Circular Coin-Passing Game

Probability · Medium · Free problem

$n \geq 2$ players labeled

$ through $n$ sit in a circle and take turns flipping a coin that lands heads with probability $p \in (0,1)$. Player
$ flips first. If they get heads, they win. If they get tails, they pass the coin to player
$, who flips, and so on. After player $n$ flips tails, the coin returns to player
$ and the cycle repeats.

The first player to flip heads wins the game.

Find the probability that player

$ wins. Evaluate your answer for $n = 5$ and $p = 1/3$.

Hints

  1. Think about what has to happen for player 1 to get a second turn. What does the game look like at that point compared to the start?
  2. Condition on player 1's first flip. If it is tails, the coin must pass through all $n - 1$ other players without a heads before the game effectively resets.
  3. Write $\alpha = p + (1-p)^n \cdot \alpha$ and solve for $\alpha$. The cyclic structure means the game state after a full round of tails is identical to the initial state.

Worked Solution

How to Think About It: Player 1 gets the first crack at winning, which is a big advantage -- they flip before anyone else, and if they miss, they have to wait for everyone else to also miss before they get another turn. The game has a clean recursive structure: if the coin makes it all the way around without anyone hitting heads, the situation resets to exactly where it started. That means you can write one equation for player 1's win probability, condition on the first flip, and solve. The key probability to track is $(1-p)^n$ -- the chance that nobody hits heads in a full round.

Quick Estimate: With $p = 1/3$ and $n = 5$, each player has a

/3$ chance of tails. The probability all 5 players get tails in one round is $(2/3)^5 = 32/243 \approx 0.132$. Player 1 wins on the first flip with probability
/3 \approx 0.333$. If we ignore second-round effects, player 1 wins at least $0.333$ of the time. For a rough upper bound, player 1 gets a geometric series of chances: $p + p(1-p)^n + p(1-p)^{2n} + \cdots = p/(1-(1-p)^n)$. Plugging in: $p = 1/3$, $(1-p)^n = 32/243$, so the denominator is
- 32/243 = 211/243$. The answer is $(1/3)/(211/243) = 243/(3 \times 211) = 81/211 \approx 0.384$. That makes sense -- player 1 has about a 38% chance, which is better than the naive
/5 = 20%$ share because going first is a significant advantage.

Approach: Use the law of total probability, conditioning on the outcome of player 1's first flip, and exploit the cyclic structure of the game.

Formal Solution:

Let $\alpha$ denote the probability that player 1 wins. On player 1's turn, two things can happen:

  • With probability $p$, player 1 flips heads and wins immediately.
  • With probability
    - p$, player 1 flips tails and passes the coin to player 2.

For player 1 to get another chance, all remaining players

, 3, \ldots, n$ must also flip tails. Each flips tails independently with probability
- p$, so the probability of a full round of tails (after player 1's miss) is $(1-p)^{n-1}$. If that happens, the game resets to its initial state and player 1's win probability from that point is again $\alpha$.

If any player

$ through $n$ flips heads during the round, that player wins and player 1 loses.

By the law of total probability:

$\alpha = p + (1 - p) \cdot (1 - p)^{n-1} \cdot \alpha = p + (1-p)^n \cdot \alpha$

Solving for $\alpha$:

$\alpha - (1-p)^n \cdot \alpha = p$

$\alpha = \frac{p}{1 - (1-p)^n}$

This is recognizable as a geometric series: player 1 wins on round $k$ (for $k = 0, 1, 2, \ldots$) with probability $p \cdot ((1-p)^n)^k$, and summing gives the same formula.

Plugging in $n = 5$, $p = 1/3$:

$(1 - p)^n = \left(\frac{2}{3}\right)^5 = \frac{32}{243}$

$\alpha = \frac{1/3}{1 - 32/243} = \frac{1/3}{211/243} = \frac{243}{3 \times 211} = \frac{81}{211}$

Answer: The probability that player 1 wins is $\dfrac{p}{1 - (1-p)^n}$. For $n = 5$ and $p = 1/3$, this gives $\dfrac{81}{211} \approx 0.3839$.

Intuition

This problem is a clean example of exploiting self-similarity in a stochastic process. The game has no memory beyond whose turn it is, so after a full round of tails the state is identical to the starting state. That lets you collapse an infinite sequence of rounds into a single equation. This trick -- writing a value in terms of itself and solving -- shows up constantly in quant work: pricing perpetual options, computing expected hitting times for random walks, and valuing any asset with a repeating payoff structure.

The result also illustrates the first-mover advantage in sequential games. Player 1's win probability $p/(1-(1-p)^n)$ is always strictly greater than

/n$ (the "fair share") because they get to flip before anyone else. As $n$ grows, the advantage shrinks because the chance of a full round of tails becomes tiny and player 1 rarely gets a second turn -- but the first-flip edge never disappears. In trading, this is analogous to the advantage of being first to act on information: even a small priority edge compounds into a meaningful probability advantage.

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