Originally posted by bbarrSimultaneously saying A>B and C>D entails a contradiction. Otherwise, classical probability is totally unable to deal with this question because the distributions are not given, or even hinted at.
Don't you mean that choosing A over B [b]or C over D is arbitrary?[/b]
Originally posted by royalchickenRoyal don't spoil the fun . . .
Saying A better than B is saying P(red) > P(blue) while C over D is saying P(red,green) < P(blue,green). Since P(any two colors simultaneously) = 0, this is equivalent to saying P(blue) > P(red), a truly Feivelian construction.
Originally posted by telerionWithout doing the calculations and with a fairly obvious assumption, I'd guess that (B,D) is the worst choice as it maximises variance, and the others are equally good. However, if we're allowed to know the result of the first gamble before picking the second, then I'd go B, followed by D if I lost the first gamble and C if I won it.
Answer the following two questions.
Please limit your answers to the options provided. No explainations from you needed at this point. Just a one-letter response is sufficient. Please do not post questions. Instead message them to me.
Ok here goes.
The following description applies to both questions.
There is an urn that contains 300 balls. 100 ...[text shortened]... rticipation.
I'll keep a tally and fill every interested party in on this in a day or two.
Originally posted by royalchickenGo go gagdet Bayesian probability!
Simultaneously saying A>B and C>D entails a contradiction. Otherwise, classical probability is totally unable to deal with this question because the distributions are not given, or even hinted at.
I think alternative systems for making such a decision exist, I just don't know what they are.
Originally posted by royalchickenYeah going through some of that right now.
My parents and I are about to pay large sums to have a guy in a cardigan explain that to me .