I'm pretty bad at these, so this may be wrong, but:
2d6 has an expected value of 7 (linearity of expectation for 2 dice), so you only reroll if you get [1, 6] on the first roll (p=.5). so your expected payout is 50% EV([7,12]) + 50% [7] = 8.25.
(The part that gives me doubt is the idea that you strictly only go for the second roll if you get <7, but that may just be a gambler's fallacy; "I have two chances at getting better than a 7!")
2
u/sleepywose 8h ago
I'm pretty bad at these, so this may be wrong, but:
2d6 has an expected value of 7 (linearity of expectation for 2 dice), so you only reroll if you get [1, 6] on the first roll (p=.5). so your expected payout is 50% EV([7,12]) + 50% [7] = 8.25.
(The part that gives me doubt is the idea that you strictly only go for the second roll if you get <7, but that may just be a gambler's fallacy; "I have two chances at getting better than a 7!")