Roulette - FAQ

You're right the chances of winning are 24/38, or about 63%. However you have to risk $200 to only win $100. If you want to increase your chances of winning even more then bet on any 35 numbers. The probability of winning will be 92%.

As long as you stay away from the 0-00-1-2-3 combination, the house edge on any combination of bets is always exactly 1/19, or 5.26%. There are ways to increase your probability of winning, but at the cost of winning less relative to your total wager.

The probability of getting any number three times out of 4 is 38*4*(1/38)³*(37/38) = 1/5932. However, if you play long enough you almost can't help but notice unusual events like this. This does not nearly rise to the level of being suspicious. Cheating does occur in real casinos. It is usually a rogue dealer who is caught by casino security. There have been some strong cases of cheating made against online casinos but no governmental authority has ever convicted anyone to the best of my knowledge.

See my Ten Commandments of Gambling. The sixth commandment is "never hedge thy bets." About your roulette question, the probability of losing all ten bets by betting one at a time is (37/38)¹⁰=76.59%. The probability of losing all ten bets by betting them all at once on different numbers is (28/38)=73.68%. By hedging, or betting ten numbers at once, you lower your probability of a total loss but also limit your maximum win to $26. The player betting one at a time could win up to $350. Both these methods have the same total expected return of 94.74%.

There are 30 ways to win $15, 6 ways to lose $135, and 2 way to win $45 (not $40). The expected return of this combination of bets is ((30/38)*15 + (6/38)*-135 + (2/38)*(45))/135 = -.0526, or 5.26%, the house edge on any one bet or combination of bets as long as the dreaded 0-00-1-2-3 combination is avoided. In your observations you have likely seen fewer than expected 19-24 occurrences, which accounts for the illusion that this method is winning.

You're just rehashing the gambler's fallacy. If the ball landed in odd 100 times in a row on a fair wheel the odds that the next spin would be even are still the same as every spin, 47.37% on a double zero wheel. So it does not help that you can spin without betting. The ball does not have a memory.

I've heard of both these techniques being used. I don't know much about devices to clock wheels, except they are known to exist and be used from time to time. Here in Nevada such a device would be highly illegal. Taking advantage of biased wheels I hear a lot more about. It has been done lots of times. I think casinos with old wheels are the most vulnerable target. I've been saying for years that I think Argentina is a ripe target for that.

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. So, you have 99.52% of winning one unit, and 0.48% of losing 255 units.

You can expect a repeat once every 37 pairs of numbers. So, with 36 numbers we have 35 pairs of numbers. So, the expected number of repeats is 35/37 = 0.9459.

Show me someone who is winning at a fair game of roulette and I'll show you somebody who is just lucky, and will likely lose it all back. You can only skillfully beat roulette by advantage play, like exploiting a biased wheel, or clocking the wheel.

Casinos don’t need to resort to such tactics to win. Furthermore casinos have nothing to fear from progressive bettors. Most of the time progressive bettors win, but the few that hit their bankroll limits pay for all the winners and then some more for the casino. Furthermore it would take a great deal of skill to deliberately spin a ball into a specific section. I don’t believe it can be done with any marked degree of accuracy.

Thanks for the compliment. First of all the standard deviation on any even money bet is 0.998614 and on a single number is 5.762617. The probability of coming out ahead by flat betting even money bets over 1, 100, and 10000 spins is 0.473684, 0.265023, and 0.00000007 respectively. The probability of coming out ahead by flat betting single number bets over 1, 100, and 10000 spins is 0.0263158, 0.491567, and 0.18053280 respectively. It seems you are trying to argue that single number bets are better because of the higher probability of finishing ahead over multiple bets. This is true, however the probability of a substantial loss is also much greater. Over a session the expected results always fall somewhere on a bell curve. With low volatility bets like red or black that bell curve is sharp and doesn’t stray far from a small loss. With high volatility bets like single numbers the bell curve is wide, allowing for a much wider range of net results, both good and bad.

He is just lucky. As I have said thousands of times, no betting system can pass the test of time.

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.

Not quite. You would have a 12/38 chance of winning 3 units, 12/38 of breaking even, and 14/38 of losing 3 units. The expected value is [(12/38)*3 + (12/38)*0 + (14/38)*-3]/3 = (-6/38)/3 = -2/38 = -5.26%. This will be true of any combination of bets as long as you avoid the dreaded 5 number combo (0/00/1/2/3). If you only play for one spin and want to maximize your probability of winning then bet equally on 35 of the numbers. You’ll have a 92.11% chance of winning 1 unit and a 7.89% chance of losing 35 units.

