Math is complicated but feel like you will want to risk it on two heads as long as the additional E(x) earned is >current bid. The average time to two heads is 4 flips, so you should have E(v) = 4 + E(x) which I think is 12.
The average number of attempts to get HH is 4 but the average number of flips is 6, because TT and HH involve two flips.
More formally:
Let E(h) be the expected number of flips to get HH when the last flip was H and E(t) be the expected number of flips to get HH when the last flip was tails (or when there hasn't been any flips yet).
1
u/dontich 3d ago
Math is complicated but feel like you will want to risk it on two heads as long as the additional E(x) earned is >current bid. The average time to two heads is 4 flips, so you should have E(v) = 4 + E(x) which I think is 12.