Breaking Even in a Coin Game
Lewis and Lilly are playing a game with a fair coin. They each start with $\$5$. The coin will be tossed up to $10$ times. On each toss, if it lands heads, Lewis pays Lilly $\$1$; if tails, Lilly pays Lewis $\$1$. The game ends early if either player's balance hits $\$0$. Otherwise, it continues for all $10$ tosses.
What is the probability that both players finish the game with exactly $\$5$ -- i.e., they break even?
Hints
- Think of Lewis's fortune as a random walk. What must be true about the total number of heads and tails for them to break even?
- You need exactly $5$ heads and $5$ tails, so start with $\binom{10}{5}$ balanced sequences. Now ask: which of these sequences cause the walk to hit $0$ or $10$ before step $10$?
- Check when the walk can reach the boundary. Lewis hits $\$0$ at step $k$ when the number of heads equals $(5 + k)/2$. Given exactly $5$ heads total, which values of $k$ are even feasible?
Worked Solution
How to Think About It: This is a symmetric random walk on $\{0, 1, \ldots, 10\}$ with absorbing barriers at $0$ and $10$, starting at $5$. Lewis's fortune is the walk position. We want the probability that the walk returns to $5$ after exactly $10$ steps without ever hitting $0$ or $10$ along the way. The first instinct: count the paths of length $10$ that end at $5$, subtract the ones that touch a barrier.
Quick Estimate: Total outcomes: $2^{10} = 1024$. Paths ending at $5$ need exactly $5$ heads and $5$ tails: $\binom{10}{5} = 252$ such paths. Now, how many of those hit $0$ or $10$? The walk starts at $5$ and moves $\pm 1$ each step. To reach $0$ or $10$ from $5$, you need $5$ consecutive moves in the same direction. That is extremely constrained -- the only way is $5$ straight heads or $5$ straight tails right from the start (with the opposite $5$ filling the rest). So we expect very few bad paths -- maybe just $2$. That gives roughly $250/1024 \approx 0.244$, about a $1$ in $4$ chance.
Approach: We count the sequences of $10$ fair coin flips with exactly $5$ heads that never cause Lewis's or Lilly's balance to hit $\$0$, then divide by $2^{10}$.
Formal Solution:
Let $H_k$ denote the number of heads in the first $k$ tosses. Lewis's fortune after $k$ tosses is:
$$S_k = 5 - H_k + (k - H_k) = 5 + k - 2H_k$$
The game ends if $S_k = 0$ (Lewis broke) or $S_k = 10$ (Lilly broke) at any step $k \leq 10$.
*When can Lewis hit $0$?* We need $5 + k - 2h = 0$, i.e., $h = (5 + k)/2$. This requires $k$ odd. Checking each odd $k$ with the constraint that we need exactly $5$ heads total in $10$ tosses:
- $k = 1$: $h = 3$ -- impossible (only $1$ toss so far)
- $k = 3$: $h = 4$ -- impossible (only $3$ tosses)
- $k = 5$: $h = 5$ -- possible. All $5$ heads come in the first $5$ tosses: HHHHHTTTTT
- $k = 7$: $h = 6$ -- impossible (only $5$ heads total)
- $k = 9$: $h = 7$ -- impossible
*When can Lilly hit $0$?* We need $S_k = 10$, i.e., $h = (k - 5)/2$. Again $k$ must be odd and $k \geq 5$:
- $k = 5$: $h = 0$ -- possible. All $5$ tails come first: TTTTTHHHHH
- $k = 7$: $h = 1$ -- need $4$ heads in last $3$ tosses, impossible
- $k = 9$: $h = 2$ -- need $3$ heads in last $1$ toss, impossible
So exactly $2$ sequences out of $\binom{10}{5} = 252$ hit a barrier. The valid count is $252 - 2 = 250$.
Answer:
$$P(\text{break even}) = \frac{250}{1024} = \frac{125}{512} \approx 0.2441$$
Intuition
This problem is a classic random walk with absorbing barriers, which shows up everywhere in finance -- gambler's ruin, barrier option pricing, drawdown analysis. The key insight is that the absorbing barriers barely matter here. Out of 252 balanced paths (5 heads, 5 tails), only 2 hit the boundary. Why so few? Because the walk starts at the midpoint of a 10-wide interval and only runs for 10 steps. Reaching a barrier from the center requires 5 consecutive moves in the same direction, which is extremely unlikely given the constraint that the walk must also return to center. In general, when the number of steps roughly equals the width of the corridor and you start near the middle, absorption is rare and the "free" walk is a good approximation.
The enumeration technique here -- count all paths with the right endpoint, then subtract the bad ones -- is the ballot problem / reflection principle in disguise. For more complex barrier problems (longer walks, asymmetric start, two-sided barriers), you would use the reflection principle or the transition matrix of the absorbed walk. But for small cases like this, direct counting is fastest and least error-prone.