7500 spins? Is that it? Anyone can show a profit of 7.94% of total money bet over 7500 spins if they bet aggressively. Same is true about 15000 spins. Most systems are designed to have a lot of small wins and small number of large losses. A system requiring a huge bankroll can easily go 15000 spins and show a profit. Eventually the losses will come in and it won't pass the test of time. The big losses might also come at the beginning. The true way to put a system to the test is to play it over billions of trials. My opinion about these systems is the same as all systems, they are worthless. I have no problem with you trying them out but I do have a problem with anyone putting one dime in the pockets of those selling them.

Note: See the follow up to this question in the next column.

You are correct that option B has the greater probability of success, although the goal and the capital are the same. The reason is the average amount bet in option B is less, thus your money is exposed to the house edge less, thus the probability of winning increases. The amount bet in option A is always $500. The average amount bet in option B is (12/37)*125 + (25/37)*(12/37)*(125+187.5)+ (25/37)*(25/37)*(125+187.5+187.5) = 337.29.

When I was on the Vegas Challenge, with a few minutes to go, I had about $8,000 and needed to get to at least $24,000. So I split my bankroll into four piles of $2000 each and bet each one on a 4-number combination, each of which would have paid $22,000. This way I was not necessarily exposing my entire stake to the house edge, which increased my probability of winning.

I measure the value of a bet to be the expected return, not the probability of winning. So betting on all 38 numbers has a house edge of 2/38 = 5.26%, the same as for one number or any number of numbers covered. Although betting all 38 numbers has a 0% chance of showing a net win, the down side is losing only 5.26% of your total bets. If forced to bet and you want to minimize variance then you should bet all 38 numbers. A practical example is if you had promotional chips you had to bet and you don’t want to gamble bet get your exact expected value out of them. So to answer your question there is no optimal range of numbers. All ranges are equal in expected value.

First, let me say this guy was a fool. He bet $138,000 on a normal American roulette wheel which has two zeros and a house edge of 5.26%. This amounted to an expected loss of $7,263. However had he taken a 10 minute ride to the Bellagio, Mirage, or Aladdin he could have made the bet on a single zero wheel which follows the European rule of giving half an even money bet back if the ball lands in zero. He planned to make an even money bet anyway. So, at these wheels with full European rules his house edge would have been only 1.35%, for an expected loss of only $1865.

To answer your question, if forced to make just one even money type bet I would have chosen the banker bet in baccarat with a house edge of 1.06%.

Risk of ruin questions are mathematically complicated. Unless it is a simple win/lose game I would recommend doing a random simulation on a computer.

There are lots of single zero wheels in Atlantic City. Most of the casinos there have them, but at a $25 minimum.

1. No
2. A single-zero wheel is used at Casino on Net. So the probability of going 16 times with a zero is (25/37)¹⁶ = 0.1887%.
3. (25/37)¹⁷ = 0.1275%.

The chances of any number occurring exactly 5 times in 10 spins in a double-zero roulette game could be closely approximated by 38*combin(10,5)*(1/38)⁵*(37/38)⁵ = 1 in 359275.

Thanks. You make a valid point. The house edge in 6 to 5 blackjack is 1.44% under the usual rules, while double zero roulette is 5.26%. That is 3.7 times as bad. However I have learned through the years that it is almost hopeless getting players to leave a game they like, regardless of how bad the house edge is. So the best I can do is advise them how to play their game of choice. For blackjack players there is still no shortage of 3 to 2 games out there. Playing 6 to 5 is giving the casino an extra 0.8% advantage for no reason at all. I also stress the importance of looking for single-zero roulette if you are a roulette player. So I see no inconsistency.

There are usually two minimums in roulette. For example: $5 outside, $1 inside. Outside bets are all even money bets, column bets, and dozen bets. Inside bets are those on the numbers, including groups of 2, 3, 4, 5, and 6. In this case the minimum on outside bets is $5 and $1 on inside bets. However, you must bet at least $5 in total on inside bets or make none at all.

I’m on your side. If this could be done then dealers could easily conspire with players and share in the profits. Yet I never hear of this happening. A good test would be to get somebody who claims to be able to influence the roll and have him attempt to land it in a particular half of the wheel as many times as possible over 100 spins. The more times he makes it the greater weight his claim will have. The table below shows the probability of 50 to 70 successful spins. For example, the probability of 60 or more successful spins is 2.8444%. Common confidence thresholds in statistics are the 90%, 95%, and 99% levels. To beat a 90% confidence test, in which the probability of failing given random spins is 90%, the number of successful spins would need to be 57 or more. To beat a 95% test the number would need to be 59 or more, and at 99% the number would need to be 63 or more.

