Probability of Exactly One Win Before Bankruptcy
Fred is playing a game that costs $\$1$ per round. Each round, he wins with probability $p$ (where $0 < p < 1$) and loses with probability $1 - p$, independently. If he wins a round, he receives $\$k$ (an integer with $k > 1$), but he does not get his $\$1$ entry fee back. If he loses, he simply loses his $\$1$.
Fred starts with $\$s$ in his bank (an integer with $s \geq 1$). He keeps playing until he goes bankrupt (hits $\$0$).
Find the probability that Fred wins exactly once before going bankrupt.
Evaluate your answer for $s = 3$, $k = 4$, and $p = 1/3$. Round to the nearest ten-thousandth.
Hints
- Fred runs out of money after $s$ rounds if he never wins. So his single win must happen early enough -- when exactly is the deadline?
- If the win happens in round $j$, trace Fred's bankroll to figure out how many total rounds he plays. You should find the total does not depend on $j$.
- Every valid outcome has exactly one win (in positions 1 through $s$) and $s + k - 1$ losses. There are $s$ such outcomes, each with probability $p(1-p)^{s+k-1}$.
Worked Solution
How to Think About It: The key structural observation is about *when* the single win can happen. Fred starts with $s$ dollars and loses $\$1$ per round, so if he never wins, he is bankrupt after round $s$. That means his one win must occur in one of the first $s$ rounds -- otherwise he is already broke. Once you see that constraint, the rest is just counting: how many total rounds does he play, and how many ways can that single win be placed?
Quick Estimate: With $s = 3$, $k = 4$, $p = 1/3$, Fred's one win must come in rounds 1, 2, or 3. The total game length is $s + k = 7$ rounds (we will verify below). So we need exactly 1 success in 7 trials, but restricted to the first 3 positions. Each valid configuration has probability $p(1-p)^6 = (1/3)(2/3)^6 = 64/2187$, and there are 3 such configurations, giving $3 \times 64/2187 = 192/2187 \approx 0.0878$. That is our target.
Approach: Count the total rounds played given exactly one win, then compute the probability by summing over the valid win positions.
Formal Solution:
Suppose Fred's single win occurs in round $j$, where $1 \leq j \leq s$. Track his bankroll:
- Before round $j$: he has lost $j - 1$ rounds, so his bankroll is $s - (j - 1) = s - j + 1$.
- He pays $\$1$ to play round $j$ and wins $\$k$, so his bankroll becomes $s - j + k$.
- He then loses every subsequent round. Starting from $s - j + k$ dollars, he goes bankrupt after exactly $s - j + k$ more losing rounds.
The total number of rounds played is $j + (s - j + k) = s + k$, regardless of when the win occurs. This is an important simplification: every valid outcome has the same length.
Since the game lasts $s + k$ rounds with exactly one win and $s + k - 1$ losses, each such outcome has probability:
$$p \cdot (1 - p)^{s + k - 1}$$
The win must fall in one of the first $s$ rounds (otherwise Fred goes bankrupt before winning). There are $s$ such positions, so the total probability is:
$$P(\text{exactly one win before bankruptcy}) = s \cdot p \cdot (1 - p)^{s + k - 1}$$
Plugging in $s = 3$, $k = 4$, $p = 1/3$:
$$3 \cdot \frac{1}{3} \cdot \left(\frac{2}{3}\right)^{6} = 1 \cdot \frac{64}{729} = \frac{64}{729} \approx 0.0878$$
Answer: The probability of exactly one win before bankruptcy is $s \cdot p \cdot (1 - p)^{s+k-1}$. For $s = 3$, $k = 4$, $p = 1/3$, this equals $64/729 \approx 0.0878$.
Intuition
This problem is really about a constraint on timing. The win is useless if it comes too late -- Fred must win before his bankroll hits zero, which gives him exactly $s$ chances. Once you impose that constraint, a nice simplification emerges: regardless of when the single win occurs, the total game length is always $s + k$ rounds. That is because the win injects $k$ dollars of "fuel" into the bankroll at the cost of one round, and the net effect on total game length is the same no matter when it happens. So you are just counting arrangements of one success in $s + k$ trials, restricted to the first $s$ positions.
This pattern shows up frequently in gambler's ruin and first-passage problems. The broader lesson is that when you have a process with a fixed payoff structure, the total path length often depends only on the number of wins and losses, not their order. Recognizing that symmetry early saves you from a painful case-by-case analysis and lets you write down the answer in one line.