logic problem (fixed wording cuz i suck)

[yoshi_yoshi]

[ QUOTE ]
answer:
Pair 1 is a
Pair 2 is a, due to the calcelation principle, meaning if in 2 decisions there is a common element that would produce the same outcome it would be ignored.

[/ QUOTE ]

why?

if a is correct because you want to reduce variance, and the 'trick' you mentioned above is the calcelation principle, that's hilarious.

[Sephus]

[ QUOTE ]
answer:
Pair 1 is a
Pair 2 is a, due to the calcelation principle, meaning if in 2 decisions there is a common element that would produce the same outcome it would be ignored.

[/ QUOTE ]

i'm not an expert on probability theory, but for "pair 2" it seems to me that you will win 2/3 of the time if you choose a, and 2/3 of the time if you choose b.

do you think by choosing a you win more than 2/3 of the time?

[Sephus]

[ QUOTE ]
[ QUOTE ]
answer:
Pair 1 is a
Pair 2 is a, due to the calcelation principle, meaning if in 2 decisions there is a common element that would produce the same outcome it would be ignored.

[/ QUOTE ]

why?

if a is correct because you want to reduce variance, and the 'trick' you mentioned above is the calcelation principle, that's hilarious.

[/ QUOTE ]

i dont think you can reduce variance. both a and b have the same probability of winning you the $100 on each trial, right? the added "step" of uncertainty doesn't affect variance.

[w_alloy]

[ QUOTE ]

let B the probability of choosing a black ball and (1-B) be the probability of choosing a yellow ball.

In Pair 1, if you choose red you get 100*(1/3). if you choose black you get 100*(B). since the proportion of B is unknown, it ranges from either 0 to 60. in this case, if the distribution of black balls is randomly chosen uniformally over the range, then you end up with 30, or 30/90 which will give you the same payoff as red.

in Pair 2

holding (1-B) constant, you get

100*(1/3)+(1-B)*100.

if you choose b) you get 100*B+100*(1-B).

again, if the distribution is randomly chosen to be uniform accross the range you will end up with the same thing. we can make assumptions and set limits like if B<30 then 1-B >30 so a red or yellow ball is more likely than a blakc or yellow ball.

in any case, the answer is, assuming equiprobable outcomes for the proportions of yellow:black that you are indifferent between the options.

Barron

[/ QUOTE ]

I was about to post something very similiar to this but with poorer wording. The OP's answer seems very obviously wrong. This seems very simple.

[MINETZ]

I fully agree that the answer seems wrong. My girlfriend gave me this question and I would not believe the answer, stupid account book...

[DcifrThs]

[ QUOTE ]
I fully agree that the answer seems wrong. My girlfriend gave me this question and I would not believe the answer, stupid account book...

[/ QUOTE ]

the problem as you stated it and the answer as you gave it do not match.

unless there is a priori info on the distributions then the choices int he pairs yeild exactly the same expectation given the assumptions i made.

also, i dont know what the "cancellation" principle is but i looked it up on my most trusted math site "mathworld" and it wasn't even there!! there was Anomalous Cancellation and Cancellation but the former involves looking at tables of factors and prime #s and the latter involves taking Mods of numbers to get an equivalent answer with different bases of the Modular arithmetic.

so if mathworld doesn't list this cancellation principle, i dont think its relavent or even exhists in the world of mathematics.

Barron

[d10]

This isn't really a logic problem. This problem was presented in "Risk, Ambiguity, and the Savage Axioms" (Ellsburg 1961) to demonstrate that given two situations with equal expected values, people generally show a tendency to avoid ambiguity. Most people would choose (a) out of Pair 1, and (b) out of Pair 2.

[Benholio]

[ QUOTE ]
I fully agree that the answer seems wrong. My girlfriend gave me this question and I would not believe the answer, stupid account book...

[/ QUOTE ]

You need to get ahold of this book, I don't think anyone is convinced that your answer is correct. Or maybe expand on your earlier answers. Starting with question #1, why is a) better than b)?

[Benholio]

[ QUOTE ]
This isn't really a logic problem. This problem was presented in "Risk, Ambiguity, and the Savage Axioms" (Ellsburg 1961) to demonstrate that given two situations with equal expected values, people generally show a tendency to avoid ambiguity. Most people would choose (a) out of Pair 1, and (b) out of Pair 2.

[/ QUOTE ]

This makes sense.

[CardSharpCook]

it is a variance question. In each question the EV is neutral, but in question 1 you can pick a known 1/3 chance or an unknown chance that averages to 1/3. In question 2, a known 2/3 chance of getting paid or an unknown chance that averages to 2/3.