# Roulette - General Questions

"Anonymous" .

In roulette any bet or combination of bets carries a high house edge. The more likely you are to win the more you will have to risk relative to the reward. If you do this 10 times the probability of showing a profit is 46.42%. At 100 times the probability drops to 24.6%.

Bob

Start the same way as I explained with even money bets. Each bet should be the sum of the left and right numbers. However, following your two column strategy you should add double the amount to the right if you lose.

Jon M. from Danville, New Hampshire

I played it in practice mode and it seems to be a legitimate no-zero roulette wheel. There is no system that can either beat or lose to this game in the long run. The more you play the more the ratio of the net win to the total amount bet will get closer to zero.

**Update**: This casino has since closed.

1) Regarding the basic blackjack strategies, you have for different online casino groups. Specifically, two different ones that both use single deck: Microgaming and Unifed Gaming. I cannot understand why you list 11 VS 10 as a hit for Microgaming, but as a double for Unified. Since they both use single deck, it seems the same strategy should be used here. I lose more often than win when I double this.

2) In Roulette, it seems to me that your odds would be better to bet equally on both red, and the 3rd column, or black and the 3rd column. The 3rd column has, I believe, 8 reds and only 4 blacks. Conversely, the first column has more blacks. Does betting like this lower the house edge?

Brian from Pennsylvania, USA

If you double on 11 at a Microgaming casino and the dealer gets a blackjack you will lose the total amount bet. At Unified Gaming the blackjack would be turned over immediately if the up card were a ten so there is no risk of losing to a blackjack when doubling in this situation.

All combinations of bets in roulette yield the same expected return, assuming the dreaded five-number combination is avoided. You're right that the third column has eight reds and four blacks. The probability of winning 3 units is 8/38, 1 unit is 4/38, breaking even is 10/38, and losing 2 units is 16/38. The combined expected per unit bet is return is (1/2)*(3*8 + 1*4 + 0*10 + 2*16)/38 = -2/38. Betting on black and the third column the probability of winning 3 units is 4/38, 1 unit is 8/38, breaking even is 14/38, and losing 2 units is 12/38. The expected return is (1/2)*(3*4 + 1*8 + 0*14 + 2*12)/38 = -2/38. Both combinations weight the various outcomes differently, but they average to the same number.

Scott from Elmhurst, Illinois

Once you have hit n numbers, the probability of getting a new number on the next spin is (38-n)/38. If the probability of an event is p, then the expected number of trials before it happens is 1/p. Thus the expected number of spins to get a new number, given that you already have n, is 38/(38-n). For example once you have hit 20 numbers the expected number of spins to get the 21st is 38/18=2.11. So the answer is the sum of the expected number of spins at each step: (38/38)+(38/37)+(38/36)+...+(38/1)=160.66.

Doug from Eugene, USA

(18/38)^{18} =~ 1 in 693745.

Oh, and the man behind me in the white coat wants to say that he loves your web site and is grateful for all the work you have put in to show the math. He says it saved him a lot of time and money. Keep up the good work.

Iggy

I’m happy to have helped the main in the white coat. My roulette advice is limited to games on earth, bribery is recommended on your planet.

Bill K.

The probability of your number hitting exactly x times is combin(1000,x)*(1/38)^{x}*(37/38)^{1000-x}. The following table shows the probability of all number of hits from 0 to 6 and the total.

### Wins in 1000 Roulette Bets

Number | Probability |

0 | 0.00000000000262 |

1 | 0.00000000007078 |

2 | 0.00000000095556 |

3 | 0.00000000859146 |

4 | 0.00000005787627 |

5 | 0.00000031159330 |

6 | 0.00000139655555 |

Total | 0.00000177564555 |

So the answer is 0.00000177564555, or 1 in 563175. I hope this didn’t happen at an Internet casino.

You may wonder why I didn’t use the normal approximation as I did with the coin flipping problem above. That is because it doesn’t work well with very high and very low probabilities.

"Anonymous" .

You accessed my strategy quite well. I intended to bet the minimum through most of the hour to avoid the house edge grinding me down. The producer was not happy with such a boring strategy so I tried to act more excited than I normally would be. The reason I picked a single number as opposed to an even money bet is that I wanted volatility. I knew if I ended close to my starting point of $10,000 I would probably lose. So I wanted a bigger shot at getting ahead. I stuck with 23 the entire time, although towards the end I added 5 as well. I’m Glad you liked the show.

"Anonymous" .

No. Dealers are taught only the basics and nothing nearly that skillful. In fact if a dealer had that control he could simply get an accomplice to bet wherever he planned for the ball to land and they could easily make millions.

"Anonymous" .

The probability of losing any one spin is 1-(8/37) = 78.38%. So the probability of losing 15 spins is .7838^{15} = 2.59%.

