The Gambler’s Ruin

You hold some units of money and the casino holds some more. Each round you stake one unit on a bet: you win it with a fixed chance, say a half, and otherwise lose it. Your bankroll goes up by one when you win and down by one when you lose, and the casino's moves the other way. The game stops the first time either of you has nothing, so it stops when your bankroll reaches 0, which is ruin, or when it reaches the total of both bankrolls, which is the casino's ruin.

The picture follows twenty games from the same start. Each thin line is one game, a bankroll wandering up and down between two walls until it touches one. Lines that touch the lower wall turn red and lines that touch the upper wall turn blue. Count them. At the opening setting, ten units each and a fair coin, about half should end in each colour, and the exact chance of ruin is printed beside the picture: 50%.

Now give the casino more money and keep the coin fair. Nothing about a single round has changed, and yet the red lines multiply.

A fair bet is worth exactly nothing on average: you are as likely to win a unit as to lose one. So over any number of rounds your expected bankroll is the one you started with. That sounds like safety, but an average can be made of very different endings. If you start with 10 units against a casino holding 90, you finish either with 0 or with 100. The only way the average can stay at 10 is for you to finish with 100 about one time in ten, and with 0 the other nine.

That is the whole effect for a fair bet. Your chance of ending ruined is the casino's share of the money, and your chance of taking everything is your own share. The walls are absorbing: once the bankroll touches one it stays there, so every game ends in one of two ways and the only question is how the probability splits between them.

The picture of the chance of ruin shows the split for every starting bankroll with the same total. With a fair bet it is a straight line sloping from certain ruin at 0 units to no ruin at the total.

Whether a bet is fair is a statement about a single round. Who ends up with everything is a statement about the whole game, and that depends on the walls. A gambler with a modest bankroll against a large casino is close to the lower wall and far from the upper one, so a random walk with no lean either way is far more likely to reach the near wall first.

If the bet tilts toward the casino, the tilt works on top of that. A bet that wins 45% of rounds loses only a tenth of a unit per round on average, which looks like nothing, yet against an equal bankroll of ten units it ruins you 88% of the time, because the tilt acts on every one of the many rounds that the game lasts. A tilt of one point, 49% against 51%, ruins a gambler with 50 units against 50 about 88% of the time too.

So both things matter, and they trade off. Fair against a richer casino ruins you almost as surely as a slightly unfair bet against an equal one: 90% for ten units against ninety, and 88% for ten against ten at 45%. The laboratory lets you move each separately and see the other change.

Let a gambler hold $i$ units and the casino $N-i$, so the total is $N$. Each round the gambler wins one unit with probability $p$ and loses one with probability $q=1-p$. Let $r_i$ be the probability that the bankroll reaches $0$ before it reaches $N$ when starting from $i$, and let $D_i$ be the expected number of rounds until it reaches either. Put $\rho =q/p$.

For $p\ne \frac{1}{2}$,

and for the fair bet $p=\frac{1}{2}$,

The expected final bankroll is $N(1-r_i)$, which equals $i$ exactly when the bet is fair.

Against a casino with no limit, letting $N\to \infty$, the ruin probability tends to $1$ when $p\le \frac{1}{2}$ and to ${\rho }^i$ when $p>\frac{1}{2}$.

Look at the first round. With probability $p$ the bankroll moves to $i+1$ and with probability $q$ it moves to $i-1$, and after that the game starts afresh from there, so for $0<i<N$

Since $p+q=1$ this is $p(r_{i+1}-r_i)=q(r_i-r_{i-1})$. Write $d_i=r_{i+1}-r_i$. Then $d_i=\rho d_{i-1}$, so $d_i={\rho }^i{d}_0$. The differences add up to $r_N-r_0=-1$, so $d_0(1+\rho +\cdots +{\rho }^{N-1})=-1$. For $\rho \ne 1$ this gives $d_0=-(1-\rho )/(1-{\rho }^N)$, and then

For $\rho =1$ every $d_i$ is the same, equal to $-1/N$, and $r_i=1-i/N$.

One thing needs saying: the game must end. From any bankroll, $N$ wins in a row reach the upper wall, and that has probability at least $p^N>0$. So the chance that the game is still going after $kN$ rounds is at most $(1-p^N{)}^k$, which tends to 0, and ruin and breaking the bank are the only two endings. That is why $1-r_i$ is the chance of taking everything.

For the duration, the first round gives $D_i=1+p{D}_{i+1}+q{D}_{i-1}$ with $D_0=D_N=0$. For the fair bet, substitute $D_i=i(N-i)$: $1+\frac{1}{2}(i+1)(N-i-1)+\frac{1}{2}(i-1)(N-i+1)=1+iN-i^2-1=i(N-i)$. For $p\ne \frac{1}{2}$, $i/(q-p)$ solves the equation with the $1$, since $p(i+1)+q(i-1)=i-(q-p)$, and the solutions of the same equation without the $1$ are $A+B{\rho }^i$. The two boundary values fix $A$ and $B$ and give the formula above.

The check on the fair bet's average: the gambler finishes with $N$ units with probability $i/N$ and with $0$ otherwise, so the average is $N\cdot i/N=i$, the starting bankroll.

The kernel evaluates each formula in exact rational arithmetic. The tests compare it, case by case, with the first-step equations solved as a linear system, with an exhaustive enumeration of every sequence of rounds up to a length, and with seeded simulations within four standard errors for the ruin share and five for the mean length. These are checks on the code. The formulas themselves are proved above.

The model is the simplest one. The stake is always one unit, the chance of winning never changes, and the walls are fixed. A gambler who stakes a larger share of the bankroll when ahead, or who may stop at any time, is playing a different game, and the page says nothing about it. What carries over, for fair bets, is that no rule for choosing stakes changes the average final bankroll. That is the optional stopping theorem for martingales, a standard result that is cited rather than proved here.

The expected length of a game is an average and a misleading one. For ten units against ten at a fair bet it is 100 rounds, yet individual games range from ten rounds to several hundred, and a few run far longer. The twenty seeded games show that spread directly. A game that has not ended yet is still exposed to both walls, so a long game is not a safer one.

Real casinos do not offer fair bets. In the laboratory an edge of a few points is harmless to a short session and ruinous to a long one, which is why the exact curves matter more than a twenty-game sample. Against a casino with no upper limit, even a favourable bet has a positive chance of ruin that does not depend on how long you keep playing, which the dotted line shows: at 55% and a bankroll of ten units it is about 13.4%.

The page proves the formulas for the model it states and checks them against independent computations. It does not claim that any betting system beats a casino, and it does not model a bankroll that can go below zero.