I tried to find that game but the site was down when I checked. However, assuming such a game did exist, the answer is no. No system could be expected to beat it, nor lose to it, over the long-run. The expected value of every system would be exactly zero.

Regardless of what the first number is, the probability the second spin only matches it is (1/37)*(36/37). The probability only the third spin matches it is (36/37)*(1/37). The probability neither spin matches, but the second and third match each other, is (36/37)*(1/37). The probability both second and third spins match it is (1/37)*(1/37). Add up all this and you get 3*(1/37)*(36/37)+ (1/37)*(1/37) = 7.962%.

I tried to sign up for an account there to check this out but they block U.S. players. I’m told the minimum bet is £2 and the maximum is £50. Even in a zero house edge game like no-zero roulette there is still no betting system that will get above, or below, that 0% figure. No matter what you do the more you do of it the closer the actual house win will get to 0%.

I saw a television show about him once, and I applaud what he did. What I define as a “system” is a betting pattern, such as the Martingale, applied to a game with a house advantage, such as a fair roulette game. What Gonzalo Garcia-Pelayo successfully did was survey how often the ball landed in each number, in an effort to find, and then exploit, biased roulette wheels. This I would call a strategy, as opposed to a system. There are lots of profitable strategies for beating the casinos, but zero profitable betting systems.

No.

Thanks for the kind words. I think I’ve answered this one before, but no, even with zero house edge there is still no betting system that can win, over the long run.

This is far from the first time I have heard this claim, and I am very skeptical of it. Most dealers also believe myths like a bad third-baseman will cause the other players to lose in blackjack, so as a group they are not the most skeptical bunch. What I think is happening is they remember the times they were successful at attempting to control the spin, and conveniently forget the times they were not. Much like they remember the times the third-baseman took the dealer’s bust card, but forget the times he saved the table.

If dealers really could do this, it would be easy to have a confederate play, causing him to win, and causing other players to lose, to make up for it. As long as they were following proper procedures for the spin, and didn’t appear with the confederate in public, it would all look completely legitimate. Yet, you never hear about this happening. I suppose the believers could say that those doing it are just keeping a low profile, but that is what believers in worthless betting systems say too. If this were as easy as the roulette dealers where you work claim, the cheating problem as a result would be rampant.

The probability that any given number will not have hit is (37/38)²⁰⁰ = 0.48%.

With 38 numbers, we could incorrectly say that the probability that any one of them would not be hit is 38 × (37/38)²⁰⁰ = 18.34%.

The reason this is incorrect is it double counts two numbers not being hit. So we need to subtract those probabilities out. There are combin(38,2) = 703 sets of 2 numbers out of 38. The probability of not hitting any two given numbers is (36/38)²⁰⁰ = 0.000020127. We need to subtract the probability of avoiding both numbers. So we are at:

38×(37/38) ²⁰⁰ - combin(38,2)×(36/38) ²⁰⁰ = 16.9255%.

However, now we have canceled out the probability of three numbers not hitting. For any given group of three numbers, we triple counted the probability of any single number not being hit. We then triple subtracted for each way to choose two numbers out of the three, leaving with zero for the probability that all three numbers were not hit. There are combin(38,3)=8,436 such groups. Adding them back in we are now at:

38×(37/38) ²⁰⁰ - combin(38,2)×(36/38) ²⁰⁰ + combin(38,3)×(35/38)²⁰⁰ = 16.9862%.

Yet, now we have over-counted the probability of four numbers not hitting. For each of the combin(38,4)=73,815 groups of four numbers, each was originally quadruple counted. Then we subtracted each of the combin(4,2)=6 groups of 2 out of the 4. Then we added back in the 4 groups of 3 out of the 4. So, for each union of four numbers, it was counted 4 − 6 + 4 = 2 times. To adjust for the double counting we must subtract for each group. Subtracting them out we are now at:

38×(37/38) ²⁰⁰ - combin(38,2)×(36/38) ²⁰⁰ + combin(38,3)×(35/38)²⁰⁰ - combin(38,4)×(34/38)²⁰⁰ = 16.9845%.

Continuing in the process we would keep alternating adding and subtracting, all the way until missing 37 numbers. Thus the probability of at least one number never being hit is:

Sum i=1 to 37 [(-1)⁽ⁱ⁺¹⁾ × combin(38,i) × ((38-i)/38)³⁸] = 16.9845715651245%

Here are the results of a random simulation of 126,900,000 such 200-spin experiments.

