FreeProbabilityEasy

Passing The Coin I

Asked atVirtu Financial

Five players, numbered $1$ to $5$ , take turns with a biased coin that shows heads with probability $\tfrac13$ . Player $1$ goes first, then $2$ , $3$ , $4$ , $5$ , then back to $1$ , and so on. On your turn you flip once: heads and you win the game; tails and the coin passes to the next player.

What is the probability that player $1$ wins?

Solution

Try it yourself first. Open a hint if you get stuck.

Hint 1

Player $1$ either wins on the first flip, or all five players flip tails and the coin comes back to player $1$ . What does the game look like at that point?

Hint 2

Let $x$ be player $1$ 's chance of winning. After a full round of tails, player $1$ 's chance from there is $x$ again. Write one equation for $x$ .

Show solution

Let $x$ be the probability that player $1$ wins.

Method 1: back to square one.

On player $1$ 's first turn, one of two things decides what happens next:

Heads (probability $\tfrac13$ ): player $1$ wins immediately.
All five players flip tails, one each (probability $\left(\tfrac23\right)^5$ ): the coin is back with player $1$ , and nothing else has changed. Player $1$ is back to square one, so from here they win with probability $x$ again.

In every other case, one of players $2$ to $5$ flips heads first and player $1$ loses. So

x = \frac13 + \left(\frac23\right)^5 x.

Solve, writing $\left(\tfrac23\right)^5 = \tfrac{32}{243}$ :

x\left(1 - \frac{32}{243}\right) = \frac13 \quad\Longrightarrow\quad x \cdot \frac{211}{243} = \frac13 \quad\Longrightarrow\quad x = \frac{81}{211}.

The same answer as a series. Player $1$ can win on their 1st turn, their 2nd turn (after one full round of tails), their 3rd turn (after two rounds), and so on:

x = \frac13 + \left(\frac23\right)^5 \frac13 + \left(\frac23\right)^{10} \frac13 + \cdots = \frac13 \cdot \frac{1}{1 - (2/3)^5} = \frac{81}{211}.

This is what you get by substituting the equation above into itself again and again. The one-line equation is quicker and harder to get wrong.

Answer:

x = \frac{81}{211} \approx 0.38389.

Method 2: compare the players. This gives all five players' chances at once, and it is the idea Passing The Coin II uses.

Write $r = \tfrac23$ and $P_k$ for the chance that player $k$ wins. For player $2$ to win, player $1$ must first flip tails (probability $r$ ). After that, player $2$ holds the coin at the start of their turn, which is exactly player $1$ 's situation at the very start. So $P_2 = r\,P_1$ , and in the same way $P_3 = r\,P_2$ , and so on:

P_1,\; P_2,\; P_3,\; P_4,\; P_5 \;=\; P_1,\; rP_1,\; r^2P_1,\; r^3P_1,\; r^4P_1.

Someone wins with probability $1$ , so these add up to $1$ :

P_1\left(1 + \tfrac23 + \tfrac49 + \tfrac{8}{27} + \tfrac{16}{81}\right) = P_1 \cdot \frac{211}{81} = 1 \quad\Longrightarrow\quad P_1 = \frac{81}{211}.

Going first is worth a lot: player $1$ wins about $38\%$ of the time, more than twice as often as player $3$ and five times as often as player $5$ .

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