Ask the Wizard: Martingale Betting System

The Martingale Betting System

If someone follows such a Martingale system in blackjack, what is the probability of being able to win $200 per day or lose the entire $5000? Also, does increasing the amount available for total wagering increase the likelihood of winning the $200.

If you had a game with no house edge the probability of winning $200 with $5000 to risk, using any system, would be 5000/(5000+200) = 96.15%. The general formula for winning w with a bankroll of b is b/(b+w). So the larger the bankroll the better your chances. The house edge will lower the probability of success by an amount that is hard to quantify. For a low house edge game like blackjack the reduction in the probability of success will be small. It would take a random simulation to know for sure. Forgive me if I don't bother with that. VegasClick did a small simulation about the probability of success with the Martingale. Jan. 9, 2005

I disagree with your answer about using the Martingale with an infinite bankroll (May 22, 2004 column). If I had an infinite amount of money and time, and the casino would take any bet, then could I ensure a profit by playing the Martingale (doubling after every loss until I win) on a fair bet on the toss of a coin? The question writer proposes a random walk on a fair bet. The expected value is indeed zero, as you say. But the probability of ever being ahead is 1, as long as you are willing to quit after being ahead some finite amount. The probability of eventually achieving that finite amount with an infinite bankroll and infinite time is 1.... for ANY finite level of winnings. Even if the game is unfair, infinite bankrolls can ensure that eventually you can receive a positive result... and then quit. Pick a level of winning you want... $1 million. Bet a million. If you lose, bet $2 million. If you lose again bet $4 million. In an infinite number of flips, even with the game as unfair as you like, you will eventually win. Take your $1 million and go home. Come back tomorrow and repeat. - Jonathan Falk, actuary

I had a feeling one of my fellow actuaries might disagree with me on this one but I stand by my answer. I see this as a question of expected value rather than probability. The writer used the word "ensure", which is related to the word insurance. An insurance policy would have a fair cost of 1, which is simply the product of the probability (1/2^infinity) and amount covered (2^infinity). As I said in my original reply, 2^infinity/2^infinity = 1. So the player would give up his one unit win to pay for the insurance policy. You might argue that the insurance company would never have to pay because they could claim an infinite number of flips have not occurred yet, but I'm assuming a timeless quality in the question. If we did consider time I would be even more right because the player would never live long enough to play an infinite number of flips, and any finite number of losses is definitely possible. May 30, 2004

If I had an infinite amount of money and time, and the casino would take any bet, then could I ensure a profit by playing the Martingale (doubling after every loss until I win) on a fair bet on the toss of a coin?

No. Some might argue that it would take an infinite number of losses to lose in this situation, which would be impossible. The truth is that 0.5^infinity approaches 0 but does not equal zero. If this did happen you would lose $2^infinity. The expected return of this strategy is thus 1- $2^infinity * 0.5^infinity = $1 - 1 = 0. Another more graceful way to look at is that as your bankroll increases the expected value still remains unchanged at zero. So the limit of the expected value as the bankroll approaches infinity is zero. In other words an increasing bankroll doesn't help your odds, even if it goes to infinity. May 22, 2004

I just left a discussion in which we all agreed the Martingale System is not good use. My discussion mates used an unlimited bankroll to make the theory work to their advantage by saying the ONLY thing wrong with the theory is the table limits set by the casinos. I conceded that point, that, yes, the table limits will stop this system. However, I went one step deeper by saying that the system will fail because Martingale assumes that on an unlimited bankroll you will win one time to make the 1 unit profit sought after. I disagree. Albeit the house edge is small on certain types of wagers in a casino, there is no guarantee that the win will ever occur. Granted, losing 1,000 or 1,000,000 times in a row at something is very unlikely, but it is possible. If, like I did, it is assumed that an unlimited bankroll is at our fingertips and there is no table limit - meaning doubling to infinity is possible, can/does the system still fail because the "theory assumed win" is not a guarantee? - Eric Rissmiller from Ephrata, Pennsylvania

