Passing The Coin II

After player $1$ flips tails, the coin goes to player $2$. How does the game from that moment compare with a brand-new game?

Player $3$ is in the "goes first" seat once players $1$ and $2$ have both flipped tails. What is that probability, and what is the first player's chance of winning (Passing The Coin I)?

Step 1: The game has no memory, so you can relabel the players. Suppose player $1$ flips tails. The coin goes to player $2$, and from that moment the order is $2, 3, 4, 5, 1, 2, \dots$ The earlier tail changes nothing about the coin. So this is exactly a brand-new game in which player $2$ sits in the "goes first" seat. Rename player $2$ as "player $1$", player $3$ as "player $2$", and so on, and you can't tell it apart from a fresh start.

Step 2: Get player 3 into the "goes first" seat. Player $3$ is in that seat once players $1$ and $2$ have both flipped tails, which happens with probability

If either of them flips heads first, they win and player $3$ can't.

Step 3: Use Passing The Coin I. From the "goes first" seat, a player wins with probability $\tfrac{81}{211}$, the answer to Part I. So

The bigger picture: each player's chance is $\tfrac23$ of the previous player's. The same relabelling works for every seat. Player $k$ reaches the "goes first" seat after $k - 1$ tails, so each player is one tail further from the start than the player before them:

• player $1$: $\tfrac{81}{211}$
• player $2$: $\tfrac23 \cdot \tfrac{81}{211} = \tfrac{54}{211}$
• player $3$: $\tfrac23 \cdot \tfrac{54}{211} = \tfrac{36}{211}$
• player $4$: $\tfrac{24}{211}$
• player $5$: $\tfrac{16}{211}$

Check: $81 + 54 + 36 + 24 + 16 = 211$, so the five chances add up to $1$.

Another way: back to square one. Let $x$ be player $3$'s chance, counted from the very start. Player $3$ wins on their first turn with probability $\left(\tfrac23\right)^2 \cdot \tfrac13 = \tfrac{4}{27}$. If all five players flip tails once, the coin is back with player $1$ and the game is exactly as it started, so player $3$'s chance from there is $x$ again:

A common slip: writing $\left(\tfrac23\right)^7 x$ instead of $\left(\tfrac23\right)^5 x$. Getting back to the starting position takes one full round of $5$ tails. The first two tails, from players $1$ and $2$, are already inside $x$, because $x$ is counted from the start.

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