Martingale Betting

One unit on red, doubled after every loss until it wins. Watch a crowd of players run the system for a thousand sessions, then change the bankroll and see what a deeper ladder buys.

Worked example — Martingale Betting

Edge on each unit staked	2.70%
Bets the bankroll covers, k	10
Sessions that end one unit up	99.8725%
Sessions that lose the whole ladder	0.1275%
Expected result of a session, E(R)	−0.3056
Expected staked in a session, E(S)	11.308
Result per unit staked	−2.7027%
Players with no lost session in 1,000	27.92%

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Martingale Betting

The martingale betting system stakes one unit on an even-money chance, doubles the stake after every loss and stops at the first win, which nets one unit. With a bankroll that covers k bets, 2^k − 1 units, a session loses all of it with probability q^k; its expected result is 1 − (2q)^k and its expected stake is the sum of (2q)ⁱ for i from 0 to k − 1, and the ratio of the two is p − q exactly. In probability a martingale is a process whose expected next value, given everything so far, is its present value: the mathematics of a fair game.

What each symbol means

B: the bankroll, in units of the first bet
k: the bets the bankroll covers, the largest k with 2^k − 1 ≤ B
q: the chance of losing one bet, 1 − p
R: the result of one session: +1, or −(2^k − 1)
S: the total staked in one session
X: the number of lost sessions among n

Assumes: Every session starts again at one unit with the bankroll topped up, and every bet pays even money and loses the whole stake on zero. A table limit cuts the ladder exactly as a smaller bankroll does.

Doubling wins almost every session and loses the house edge on every unit it stakes 🖖

Press 1,023 units, 1,000 sessions. Bet 1 unit on red, double after each loss, stop at the first win: whatever came before, that win nets one unit, because each bet is one more than everything lost before it. With 1,023 units you can afford ten bets, 1 + 2 + … + 512, so a session ends one unit up unless red fails ten times running, which happens in 0.1275% of sessions. Then all 1,023 are gone. The grey dashed line is the system as it feels from the inside, a unit a session. The expected result of a session is 0.998725 × 1 − 0.001275 × 1,023 = −0.3056. A session stakes 11.308 units on average, since most end on the first or second bet and a few climb the ladder, and the ratio of the two, taken before rounding, is −2.7027%, which is −1/37, the edge of a single bet on red, to every digit. It holds at every bankroll, because a session is only a string of bets and each loses 1/37 of its stake on average, whatever its size. Press A fair coin and the expected result is exactly 0. This is Doob's optional stopping theorem, and Kyle Siegrist's textbook gives the gambler's version of it plainly: if the games are fair or unfavorable, "no stopping time, regardless of how cleverly designed, can help the gambler". A session that is over within ten bets is the plainest case. The name belonged to gamblers first. Roger Mansuy finds it in the abbé Prévost's dictionary of 1750, doubling the stake after each loss at faro "in order to quit with a sure profit, provided that he wins once", and Jean Ville carried the word into probability in 1939.

Almost every session wins, and after 1,000 sessions most players are behind 🖖

Stay on 1,023 units, 1,000 sessions and look at the crowd: 400 players, each with the bankroll topped back up after a loss. Every line climbs a unit a session and falls 1,023 at a lost one. Lost sessions are rare, 0.1275% each, but a player plays a thousand of them. After 1,000 sessions 27.92% have never lost one and stand at +1,000. Everyone else has lost at least once, and a single loss already leaves a player at 999 − 1,023 = −24, so 72.08% are behind. The median player's first lost session is the 544th; press 100 sessions and 88.02% are still unbeaten. The average after 1,000 sessions is 1,000 × (−0.3056) = −305.6, the purple dashed line, and the lower panel shows that no player stands there. The totals can only be 1,000, −24, −1,048 and on down in steps of 1,024, since each lost session costs the 1,023 and the unit it did not win. What the system does is pack the loss into rare lumps of 1,023, which is why it feels like winning for so long.

