# Die rolling expected value game

The game I have in mind goes as follows. You roll can roll a die up to $n$ times, on the $n$th time, you are forced to keep the number on the dice. Is there an explicit formula for calculating the Expected Value of the game. Can this formula be modified for varying sided die (e.g. 200 sided dice).

• What is the criteria for choosing to roll/not roll the die in the next round? – Karn Watcharasupat Nov 18 '17 at 5:11
• Well that's up to you. You are presumably playing optimally to maximize your score. But you can keep rerolling to get a higher score until the nth time, at which you are forced to take what you get on the nth roll – Nujra Nov 18 '17 at 5:12
• If you can choose whenever to stop then the previous values do not matter at all. The expected value of an $s$-sided die is just $(1+2+\ldots+s)/s=(s+1)/2$. – Karn Watcharasupat Nov 18 '17 at 5:13
• thats the EV of just rolling once, but with this game you could also do a reroll, which effects the EV of the game, meaning its higher. For example, if $n$ = 2, then we would reroll on a 3 or lower, and the EV of the second roll is fixed at 3.5. So with 1/2 chance we take 3.5 and with 1/2 chance we take (4+5+6)/3 – Nujra Nov 18 '17 at 5:14
• I would like to close this as a duplicate of this question which was for a d20 but the technique is valid for any size die. I did not find an explicit formula for the value. Does that do what you are looking for? This question involves a six sided die but has less detail. – Ross Millikan Nov 18 '17 at 5:25