The ratio of times at least one number was not hit was 0.169833.

Sadly, ignorance can go pretty high up the ladder. I don’t dispute that an expert can clock the wheel on a very slow spin. However, that issue aside, changing dealers does not change the odds. There is no such thing as a lucky or an unlucky dealer. Superstition is a difficult thing to let go of. As I have said many times, the more ridiculous a belief is, the more tenaciously it tends to be held.

If my terminology is correct, "clocking a wheel" means to predict where the ball will land judging by the ball speed, ball location, and wheel speed. It sounds like what you are doing is exploiting a biased wheel, which is a different advantage play. As long as we’re on the topic, a third advantage play is exploiting "dealer signature," where the croupier is so consistent that the ball and wheel speed are nearly the same every spin. This allows the player to predict where the ball will land based on ball location and past results.

To answer your question, the expected number of times you should have hit your number is 8672/37=234.38. The variance is 8672×(1/37)×(36/37)=228.04. The standard deviation is the square root of the variance, or 15.10. You had 278-234.38=43.62 more hits than expected. That is (43.62-0.5)/15.10 = 2.8556 standard deviations. The reason for subtracting 0.5 is hard to explain. Suffice it to say it is an adjustment factor for using a continuous function to estimate a discrete function. Doing a Gaussian approximation, the probability of hitting your number that many times, or more, is 0.21%. So, there is a good chance you found a biased wheel. However, there is still a 1 in 466 chance it was just good luck.

The math works out more easily if he bet on the Player. I work out a similar problem in roulette at my mathproblems.info site, problem number 116. For even money bets, the general formula is ((q/p)^b-1)/((q/p)^g-1), where:

b = starting bankroll in units.
g = bankroll goal in units.
p = probability of winning any given bet, not counting ties.
q = probability of losing any given bet, not counting ties.

Here the player starts with $12 million, or 60 units of $200,000, and will play until reaches 120 units or goes bust. So in the case of the Player bet the equation values are:

b = 60
g = 120
p = 0.493175
q = 0.506825

So the answer is ((0.506825/0.493175)⁶⁰-1)/(( 0.506825/0.493175)¹²⁰-1) = 16.27%.

It is much more complicated on the Banker bet, because of the 5% commission. That would result in the distinct possibility of the player overshooting his goal. If we add a rule that if a winning bet would cause the player to achieve his goal, he could bet only what was needed to get to $12 million exactly, then I estimate his probability of success at 21.66%.

A simpler formula for the probability of doubling a bankroll is 1/[1+(q/p)^b].

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

Counting the first trial, I show the mean is 8.408797, the median is 8, and the mode is 7.

The probability of two numbers without a repeat is 37/38 = 97.37%.

The probability of three numbers without a repeat is (37/38)×(36/38) = 92.24%.

The probability of four numbers without a repeat is (37/38)×(36/38)×(35/38) = 84.96%.

Following this pattern, the probability of no repeats in 8 numbers is (37/38)×(36/38)×(35/38)×...×(31/38) = 45.35%.

So the probability of a repeat within 8 numbers is 100% - 45.35% = 54.65%.

I suspect most people would estimate that that probability of a repeat within 8 numbers would be less than that. If you’re not above taking advantage of your math-challenged friends, propose a bet that it will take 8 or fewer numbers for at least one to repeat. So you would be betting on 8 or fewer, and your friend 9 or more. If he/she balks, then offer to take 7 or over, which would have a 55.59% chance of winning. Basically, whichever side covers the median of 8 is likely to win.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

No, he is not correct. The probability of Andy's single bet strategy is 1/38 = 2.6316%.

After much trial and error, I devised my "Hail Mary" roulette strategy, which will increase the odds of turning $30 into $1,000 to 2.8074%.

Wizard's "Hail Mary"strategy for roulette:

This strategy assumes that bets must be in increments of $1. In all bet calculations, round down.

Let:
b = Your bankroll
g = Your goal

1. If 2*b >=g, then bet (g-b) on any even money bet.
2. Otherwise, if 3*b >=g, then bet (g-b)/2 on any column.
3. Otherwise, if 6*b >=g, then bet (g-b)/5 on any six line (six numbers).
4. Otherwise, if 9*b >=g, then bet (g-b)/8 on any corner (four numbers).
5. Otherwise, if 12*b >=g, then bet (g-b)/11 on any street (three numbers).
6. Otherwise, if 18*b >=g, then bet (g-b)/17 on any split (two numbers).
7. Otherwise, bet (g-b)/35 on any single number.

