You have $100. You bet $1 at a time on red at European roulette and keep playing until you have $200 or run out of money. Red comes up in almost half the spins. What is the chance you leave the casino with $200? Lock in your answer, then play yourself, run thousands of sessions and see how the stake, the target and the casino's wealth change the odds.
A player with $100 bets $1 at a time on red at European roulette until they have $200 or nothing. Red comes up with chance 48.6%, almost half, yet only 1 player in 224 doubles their money: 0.45%. The house edge is just 2.7 cents per dollar, but over 3,667 bets it eats nearly everything. A single $100 bet would give 48.6%. Here you play yourself, run thousands of sessions and see why in a casino boldness beats caution.
You have made up your mind: either I double my money or I lose it all, and then I go home. You play as cautiously as possible: the smallest stake on an almost even game.
European roulette has 37 pockets: 18 red, 18 black and one green zero. The ball lands in any of them with equal chance, so red comes up with chance 18/37 ≈ 48.6%.
Each time you put $1 on red. Red comes up: the casino returns your stake and pays $1 more. Black or zero: the stake is lost. So it is +$1 or −$1.
You play until you have exactly $200 or have lost your last dollar. There is no other reason to leave: no tiredness, no closing time.
What is the chance that you go home with $200?
"Nearly 50-50, say 45%"
In fact the chance is only 0.45%, 1 in 224. Each bet is almost fair, but there are 3,667 of them on average, and over that time the casino's small edge pulls the money down harder than luck swings it about. A single $100 bet on red would give 48.6%.
What happens if you change the stake is taken apart in rounds 2 and 5.
Pick an option before reading on. Start with $100, bet $1 on red each time, play to $200 or to zero.
Reminder: red comes up in 18 cases out of 37, a chance of 48.6%.
A breakdown of your answer will appear here.
Hard to decide? Open the hints one at a time: the easy one first, then stronger ones.
Don't look at a single bet: it really is almost fair. Ask yourself how many such bets it takes to reach $200 or zero.
In a fair game without the zero the answer would be exactly 50%. Here every bet takes 1/37 of a dollar on average. How much is that over a few thousand bets?
A loss is 19/18 times as likely as a win, and $200 takes 100 more wins than losses. What happens if you multiply 19/18 by itself 100 times?
Start with $100, the target is $200. Spin one at a time or play the session to the end. Then change the stake: a cautious $1 or a bold $100 all at once? The counter on the right shows how many sessions ended at $200.
Every player starts with $100 and bets $1 on red until $200 or zero. Switch to a fair coin to see the same game without the zero: there exactly half reach $200, but the sessions last almost three times longer.
Each bar counts the sessions that ended in that range. The tail on the right is long: some players wander around the middle for tens of thousands of bets.
How much money a player had at their best moment before the session ended. Many were ahead at some point, but the rules do not let them leave ahead.
A single bet is almost fair. What makes the game unfair is how many such bets it takes to reach $200.
Each bet takes 1/37 of a dollar on average. Over n bets that is n/37, while the random spread around that average is about √n. Early on the spread is bigger, and a player can climb. But the drift grows faster: after 3,700 bets it reaches $100, while the spread is only about $61.
In a fair game the average amount in your pocket does not change, however long you play. At the end it is either $200 or $0, so 200 · P = 100 and P = 1/2. In general, the chance of getting from i to N is i/N. And the average length is i · (N − i) = 100 · 100 = 10,000 bets. Add 1 pocket out of 37 and instead of 50% you are left with 0.45%.
The casino takes on average 1/37 of every dollar wagered, whatever the strategy. So the average loss is 2.7% of the total of all bets. The cautious player pushes the same money back and forth 3,667 times and loses $99.11 on average. The bold one bets $100 just once and loses $2.70 on average. That is where the gap in the odds comes from.
A loss is 19/18 times as likely as a win. To reach $200 you need 100 more wins than losses; to reach zero, 100 more losses. Those 100 steps multiply the tilt: (19/18)100 ≈ 223, like 5.6% interest over 100 years. So zero is about 223 times likelier than $200, and the chance of $200 is 1/(223 + 1) = 1/224.
The player has i dollars, the target is N, a bet is won with chance p and lost with chance q = 1 − p. The chance of reaching N obeys a simple equation: after one bet you are either a step up or a step down.
Its solution, where r = q/p. For a fair game r = 1 and the formula becomes i/N.