If you did have an infinite bankroll then yes you could keeping doubling until you won a single net unit. However you in fact would not be adding to your bankroll because infinity+1 = infinity. In fact I allege that it would be impossible to affect the bankroll making any combination of finite bets because infinity +/- a constant still equals infinity. I would invite differing opinions only from those holding a masters degree or higher in mathematics. Feb. 4, 2003

I read your topic in Roulette on the Martingale method. I have tried this method a few times on the computer and I have been up $500. Then I went to the casino and lost over $1000. Because black came up 8 times in a row. But I'm just starting to learn baccarat. I was trying it on the computer and again I have been up $500, by betting on the banker. Starting at $20 then going to $40 then $80 and so on. I was up $500 even with paying the 5%on each hand. Do you think this method would work in a casino? I thought I would ask before I go and lose another $1000. Like I said black came up 8 times in a row. But do you think that the player hand would win 8 times in a row? Plus this game is good because a tie is a push, where in roulette 0, or 00, is a loss. - Andrew Jackson from Maitland, Canada

The Martingale is dangerous on every game and in the long run will never win. However it is better to use in baccarat than roulette, just because of the lower house edge. The probability of the player winning 8 times in a row is 0.493163^8 = 1 in 286. Also keep in mind you could win a hand late in the series and still come out behind because of the commission. For example if you started with a bet of $1 and you won on the 7th hand you would win $60.80 ($64*95%), which would not cover the $63 in previous loses. Aug. 4, 2002

I play the negative system in black jack meaning I double every time I lose until I Win. I wanted to what the odds are of losing 4,5,6,7,8,9 hands in a row? How many hands should I expect to play till I lost 8 hands which is my stopping point? - Jay from New Haven, Connecticut

The name for this system is the Martingale. Ignoring ties the probability of a new loss for a hand of blackjack is 52.51%. So the probability of losing 8 in a row is .5251⁸ = 1 in 173. Aug. 4, 2002

What is the expected gain of the Grand Martingale system in the game of roulette? - Jane from Dayton, USA

The expected loss is 5.26% of total money bet. This is true of ANY betting system based on American roulette rules. Dec. 4, 2001

When playing the Martingale double up system against the single zero roulette wheel on any one of the even chances. I figured that you will lose one time in every 248 sessions: meaning a session that runs to completion with either a win of one unit or a loss of 255 units. Am I figuring correctly, if not could you please give the correct odds? - Jack Thompson from Neenah, Wisconsin

If the maximum loss is 255 units then you can bet up to 8 times. The probability of losing eight bets in a row is (19/37)^8=.004835, or 1/207. Feb. 10, 2001

Q: I've always thought that the major flaw,(but certainly not the only one),in the Martingale system was that the return was simply too small to justify the risk. My question is, would a tripling of the bet which would yield a profit per win of approx. 50% of the winning hand justify the system. In other words would a bankroll of 1093 units playing through 7 betting levels produce an overall profit exceeding its loss, or is it doomed to eventual failure the same as the standard Martingale? Jack Barlow of Desoto, Texas

A: Every betting system based on a negative expectation game like craps is doomed to eventual failure. By tripling your bets you will have bigger single wins but you will reach your bankroll limit faster and have more losses. July 16, 2000

Q: I have a thick, and I mean thick, friend who is intoxicated with having won a fair amount betting player only in baccarat. He plays $10 units and does the dumb play that 9 straight banks won't happen. He escalates to a risk of $3,980 by going 10 10 30 60 120 250 500 1000 2000. How can I find solid mathematical evidence to try to convince him to stop? - Charger

A: This is a close variation of the Martingale betting system in which the player doubles after every loss. Usually the Marginale player will win but occasionally will have more consecutive losses than he can handle and suffer a major loss. Assuming your friend is betting on the player the probability that any given bet will begin a streak of 9 losses in a row is (2153464/(2153464+2212744))⁹ =~ .001727, or 1 in 579, assuming ties are ignored. There is more information available about the folly of the Martingale in my section on betting systems. However the more ridiculous a belief is the more tenaciously it tends to be held. Usually it takes a big loss to convince a believer in betting systems to stop. July 2, 2000

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