(a)-P(b)), which is > 1/2 by
assumption.
Jonathan Dushoff
dushoff at eno.princeton.edu
Nis Jorgensen
(Jonathan Dushoff) wrote:
>
>: #1. 1/2 * ((1-P(a)) + P(b)) = 1/2 + (P(b)-P(a)) <-- that doesn't
>: follow.
>
>Sure it does.
No it doesn't
1/2 * ((1-P(a)) + P(b)) = 1/2 + 1/2 * (P(b)-P(a))
which leads to the same conclusion.
It looks to me like the puzzle statement is wrong thoug. Instead of
"I pick two numbers, randomly, and tell you one of them."
it should start
"I pick two numbers and tell you one of them, chosen randomly."
--
Nis Jorgensen
Amsterdam
Please include only relevant quotes, and reply below the quoted text. Thanks
Siqi Chen
the solution said. The correct strategy should be guess high with
probability F(y), where F(y) is any CPF function and y is the number
shown.
It said in the given answer that the optimal strategy is to guess low
with probability F(y) and high with probability 1 - F(y).
Now, to find the value for the probability of a correct guess using
our
method, we use the following, correct?:
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.
The method given, as you can see, yields a worth than .5 chance of
being
correct. It's easy to see that only if we reverse our guessing
strategy, we
get a > .5 chance of being correct.
Fred the Wonder Worm
Jonathan Dushoff <dus...@gate.sinica.edu.tw> wrote:
>Siqi Chen (aznb...@yahoo.com) wrote:
>: #1. 1/2 * ((1-P(a)) + P(b)) = 1/2 + (P(b)-P(a)) <-- that doesn't
>: follow.
>
> Sure it does.
[ ... ]
>==> decision/high.or.low.s <==
> Pick any cumulative probability function P(x) such that a > b ==> P(a)
> > P(b). Now if the number shown is y, guess "high" with probability
> P(y) and "low" with probability 1-P(y). This strategy yields a
> probability > 1/2 of winning since the probability of being correct
> is 1/2 * (P(a) + (1-P(b))) = 1/2 + (P(a)-P(b)), which is > 1/2 by
> assumption.
This still doesn't fix what I assume Siqi was objecting to, namely:
1/2*(P(a) + (1 - P(b))) = 1/2 + (P(a) - P(b)) / 2
not what is written above. The conclusion is unaffected, of course.
Cheers,
Geoff.
-----------------------------------------------------------------------------
Geoff Bailey (Fred the Wonder Worm) | Programmer by trade --
ft...@maths.usyd.edu.au | Gameplayer by vocation.
-----------------------------------------------------------------------------
Siqi Chen
> This still doesn't fix what I assume Siqi was objecting to, namely:
>
> 1/2*(P(a) + (1 - P(b))) = 1/2 + (P(a) - P(b)) / 2
>
> not what is written above. The conclusion is unaffected, of course.
>
No, I disagree. See my "decision/high.or.low" thread. The conclusion
is wrong in that the correct strategy should be the exact opposite of
what was posted.
Jonathan Dushoff
: No, I disagree. See my "decision/high.or.low" thread. The conclusion
: is wrong in that the correct strategy should be the exact opposite of
: what was posted.
No, we all agree. Geoff/Fred meant that the conclusion of the
(partially) corrected version he was responding to was unaffected by the
fact that I dropped a factor of 1/2 from one of the terms.
We all agree that the archive version is backwards. The question is,
does anybody know how it can be fixed?
Siqi Chen
> (partially) corrected version he was responding to was unaffected by the
> fact that I dropped a factor of 1/2 from one of the terms.
Ahh, you're right. My fault.