# Roulette - FAQ

^{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.

^{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%.

The way it works is a player will put $75 dollars on the 1 to 18 $50 dollars on the 3rd 12 and $10 dollars on the 0-00 split for a total of $135 dollars. This covers all but six numbers (19 through 22) and will yield a 15 dollar payout every time the ball misses those 6 numbers EXCEPT when 0 or 00 hits in which case it's 40 dollars. I know it sounds nuts!!! But trust me, I'm here to tell you I have seen it win more than lose. It also works in reverse (duh). I would love to know the true odds of this system, but it's hard to tell someone that it doesn't work when they are walking off my table 2 grand richer:-)

^{8}=.004835. So, you have 99.52% of winning one unit, and 0.48% of losing 255 units.

The first one is the 3q/A-strategy found in R.D Ellison’s book "Gamble to win: Roulette", which has a verified win rate of 7.94% (7500 spins). The system was tested and developed in conjunction with" Spin roultte Gold" by Frank Scoblete and "Roulette system tester" by Eric St. Germain.

The second one is Don Young’s roulette system which is verified to beat the Roulette System Tester from Zumma Publishing(15000 spins).

Now, I must say I’m still a bit sceptical about spending money on these systems, but since they've proved themselves over the long haul, I can't really see no reason why I shouldn’t. I mean, beating these testbooks have to mean something...

What’s your opinion on these systems? And do you think I should try them out??

Thanks alot! Have nice day. Best wishes

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.

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.

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%.

- Did my waiting for 5 spins without a low number popping actually increase the chances of probability for a L to show up?
- What are the odds that a Low number won’t show up 16 times in a row?
- What are the odds that a Low number won’t show up 17 times in a row?

Thank you, I also want to thank you for that Blackjack Guide, I turned $5.00 into $100.00 using your method.

- No
- 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%. - (25/37)
^{17}= 0.1275%.

^{5}*(37/38)

^{5}= 1 in 359275.

### Probability of at Least 50 to 70Successful Roulette Spins

Wins | Probability |

70 | 0.000039 |

69 | 0.000092 |

68 | 0.000204 |

67 | 0.000437 |

66 | 0.000895 |

65 | 0.001759 |

64 | 0.003319 |

63 | 0.006016 |

62 | 0.010489 |

61 | 0.0176 |

60 | 0.028444 |

59 | 0.044313 |

58 | 0.066605 |

57 | 0.096674 |

56 | 0.135627 |

55 | 0.184101 |

54 | 0.242059 |

53 | 0.30865 |

52 | 0.382177 |

51 | 0.460205 |

50 | 0.539795 |

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.

^{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.

### Numbers Hit in 200 Roulette Spins

Numbers Hit | Observations | Ratio |
---|---|---|

31 or Less | 0 | 0 |

32 | 1 | 0.00000001 |

33 | 33 | 0.00000026 |

34 | 1812 | 0.00001428 |

35 | 68845 | 0.00054251 |

36 | 1577029 | 0.01242734 |

37 | 19904109 | 0.15684877 |

38 | 105348171 | 0.83016683 |

Total | 126900000 | 1 |

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

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.

^{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.

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.

Is Andy correct in that the best way to turn $30 into $1,000 is to put the whole $30 on a single number in roulette?

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

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

In other words, always try to reach the goal, with just one bet, if you can, without exceeding the goal. If there are multiple ways to accomplish this, then go with the one with the greatest probability of winning.

What about other games, you might ask? According to the Discovery Channel voice-over guy, "Everyone agrees that roulette is the best get rich quick scheme in the casino." Well, I don't. Even limiting ourselves to common games and rules, I find craps to be better. In particular, betting the don't pass and laying the odds.

Following my Hail Mary strategy for craps (explained below), the probability of turning $30 into $1,000 is 2.9244%. This assumes the player may lay 6x odds, regardless of the point (which is the case when 3x-4x-5x odds are allowed taking the odds). This probability of success is 0.117% higher than my Hail Mary strategy for roulette, and 0.2928% higher than the Andy Bloch strategy.

