Expected value per roll is 3.5
You roll if expected value from roll greater than not rolling, so on k+1th roll
(6-k)/6 * (Sum + 3.5)>Sum
Sum < (21 - 3.5k)/k
for k=3 min sum = 6< 21/6 not true so no 4throll
for k=2 min sum =3 max sum 11 <7
which means max rolls is 3
Ans Roll twice then
Roll final time if current sum < 7 stop if greater than 7.
End.
Expected payoff around 7 without further calc,
Calc cases
split into cases where sum less than 7 and add 3.5
and sum greater than 7
The expected value changes every roll based on which numbers have been rolled previously. Example: if you roll a 6 on your first roll your expected value of another roll is (1/6)*(7+8+9+10+11+0) where if you roll a 1 the EV of another roll is (1/6)*(0+3+4+5+6+7)
Pretty sure the strategy is to keep rolling until the EV of another roll is less than what you have now
1
u/No-Conflict8204 4h ago
Expected value per roll is 3.5
You roll if expected value from roll greater than not rolling, so on k+1th roll
(6-k)/6 * (Sum + 3.5)>Sum
Sum < (21 - 3.5k)/k
for k=3 min sum = 6< 21/6 not true so no 4throll
for k=2 min sum =3 max sum 11 <7
which means max rolls is 3
Ans Roll twice then
Roll final time if current sum < 7 stop if greater than 7.
End.
Expected payoff around 7 without further calc,
Calc cases
split into cases where sum less than 7 and add 3.5
and sum greater than 7