The game:
$2$ Players take turns trying to climb up $2$ stairs by tossing a coin. They both start on the ground floor (level $0$). Player $1$ goes first. If a heads is tossed a player climbs one step. If a tails is tossed they go back to where they were at the beginning of their turn and loose all progress made this turn. After each toss a player may choose to toss again or "bank" their progress and pass the coin onto their opponent. The winner is the first player to reach level $2.$
Find the probability player $1$ wins under optimal strategy from both players.
My attempt:
All we have to do is determine when players "bank" or keep tossing under the few states of game. I conjecture that no player should ever bank although I think I have a circular proof.
Firstly consider we are both banked at level $1$ and the game has now become the first to toss a head wins. The starting player has $\frac{2}{3}$ probability of winning. This means if your opponent is banked at level $1$ and you have just tossed a head you should not bank. Either you do bank and transition into the first to toss a head as the second player, hence have $\frac{1}{3}$ probability of winning, or you bite the bullet and toss again and have at least $\frac{1}{2}$ probability to win by tossing a head right now and winning.
The only other case to consider is if you should bank when your opponent is on level $0$ and you just tossed a head and are on level $1$.
Knowing that if you do bank, as we saw above, your opponent will never bank and will on every turn toss to win. If you do bank you transfer into a game in which you have $\frac{1}{2}$ probability of winning on your turns and your opponent has $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} $ probability of winning on their turns. However, you are the second player and so have $\frac{3}{5}$ probability to win if you transfer into this game.
If you do decide to toss again you have a $\frac{1}{2}$ probability to win right away and a $\frac{1}{2}$ probability to simply become player $2$ and the game has restarted. We need to however know the probability of winning as player $1$ or $2$ to calculate this. But we need this probability to calculate optimal strategy and to calculate the probability of winning the game as either player to find it. So we are circular?
I got around this by setting up an inequality for the threshold at which Player $1$ needs to win to do bank. Where $p$ denotes Player $1$'s probability of winning under optimal play.
We require $\frac{3}{5} > \frac{1}{2} + \frac{1}{2}\cdot(1- p) \iff p > 0.8$
Assume optimal strategy is to bank then we have Probability of player $1$ winning conditioned on the first $2$ tosses is:
$ p = \frac{1}{2}\cdot\frac{3}{5} + \frac{1}{4} \cdot \frac{2}{5} + \frac{1}{4} p $
Hence $p =\frac{8}{15} < 0.8 $ And so this contradiction tells us not to bank at $0$
So the game simply becomes first person to toss $2$ consecutive heads in which Player $1$ wins with probability $\frac{4}{7}$
Can someone find a non-circular way to do this, or clean up my method to be a bit more watertight?
Examples:
This game I thought of is inspired by, and is an incredibly simplistic version of, the board game Cant Stop.
Here is an example:
Player $1$ tosses tails so remains at level $0$ and it is now Player $2's$ turn
Player $2$ tosses heads and advances to level $1$. He then chooses to toss again, unfortunately it is a tails so they must loose all progress made on this turn and return to level $0$
Player $1$ then tosses heads, advances to floor $1$ and decides to "bank" his progress and not risk anything tossing again.
Player $2$ then tosses heads and again tries to toss again but gets tails so is returned to level $0$
Player $1$ tosses tails so remains at floor $1$ where he is banked to.
Player $2$ tosses tails and they remain on floor $0$
Player $1$ tosses heads, advances to floor $2$ and wins the game