Andy might argue that my argument above relies on an assumption of a minimum bet of $1, which is hard to find in Vegas on a live dealer game. Expecting somebody might say that, I ran through both games under an assumption of a $5 minimum and betting in increments of $5. In that case, the probability of success using my Hail Mary strategy is 2.753% in roulette and is 2.891% in craps. In both cases, greater than the 2.632% under the Andy Bloch strategy.

In all fairness, the Discovery Channel would have never put the insane rant above on the air and was surely looking for something simple that the masses would understand. Andy was surely giving them something they wanted to hear. The basic premise of his advice is that if you want to reach a certain goal, then a hit-and-run strategy is much better than letting the house edge grind you down with multiple bets. That is definitely true and something I've been preaching for 17 years.

Wizard's "Hail Mary" strategy for craps.

This strategy assumes that bets must be in increments of $1 and wins will be rounded down to the nearest dollar. In calculating bets, never bet so much that you overshoot the goal. Also, never make a bet amount that will cause you to get rounded down.

Let:

b = Your bankroll

g = Your goal

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

So, I hope Andy and the Discovery Channel are happy. I've spent days running simulations to prove them wrong.

This question was raised and discussed on my forum at Wizard of Vegas.

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.

### Roulette Question

Spins | Single Zero |
Double Zero |
---|---|---|

3 | 0.000118 | 0.000109 |

4 | 0.000455 | 0.000420 |

5 | 0.001091 | 0.001009 |

6 | 0.002094 | 0.001939 |

7 | 0.003518 | 0.003261 |

8 | 0.005404 | 0.005016 |

9 | 0.007785 | 0.007234 |

10 | 0.010684 | 0.009937 |

15 | 0.033231 | 0.031066 |

20 | 0.068639 | 0.064476 |

25 | 0.114718 | 0.108254 |

30 | 0.168563 | 0.159750 |

35 | 0.227272 | 0.216265 |

40 | 0.288292 | 0.275379 |

45 | 0.349548 | 0.335089 |

50 | 0.409453 | 0.393835 |

55 | 0.466865 | 0.450467 |

60 | 0.521017 | 0.504191 |

65 | 0.571445 | 0.554501 |

70 | 0.617922 | 0.601122 |

75 | 0.660393 | 0.643951 |

80 | 0.698930 | 0.683016 |

85 | 0.733693 | 0.718435 |

90 | 0.764897 | 0.750386 |

95 | 0.792791 | 0.779086 |

100 | 0.817638 | 0.804773 |

- There are four dice -- two green, one red, and one blue.
- If the two green dice both land on one, then the outcome of the "spin" shall be a zero.
- If the two green dice both land on six, then the outcome of the "spin" shall be a double zero.
- If any other outcome occurs with the green dice, then the 36 possible outcomes of the red and blue dice shall be mapped to the numbers 1 and 36 to represent the "spin."

How does this change the odds compared to conventional roulette?

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.

I recorded 7,456 spins in roulette. The results are as follows. I suspect the wheel is biased but am not sure if the data is conclusive enough to play it.

### Roulette Data

Winning Number |
Occurences |
---|---|

0 | 204 |

28 | 214 |

9 | 175 |

26 | 177 |

30 | 203 |

11 | 181 |

7 | 223 |

20 | 205 |

32 | 184 |

17 | 222 |

5 | 224 |

22 | 241 |

34 | 194 |

15 | 210 |

3 | 209 |

24 | 176 |

36 | 203 |

13 | 217 |

1 | 217 |

00 | 197 |

27 | 173 |

10 | 195 |

25 | 198 |

29 | 217 |

12 | 197 |

8 | 207 |

19 | 163 |

31 | 180 |

18 | 201 |

6 | 186 |

21 | 203 |

33 | 171 |

16 | 164 |

4 | 200 |

23 | 191 |

35 | 163 |

14 | 177 |

2 | 194 |

Total | 7456 |

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 Kolmogorovâ€“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.

In single-zero roulette, what is the mean and median number of spins required for every number to appear at least once?

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.

^{21}= 1 in 6,527,290. The odds must overwhelmingly favor black.

What is the expected number of spins in roulette to see five reds or five blacks in a row?