1. Bet max($1,min(b/7,(g-b)/6)) on the don't pass.
2. If a point is rolled, and you have enough for a full odds bet, then lay the full odds. Otherwise, lay whatever you can.

The general formula is:

Pr(Ball lands in 1) + Pr(Ball lands in 2) + Pr(Ball lands in 3) - Pr(Ball lands in 1 and 2) - Pr(Ball lands in 1 and 3) - Pr(Ball lands in 2 and 3) + Pr(Ball lands in 1, 2, and 3).

In double-zero roulette, for n number of spins, this comes to 3*(1-(37/38)^n)-3*(1-(36/38)^n)+(1-(35/38)^n).

The following table shows the probability of rolling all three numbers for various number of spins from 3 to 100 for single- and double-zero roulette.

The probability of the 0 and 00 winning would be 1/36 each. If bets on these outcomes paid the usual 35 to 1, then the house edge would be exactly 0%.

The probability of any other number winning would be (34/36)*(1/36) = 2.62%. Compare that to 1/38=2.63% in conventional double-zero roulette. The house edge on any bet on the numbers 1 to 36 would be 5.56%. Compare that to the 5.26% in conventional double-zero roulette. My advice in this game would be to bet the zero and double-zero only.

If anyone can confirm or deny these rules and pays, please let me know.

The following graph shows your results in sequential order on the wheel. The blue line shows your results. The red line is the number you need, 207.11, to overcome the 5.26% house edge.

A chi-squared test on this distribution comes back with a statistic of 68.1 with 37 degrees of freedom. The probability of a result this skewed or more is 1 in 725.

I don't think the chi-squared is the perfect test for this situation because it doesn't consider the ordering of the outcomes, but don't know of a better test. Some have suggested the Kolmogorov–Smirnov test, but I don't think that is appropriate. If there are any other appropriate tests, I'm all ears.

I can say if you had bet the 3-number arc around the number 5, you would have had a 10.57% profit over the spins you recorded. However, if you increased that to a 7-number arc, the advantage drops to 2.84%.

If forced to an answer in plain simple English, I would say the wheel shows evidence, but not proof beyond a reasonable doubt, that the wheel is biased. However, that bias is probably not enough to significantly and confidently overcome the house edge. Assuming the casino doesn't switch around the wheels among the tables, I would say that more data should be collected before betting large amounts of money. I'm sorry this answer is so noncommittal.

This question is raised and discussed in my forum at Wizard of Vegas.

Answering the mean is much easier, so we'll start with that. Let's go through it step by step:

• The first spin is definitely going to be a new number.
• The second spin will have a probability of 36/37 of being a new number. If an event has a probability of p, then the expected number of trials for it to occur is 1/p. In this case, the expected number of trials to get the second number is 37/36 = 1.0278.

• After two numbers have been observed, the probability that the next spin will result in a new number is 35/37. Thus, the expected number of spins after the second number to see the third is 37/35 = 1.0571.
• Following this logic, the average number of spins to see every number is 1 + 37/36 + 37/35 + 37/34 + ... + 37/2 + 37/1 = 155.458690.

The median is much more complicated. To find the exact answer, as opposed to using a random simulation, one needs to use a lot of matrix Algebra. I've discussed how to solve similar problems in other Ask the Wizard questions, so I won't go through the details again. One example of a similar question is the one on getting a 6-6 pair in the hole three times in a row, as discussed in Ask the Wizard #311. Suffice it to say that the probability of seeing every number in 145 spins is 0.49161779, and in 146 spins is 0.501522154. Thus, the median is 146.

This question is asked and discussed in my forum at Wizard of Vegas.

The same as red, 47.37% on a double-zero wheel, 18 black numbers divided by 38 total numbers.

That's true, but it doesn't matter. That’s the same probability of 20 reds followed by a black. The fact is the past doesn't matter in games of independent trials like roulette.

This is probably the most popular of all betting systems, known as the Martingale. Gamblers have been conceiving of it and using it since time immemorial. Like all betting systems, not only doesn't it beat the house advantage, it doesn't even dent it. The reason is the gambler will eventually have a bad losing streak where his bankroll isn't enough to make another double.

This is known as the anti-Martingale and is equally worthless. The times your bankroll gets grinded down to nothing will outweigh the winnings when you hit your target. Regardless of what betting system you use, or none at all, the more you play, the more your ratio of money lost to money bet will approach 5.26% in double-zero roulette.

Here is my solution (PDF).