How Long the Gambler's Ruin Game Lasts

The problem. You have $$i$ and repeatedly bet $1 on fair coin flips against an opponent who has $$(N-i)$, until one of you is ruined (reaches $0). The classic gambler's ruin question asks: what's the probability you end up with all $$N$? A less commonly asked but equally clean follow-up is: how many bets, on average, does the game last before someone is ruined? That's a genuinely different quantity, and it has its own formula worth knowing.

Setting up the recursion

Let $D_i$ be the expected number of remaining bets when you currently have $$i$ (opponent has $N-i$), with the game ending at $i=0$ or $i=N$. Each bet is one flip, after which you're at $i+1$ or $i-1$ with equal probability, so:

$$D_i = 1 + \tfrac12 D_{i-1} + \tfrac12 D_{i+1}, \qquad D_0 = D_N = 0.$$

The "+1" accounts for the bet you just placed to get there. This is a linear recursion, and the classic solved form for a fair game is:

$$D_i = i(N-i).$$

In plain English: the expected number of bets is your current stake times your opponent's stake — a simple product, peaking when the two players are evenly matched and shrinking toward zero as either side approaches having all the money (little left to fight over).

Worked example: even match vs lopsided match

Say the total pot is $N = 20$ and you start with $i = 10$ (an even match). Expected duration: $D_{10} = 10 \times 10 = 100$ bets. Now suppose you start with only $i = 2$ against an opponent with $18. Expected duration: $D_2 = 2 \times 18 = 36$ bets — much shorter, because with only $2 to lose, you're very likely to be ruined quickly, and there simply isn't much room for a long back-and-forth. Notice the game lasting the longest happens exactly when it's most evenly matched, which matches intuition: an even fight has the most "room" to swing back and forth before anyone runs out.

Worked example: checking against $N=4$, $i=2$ by hand

With $N=4$, formula gives $D_2 = 2 \times 2 = 4$ expected bets. Verify with the recursion directly: by symmetry $D_1 = D_3$. Writing out $D_1 = 1 + \tfrac12 D_0 + \tfrac12 D_2 = 1 + \tfrac12 D_2$ and $D_2 = 1 + \tfrac12 D_1 + \tfrac12 D_3 = 1 + D_1$ (using $D_1=D_3$), substitute: $D_2 = 1 + (1 + \tfrac12 D_2) = 2 + \tfrac12 D_2 \Rightarrow \tfrac12 D_2 = 2 \Rightarrow D_2 = 4$. Matches the formula exactly.

$D_i = i(N-i)$ is a downward-opening parabola in $i$, which is exactly the shape on the explorer above — peaking at the midpoint and falling to zero at both ends. Drag the curve and notice the peak always sits exactly halfway between the roots, matching where $i=N/2$.

The general technique

This is the same "set up a recursion, use boundary conditions, solve" method that answers the win-probability version of gambler's ruin, just tracking a different quantity (expected steps rather than absorption probability). The lesson that generalizes: for any absorbing random walk, both "probability of which boundary you hit" and "expected time until you hit a boundary" satisfy their own linear recursions with the same boundary conditions, and both tend to have clean closed forms for a fair (unbiased) walk.