- 24 Sep '04 01:52 / 1 editAnswer 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 of the balls are red and 200 of the balls are either blue or green (Note: This does not imply that all 200 balls are the same color. It may be that some are blue and the others green. Maybe all are green. Maybe all are blue.)

Now there are two gambles, Gamble A and Gamble B.

Gamble A: You receive $1,000 if the ball is red.

Gamble B: You receive $1,000 if the ball is blue.

Which gamble do you prefer?

Ok. Got one?

Now choose from two different gambles.

Gamble C: You receive $1,000 if the ball is not red.

Gamble D: You receive $1,000 if the ball is not blue.

Which gamble do you prefer?

Please give your first responses to the questions (i.e. Please do not reconsider any question after answering it.)

Thanks for you participation.

I'll keep a tally and fill every interested party in on this in a day or two. - 24 Sep '04 03:53 / 2 edits

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.*Originally posted by bbarr***Don't you mean that choosing A over B [b]or**C over D is*arbitrary*?[/b]

I think alternative systems for making such a decision exist, I just don't know what they are. - 24 Sep '04 04:29

Royal don't spoil the fun . . .*Originally posted by royalchicken***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.** - 24 Sep '04 16:35

Without 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.*Originally posted by telerion***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. - 24 Sep '04 16:36

Go go gagdet Bayesian probability!*Originally posted by royalchicken***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. - 24 Sep '04 17:08

Yeah going through some of that right now.*Originally posted by royalchicken***My parents and I are about to pay large sums to have a guy in a cardigan explain that to me .**

Professor: You silly students thought that after a year you had econometrics down? All the hard stuff was behind you? Forget it! Now were going to change up the statistical assumptions and start fresh!

Me: Uh . . . I had classical econometrics down?