The target is twice the start, N = 2i, so the fraction simplifies.
The average number of bets until the game ends. For a fair game D = i(N − i).
Wald's identity: the average loss equals the house edge times the average total wagered.
The key part is the factor ri. When the game is even slightly unfair, r is above 1, and r100 is already huge. So the chance falls not in proportion to the house edge but exponentially in the number of steps to the target. Halving the stake doubles the number of steps and squares the tilt.
From an exact theorem on the best strategy in an unfair game, through the logic of "the casino only has to wait", to two practical lessons: why doubling your bets does not save you, and how much to stake when the edge is yours.
| Strategy | Exact chance | Simulation | Bets |
|---|
Bold play: each time, stake everything you have, or exactly what is missing to the target if that is less. From $100 to $400 that is two bets: $100, then $200. Lester Dubins and Leonard Savage proved that when every bet is unfavourable, no strategy gives a better chance of reaching the target. Fewer bets means fewer times the casino takes its 2.7%.
In a fair game the strategy does not matter: any of them gives i/N. And if the game favours you, it is the other way round, and small bets pay. For $1 bets to a $1,000 target the chance is below 10−21, and a simulation will never see it.
A player with $100 bets $1 at a time against a casino with infinite money and plays until broke. The curves show what share of players are already broke after n bets. Exact calculation, logarithmic bets axis.
Let the target N grow to infinity. In the fair-game formula i/N the chance of winning tends to 0, so ruin is certain, with probability 1. But it takes a long time: the average length i(N − i) also grows without bound. After 10,000 bets only 32% of players are broke, after a million, 92%.
In roulette the picture is different: the player goes broke in 100 · 37 = 3,700 bets on average. The chance of ever being $100 ahead is (18/19)100 ≈ 0.45%, almost the same as in the main puzzle. The casino does not have to win every bet. It only needs enough money to outlast your luck, and time.
The martingale system: bet $1, double the stake after a loss, go back to $1 after a win. Every run ends with a $1 win, as long as you can afford to double. With $100 you survive 6 losses in a row (1 + 2 + 4 + 8 + 16 + 32 = 63), and the seventh doubled stake of $64 is out of reach. Then you bet what is left.
Six losses in a row happen in only 1.8% of runs, so mostly the money climbs in $1 steps. But reaching $200 takes about a hundred runs, and one bad run takes everything. The result: 42.1%. Better than $1 bets, but worse than a single $100 bet, just as the Dubins–Savage theorem promises.
When the game favours you, ruin becomes the main enemy: stake too much and a few losses in a row knock you out. John Kelly showed that capital grows fastest if each time you stake the share f = 2p − 1. For a 55% chance that is 10% of your capital. A stake measured as a share shrinks after losses by itself, so you never reach zero.
A stake twice the Kelly share gives almost no growth, and a bigger one leads down, although every bet is favourable. For roulette 2p − 1 is negative: the best share is zero. Edward Thorp took this principle into blackjack, and later into investing.
The earliest known mention of the problem is a letter from Blaise Pascal to Pierre de Fermat in 1656. That same year Pierre de Carcavi passed it on to Christiaan Huygens: two players throw three dice, the first scores when 11 comes up, the second when 14 does.
Huygens set the problem as the last of five in his treatise De ratiociniis in ludo aleae: each player has 12 counters, and losing a point costs a counter. A total of 11 comes up in 27 ways out of 216, 14 in only 15. The answer: the chances of winning for the player waiting for 14 and the one waiting for 11 are as 244,140,625 to 282,429,536,481, that is (5/9)12.
Abraham de Moivre, in De Mensura Sortis, solved the problem for any capitals and any odds, deriving the same ri formula as in round 4. The name "gambler's ruin" was attached to the problem much later.
That is the title of the book by Lester Dubins and Leonard Savage that proves bold play is optimal in an unfavourable casino. The idea in a nutshell: if you urgently need an amount you can only win, put as much as possible on as few tries as possible.
John Kelly, an engineer at Bell Labs, published "A New Interpretation of Information Rate", linking betting to Shannon's information theory. The story of the formula and how it was used in casinos and on Wall Street is told in William Poundstone's book Fortune's Formula.
The cautious player in our puzzle makes 3,667 bets of $1 on average, wagering 37 times what they brought. If the wheel spins once a minute, that is over 60 hours of non-stop play, and on average they go home with 89 cents.
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