[w_alloy]

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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

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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.