I guess this is a pretty simple one, but just wanted to see how others would approach it and phrase the solution when asked at an interview for example... The setup: a 6-sided fair die is rolled, and payout equal to the face shown. How much do you pay to play? Answer, $3.50 of course. The question: now you get the option for two rolls. You can stop at the first roll, or if you don't like the result of the first, you can roll again. You only get the payout of the 2nd roll if you do that. How much should you pay to play?thanks, all!

Let d_1, d_2 be what we get in rolls 1 and 2 respectively. Let p(s)=I_{d_1<=s}d_2+I_{d_1>=s+1}d_1. We need to maximize Ep(s) wrt to s, which gives s^*=3 and Ep(s^*)=4.25.

Justin's absolutely right, we should toss second time if d1 less than 4and expected payoff is $4.25. I also checked by Monte Carlo.Did you agree on a wrong answer of $3.50 in an interview?))

oh. oops, is it really not $3.50 for the simple case? I got that answer by 1/6 * (1 + ... + 6) = 21/6 = $3.50, can I have gotten even that part wrong? And then my reasoning for the 2nd problem, which is why I hope to understand how to better approach it, was: On the first roll, you will accept the result if it is 4, 5, or 6, and roll again if it is 1, 2, or 3. So the value of the first roll is at least 1/6 * (1 + 2 + 3) = $1. Add this to the average payout of the 2nd roll, which is the same as the simple case since you must accept the result. So then the total price you should pay is $1 + $3.50 = $4.50. I guess this is incorrect. Can you help me as I try to expand the notation you gave below? p(s)=I_{d_1<=s}d_2+I_{d_1>=s+1}d_1amount to pay = (prob of bad 1st roll) * (avg payout on 2nd) + (prob of good 1st roll) * (avg payout of good 1st roll) ( what is the I_ notation?) choosing good/bad break even point as 4/3= (likelihood of 1,2,3) * ($3.50?) + (likelihood of 4,5,6) * ($5)= (1/2) * ($3.50) + (1/2) * ($5) = $4.25huh, that is correct now?

supernova87ap(s) is a payoff of the game when you roll twice with decision threshold s. So if you roll s or less first time you roll the die again. p(s) is a random variable so we have to calculate it's expectation which will give the fair price of the game (for s*). I_{A} is the indicator of event A. It equals 1 if A happens and 0 otherwise.Your solution looks fine as well.

thanks for the information -- I also just wrote a little random number program to entertain myself that it works!

Ok, so what is the value if you can now have 3 throws, or four....Answer must tend towards 6 as the number of throws increases but what number do you now stop at on each throw.

QuoteOriginally posted by: thomssiOk, so what is the value if you can now have 3 throws, or four....Answer must tend towards 6 as the number of throws increases but what number do you now stop at on each throw.Quick and dirty analysis, please correct me if I'm wrong:Let E[g(n)] denote the expected value of the game that allows n max rolls. Let n>0 and assume E[g(0)]=0 (e.g. no payout if you don't play). Then,E[g(n)] = E[ g(n-1) ]*P( x<E[g(n-1)] ) + Sum( x*P(x), for x>=E[g(n-1)] ).So we have...Case n=1:E[g(1)] = E[ g(0) ]*P( x<E[g(0)] ) + Sum( x*P(x), for x>=E[g(0)] ) = 0 + 1/6*(1+2+3+4+5+6) = 3.5Case n=2:E[g(2)] = E[ g(1) ]*P( x<E[g(1)] ) + Sum( x*P(x), for x>=E[g(1)] ) = 3.5*(3/6) + (4+5+6)*1/6 = 4.25Case n=3:E[g(3)] = E[ g(2) ]*P( x<E[g(2)] ) + Sum( x*P(x), for x>=E[g(2)] ) = 4.25*(4/6) + (5+6)*1/6 = 4.67(in case notation is confusing: P( x<k ) is the probability that you roll a number x on the dice that is less than some number k).

That is indeed correct.