Solution
Optimal Dice Stopping
Show the problem again
You roll a fair six-sided die up to three times. After each roll you may stop and take the value shown as your payout, or discard it and roll again. If you reach the third roll, you must take whatever it shows. Playing optimally, what is your expected payout? Express as a fraction.
Worked solution
The answer is 14/3 ≈ 4.67. Work backwards. On the third roll you take what comes: expected value 3.5. On the second roll, keep any value above 3.5, meaning a 4, 5, or 6 (average 5); otherwise continue: EV = (1/2)(5) + (1/2)(3.5) = 4.25. On the first roll, keep any value above 4.25, meaning a 5 or 6 (average 5.5); otherwise continue: EV = (1/3)(5.5) + (2/3)(4.25) = 14/3.
Source: Standard optimal-stopping exercise from introductory probability texts. Statement written for AxiomIQ.