#1
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logic problem (fixed wording cuz i suck)
suppose an urn contains 90 balls. thirty of these balls are red, and the remaining 60 balls are either black or yellow, in unknown proportions. one balls is to be drawn from the urn, and the color of the ball will determine your payoff. for each of the following 2 pairs choose the alternative you perfer.
Pair 1 a) you will win 100 dollars if a red ball is chosen. b) you will win 100 if a black ball is chosen Pair 2 a) you will win 100 dollars if either a red or a yellow ball is chosen b) you will win 100 dollars if either a black or yellow ball is chosen |
#2
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Re: logic problem
If the yellow/black in the first problem is random, it's neutral EV.
The second one is also neutral between the two options as long as it's random. |
#3
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Re: logic problem
There is a joke here about blue balls, but I can't think of it
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#4
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Re: logic problem
no one? seems obviously but there is a trick to it, hint Account 201.
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#5
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Re: logic problem
your phrasing of these problems is terrible. but i see you are at purdue, so it is understandable.
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#6
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Re: logic problem
1. red.
2. black or yellow. i'm pretty sure i get the mean result each time, so whats the trick? |
#7
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Re: logic problem (fixed wording cuz i suck)
I really don't understand how this is a logic problem.
Pair 1 gives you $100 if a red or black ball is chosen. Pair 2 gives you $100 no matter what's chosen. .....is there something I'm missing? |
#8
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Re: logic problem (fixed wording cuz i suck)
[ QUOTE ]
I really don't understand how this is a logic problem. Pair 1 gives you $100 if a red or black ball is chosen. Pair 2 gives you $100 no matter what's chosen. .....is there something I'm missing? [/ QUOTE ] |
#9
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Re: logic problem (fixed wording cuz i suck)
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. |
#10
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Re: logic problem (fixed wording cuz i suck)
[ QUOTE ]
suppose an urn contains 90 balls. thirty of these balls are red, and the remaining 60 balls are either black or yellow, in unknown proportions. one balls is to be drawn from the urn, and the color of the ball will determine your payoff. for each of the following 2 pairs choose the alternative you perfer. Pair 1 a) you will win 100 dollars if a red ball is chosen. b) you will win 100 if a black ball is chosen Pair 2 a) you will win 100 dollars if either a red or a yellow ball is chosen b) you will win 100 dollars if either a black or yellow ball is chosen [/ 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 |
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