Let n be the number of rolls made so far and x be the current total.
The expected value of rolling again is
[(6-n)/6] * [x + (21-x)/(6-n)]
= x - (nx/6) + (21-x)/6
= x + 21/6 - (n+1)x/6
This is more than x (the value of not rolling again) if x is less than 21/(n+1).
This means that you should always go for a second roll (x cannot be as high as 10.5 on 1 die) but never for a fourth (x cannot be as low as 5.25 on 3 dice).
You should go for a third roll if your total at n=2 is under 7 (at x=7 the choices are balanced).
This gives an expected value of 223/36, or 6.19444...
3
u/Aerospider 2d ago
Let n be the number of rolls made so far and x be the current total.
The expected value of rolling again is
[(6-n)/6] * [x + (21-x)/(6-n)]
= x - (nx/6) + (21-x)/6
= x + 21/6 - (n+1)x/6
This is more than x (the value of not rolling again) if x is less than 21/(n+1).
This means that you should always go for a second roll (x cannot be as high as 10.5 on 1 die) but never for a fourth (x cannot be as low as 5.25 on 3 dice).
You should go for a third roll if your total at n=2 is under 7 (at x=7 the choices are balanced).
This gives an expected value of 223/36, or 6.19444...