From: Dan Fabulich (dfabulich@warpmail.net)
Date: Tue Jul 08 2003 - 18:36:42 MDT
Lee Corbin wrote:
> and what Dan did not quote
>
> > For the red-majority barrel, the probability of
> > drawing a red each time is 3/4, blue 1/4; so
> > the odds of 3 blues, 7 reds is p1 = (1^3)*(3^7)/(4^10).
> > For the blue-majority barrel, it's p2 = (3^3)*(1^7)/(4^10).
> > Therefore the overall probability (conditioning
> > on the 50-50 prior) is the average of those two
> > probabilities. A posteriori we know that this is
> > what happened, so the odds that it was the
> > blue-majority barrel are
>
> > p2/(p1+p2) = 27/(27+2187) = 1/82
>
> Well, I get 27/(27+1054). Is that right?
But then you corrected with
> I meant 27/(27 + 1024). And just in case I did another typo
> just then, my calculation is
>
> P(A) ~ (1/4)^7 * (3/4)^3
>
> P(B) ~ (1/2)^7 * (1/2)^3 = 1/1024
>
> and the answer is P(A) divided by P(A)+P(B). No?
Mitchell's answer is right about his problem: the case where the
distributions are 25/75 red/blue and 75/25 red/blue respectively. That's
the problem Eliezer meant to pose, as he stated in his later correction.
However, your (corrected) answer is right about your problem, the problem
posed, the case where the distributions are 25/75 red/blue and 75/75
red/blue respectively.
See how hard this is? 30 second Bayesianism indeed. It only leads to
carelessness in the end, I tell you! ;)
> > (I note, for example, that you happened to misread the problem Eliezer
> > posed; you solved a perfectly fine problem instead, but not exactly the
> > problem as posed. That problem has some handy features that which allow
> > you to simplify and solve the problem just a bit faster. [Though I sure
> > as hell wouldn't notice them unless I was working on paper.])
>
> Be sure to provide as few hints as possible. :)
Well, once I saw Eliezer correct his problem to agree with Mitchell, I
thought about it a bit more and realized that the problem Mitchell solved
was, indeed, the easier one; I was wrong.
With that said, I think the 30 second answer (which ONLY applies to
Eliezer's corrected problem, the one Mitchell solved) should be
immediately obvious by looking at Mitchell's attempt at repeatedly
adjusting his expected values. If you need a bigger hint, you can rot13
these clues... I've tried to make them in order of strength. Try
unencoding just one clue at a time.
http://tomsprograms.toms-world.org/php-rot13.php
* Abgvpr jung unccrarq nobhg fvk fgrcf vagb gur vgrengvba. Jul qvq gung
unccra?
* Fvk fgrcf va, gur rkcrpgrq inyhr jnf onpx gb bar unys naq bar unys.
Gung zrnaf gung nyy bs gur cerivbhf erfhygf pbagnvarq ab vasbezngvba.
Vg'f nf vs gurl arire unccrarq.
* Fb jung'f gur snfgrfg jnl gb fbyir gur svany cneg bs gur ceboyrz?
* Qb lbh abgvpr fbzrguvat vagrerfgvat nobhg gur trareny sbez bs Onlrf'
rdhngvba jura C(N) = bar unys? C(N|O) = C(O|N)C(N)/[C(O|N)C(N) +
C(O|~N)C(~N)]
And, finally, the rot13 ANSWER to #2 below. Please forgive my spelling
out numbers, but otherwise you'd see them straight through the rot13
encoding).
Lbh fgneg ol znxvat hfr bs n unaql gevpx nobhg Onlrf jura gur ulcbgurfvf
orvat grfgrq unf cevbe cebonovyvgl bs bar unys. Jura fbyivat guvf unaql
trareny sbez bs gur rdhngvba: C(N|O) = C(O|N)C(N)/[C(O|N)C(N) +
C(O|~N)C(~N)], va gur pnfr jurer C(N) = C(~N) = bar unys, nyy gur C(N)f
naq C(~N)f pnapry. Fb lbh'er yrsg jvgu C(N|O) = C(O|N)/[C(O|N) +
C(O|~N)].
Shegurezber, lbh unir gb abgvpr gung rirel E O cnve pnapryf vgfrys bhg va
grezf bs hfrshy vasbezngvba, orpnhfr n erq pbva fhccbegf gur "erq oneery"
ulcbgurfvf rknpgyl nf fgebatyl nf gur oyhr pbva fhccbegf gur "oyhr oneery"
ulcbgurfvf. Urapr, nyy lbh arrq gb qb vf svther bhg ubj yvxryl vg vf gung
lbh'q trg n fgevat bs sbhe erqf ba gur "oyhr oneery" ulcbgurfvf naq ba gur
"erq oneery" ulcbgurfvf, naq cyht gurz va yvxr guvf: C(oyhr oneery|Sbhe
erqf) = C(Sbhe erqf|oyhr oneery) / [C(Sbhe erqf|erq oneery) + C(Sbhe
erqf|oyhr oneery)]. Rira orggre vs lbh hfr bqqf, orpnhfr gura C(Sbhe
erqf|oyhr oneery) jvyy whfg or bar gb gur sbhegu cbjre, juvpu vf bar. Fb
abj, gur nafjre vf whfg (bar gb gur sbhegu) / [guerr gb gur sbhegu cyhf
bar gb gur sbhegu]. Zbfg crbcyr pna fbyir GUNG bar va gurve urnqf: guerr
gb gur sbhegu vf avar fdhnerq be rvtugl-bar, fb gur nafjre, nf Zvgpuryy
fubjrq, vf bar bire rvtugl-gjb.
-Dan
-unless you love someone-
-nothing else makes any sense-
e.e. cummings
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