A deeper ladder makes the system lose more per session 🖖

Press 63 units: six bets deep. Sixty-three units cover six bets, a session loses them all 1.8336% of the time, and the expected result is −0.1735 a session. With 1,023 units it is −0.3056. Press A million units: 1,048,575 units cover twenty bets, a session is lost only 1.63×10⁻⁴% of the time, and the expected result is −0.7047. The bankroll table shows where the extra loss comes from. A deeper ladder is reached rarely, but when it is reached it bets far larger amounts, so the expected stake rises, 6.420, 11.308, 26.072, and the edge takes 2.7027% of every unit of it. With the million units, 99.84% of players get through 1,000 sessions without a loss and finish 1,000 up, and the average is still −704.7, paid by the 0.16% who lose 1,048,575 at once. Casanova, recalling the Venice Carnival of 1754 in his memoirs, played a ladder six bets deep: "I never lost the sixth card. If I had lost it, I should have been out of funds" (Grinstead and Snell quote him). On a fair coin every bankroll gives exactly 0, however deep. Only an unlimited bankroll escapes: it wins one unit with certainty, at an expected stake per session that is infinite. The next rung turns the same doubling round. The St Petersburg game pays a pot that doubles with every toss, and its expected value is infinite.

Problem solved in full

Buying a clean thousand sessions 4 steps
On European red, how large a bankroll gives a 90% chance of playing 1,000 sessions without losing one, and what does it cost per session? Press 1,023 units, 1,000 sessions first: the page gives the chance for 1,023 units.

Open this problem in the tool
1. From the page, ten bets deep:
  
  $(1 - q^{10})^{1, 000} = (1 - 0.001275)^{1, 000} = 0.2792$
2. For 90% the chance of losing one session has to fall to:
  
  $(1 - q^{k})^{1, 000} \geq 0.9 \Rightarrow q^{k} \leq 1 - 0 . 9^{1/1, 000} = 0.00010535$
3. Each extra bet multiplies that chance by q = 19/37, so the ladder must be at least this deep:
  
  $k \geq \frac{ln 0 . 00010535}{ln ( 19/37 )} = \frac{- 9 . 1582}{- 0 . 6665} = 13.74 \Rightarrow k = 14$
4. Fourteen bets need 2¹⁴ − 1 units, and the expected result follows from the same formula:
  
  $B = 2^{14} - 1 = 16, 383 : (1 - q^{14})^{1, 000} = 0.9152, E (R) = 1 - (\frac{38}{37})^{14} = - 0.4526$
Answer

16,383 units, fourteen bets deep, give 91.52%. Thirteen bets, 8,191 units, give only 84.14%. The price is the expected result, −0.4526 a session instead of −0.3056, so −452.6 over the thousand sessions instead of −305.6. Sixteen times the bankroll buys the better chance and the larger expected loss together.

$B = 16, 383 : 91 . 52%, 1, 000 \times (- 0.4526) = - 452.6 (B = 1, 023 : - 305.6)$

Learning path

Fair games that ruin you

References (3)

Where the name comes from, which the first insight quotes: the abbé Prévost's dictionary of 1750 defines the martingale, at faro, as doubling the stake at each loss "in order to quit with a sure profit, provided that he wins once"; Jean Ville's 1939 thesis took the gamblers' word into probability, and Doob adopted it from there. The Provençal and harness etymologies are reported as uncertain: R. Mansuy, "The Origins of the Word 'Martingale'", translated by R. Sverdlove. Journ@l Électronique d'Histoire des Probabilités et de la Statistique / Electronic Journal for History of Probability and Statistics 5(1), June 2009.
Doob's optional stopping theorem for bounded stopping times, and the gambler's reading the first insight quotes, "If the games are fair or unfavorable, then no stopping time, regardless of how cleverly designed, can help the gambler"; with the red-and-black identity E(Xᵢ) = E(Xᵢ₋₁) + (2p − 1)E(Yᵢ), which summed over a session's bets gives the page's E(R) = (p − q)E(S): K. Siegrist, Random: Probability, Mathematical Statistics, Stochastic Processes (randomservices.org). Chapter 16, Martingales, §3 Stopping Times; and Chapter 12, Games of Chance, §9 Red and Black.
Casanova's martingale in Venice, quoted in the third insight ("I never lost the sixth card. If I had lost it, I should have been out of funds"), the doubling system set as Exercise 10 of §1.1, and Exercise 12 of §6.1, the page's 63-unit ladder on a wheel with no zero (start with 1, stop at the first win or when black has come up six times and you are down 63): C. M. Grinstead and J. L. Snell, Introduction to Probability, 2nd edition, American Mathematical Society; the CHANCE Project version dated 4 July 2006 (GNU Free Documentation License). §1.1, Historical Remarks, pp. 11–12, and Exercise 10; §6.1, Martingales and Exercise 12.

By the 42Math team · Reviewed against published sources. Corrections welcome.