Two players alternately toss a coin. The one who first gets…
Two players alternately toss a coin. The one who first gets heads wins. Determine the probability that the first player wins and the probability that the second player wins.
Explanation of Method
The first player wins if the first heads appears on toss $1,3,5,\dots$; the second if it appears on toss $2,4,6,\dots$ The first heads is on toss $k$ with probability $\left(\frac12\right)^k$. Adding over odd and even $k$ gives geometric series. A shorter argument: after two tails the game starts afresh.
Step 1: First heads on toss $k$
$k-1$ tails followed by heads: probability $\left(\frac12\right)^k$.
Step 2: The first player
\[ P_1=\frac12+\frac18+\frac1{32}+\dots=\frac{1/2}{1-1/4}=\frac23. \]
Step 3: The second player
\[ P_2=\frac14+\frac1{16}+\frac1{64}+\dots=\frac{1/4}{1-1/4}=\frac13. \]
\[ \boxed{P_1=\tfrac23,\qquad P_2=\tfrac13.} \]
The game ends with probability $P_1+P_2=1$ (it cannot go on forever).
Step 4: A shorter argument
The first player wins immediately with probability $\frac12$. Otherwise ($\frac12$) the second player is in exactly the position the first was in at the start. So $P_2=\frac12P_1$, and with $P_1+P_2=1$: $P_1=\frac23$, $P_2=\frac13$ ✓.
Step 5: General coin
If heads has probability $p$ ($q=1-p$), the same reasoning gives $P_1=\dfrac{p}{1-q^2}=\dfrac1{1+q}$ and $P_2=\dfrac q{1+q}$. The first player always has the advantage.
Result
| Player | Probability of winning |
|---|---|
| first | $\sum(1/2)^{2k-1}=2/3$ |
| second | $\sum(1/2)^{2k}=1/3$ |