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You can roll 2 die a maximum of 2 times. You will be paid the maximum value that you obtain either on the last roll (either first or second roll).
What is the expected value?
May I ask if this is this simply (6+5+4+3+2+1)/6 = $3.50
Or will it involved conditional probabilities
E.g. if the first roll is {1,2,3} then I will most likely roll again.
if it is a {4,5,6} then I will take my money because it is greater than the expected value of $3.50
EDIT: in hindsight this may be the case - expected value of 4.25???
(0.5 x 3.5) + (0.5 x 5) = 4.25
Last edited by lindah (2010-02-09 16:47:12)
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Hi lindah;
EDIT: in hindsight this may be the case - expected value of 4.25???
That is what I am getting as the optimal strategy. Stop at 4 or more on the first roll and get paid. Roll again on 3 or less.
E(stopping at 4 or more) = 4.25
I am not using your formula though.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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When I first read the question, I interpreted it as that you could roll the die twice and got paid the larger roll, without having to gamble.
If that's the case, the expected win is (1*1 + 3*2 + 5*3 + 7*4 + 9*5 + 11*6)/36 = 4.47222...
If having a second throw makes you lose the first though, then I agree with 4.25 being the best answer.
Why did the vector cross the road?
It wanted to be normal.
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Thank you for your prompt responses!!
@mathsyperson
I think the question was geared towards the second interpretation
But thank you for the first scenario as well...it gives me more food for thought
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Hi lindah;
Your welcome. I am interested in whether or not you can compute the other expectations with your formula?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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