Penney's game

aka Penney Ante

P1 and P2 decides on a sequence of 3 H and T.

We start throwing coins, the one whose sequence appears first wins.

For any P1 choice, P2 has a way to construct a choice with higher odds.

Say P1 is abc, the P2 shoudl use ~bab.

Proof:

consider the tosses 1 2 a b c, where p1 wins at the end but not earlier
p2 wants to win one step earlier, thus p2 = xab
p1 cannot win at 1 2 3
if x is b, then it could be p1’s win if a=c, this is invalid
- 1 x a b c
- a b c
it must be the case x is ~b