"Anonymous" .

First let me say how not to test any game. You should not record all your play and when you are done look for any kind of anomaly, and then write to me complaining that the casino is cheating. The correct way is to state a hypothesis for how the casino is cheating FIRST, then gather data, and finally see if the data fits your hypothesis. If you don’t know what kind of hypothesis to state I would suggest simply testing for the number of wins and losses, and then bet the same thing every time. If you do the first two steps properly and need help on the third then feel free to write me.

Jacob from Tel Aviv

In your example the casino’s hold is 75%. That is the ratio of how the value of chips players walk away with to the amount of money the dealer drops in the box. Although there are historical averages there is no way to go from house edge to hold. The hold depends in part on how long the player will circulate through his chips, and there is no way to know that.

Andy from Amsterdam

Thanks for the compliment. There is a formula for questions like this, which I explain on my site www.mathproblems.info, see problem 116.

With a bankroll of b units, winning goal of g units, probability of winning p, and probability of losing q your probability of success is ((q/p)^{b}-1)/((q/p)^{b+g}-1). In this case b=20, p=18/37, q=19/37, and g=20, 50, and 100. So for a bankroll of $200 the probability is ((19/18)^{20}-1)/((19/18)^{40}-1) = 0.253252.

For a winning goal of $500 the probability is ((19/18)^{20}-1)/((19/18)^{70}-1) = 0.045293.

For a winning goal of $1,000 the probability is ((19/18)^{20}-1)/((19/18)^{120}-1) = 0.002969.

Paul from Raleigh

It doesn’t make any difference what the past spins were. The probability of red 23 three times in a row is (1/38)^{3} = 1 in 1 in 54,872.

"Anonymous" .

The probability of three pairs and six singletons in twelve spins is combin(38,3) × combin(35,6) × combin(12,2) × combin(10,2) × combin(8,2) × fact(6)/38^{12} = 9.04%. The math gets rather messy asking about the probability that this could happen in any 12-spin span over 20 total spins. Suffice it to say that it is significantly more than 9%, more likely than not, I would guess. So these seem like very normal results to me.

A. Carter from Derby

The house edge on the specified number would be exactly 0%.

Haig from Englewood

The probability of winning is (18/38)^{3} = 10.63%. The house edge is 8×0.1063 - 1×0.8937 = 4.34%, which is less than the 5.26% on all the other bets (except the dreaded 0,00,1,2,3 combination at 7.89%).

Maff from England

For the benefit of other readers, at many casinos in Europe, if you make an even money bet in roulette and the ball lands in zero, then the wager becomes imprisoned. If the next roll would cause the wager to win, then it is returned without winnings. If the ball lands the opposite way, then the bet loses.

What happens to single-imprisoned bets if the ball lands in zero? If the casino allows only single imprisonment, then it would lose. However, some casinos allow for double imprisonment, which is what would happen. If a double imprisoned bet wins, then it returns to being single imprisoned. If the casino allows only up to double imprisonment, then a double imprisoned bet would lose on another zero. By the same logic, if a casino allows triple imprisonment, then a double imprisoned bet would become triple imprisoned on another zero.

The house edge under triple imprisonment is 1.370120%. Before I explain how I arrived at that, let’s do single and double imprisonment first. Assume the first bet is on red.

Let Z = probability of a zero = 1/37.

Let R = probability of a red number = 18/37.

**Single imprisonment**

pr(push) = ZR = 0.0131482834.

pr(win) = R = 0.4864864865.

pr(loss) = 1-pr(push)-pr(win) = 0.5003652301.

Expected value = pr(win)-pr(loss) = -0.0138787436.

**Double imprisonment**

The player will push if the first spin is a zero, then the player can repeat ZR from 0 to an infinite number of times, and then get a red. In other words, the ways to push are:

ZR, Z(ZR)R, Z(ZR)(ZR)R, Z(ZR)(ZR)(ZR)R, ...

pr(push) = 0.013323464

pr(win) = 18/37 = 0.4864864865.

pr(loss) = 1-pr(push)-pr(win) = 0.5001900494.

Expected value = pr(win)-pr(loss) = -0.0137035629.

**Triple imprisonment**

First, let’s find the probability p_{1} that a single imprisoned bet becomes triple imprisoned with two more zeros, and then eventually rises back up to being single imprisoned again. This can happen as follows:

ZZRR, ZZ(RZ)RR, ZZ(RZ)(RZ)RR, ZZ(RZ)(RZ)(RZ)RR, ...

In other words, the bet can bounce between triple and double imprisonment up to an infinite number of times.

Second, let p_{2} = the probability that a single imprisoned bet reaches the first or second level of imprisonment and then returns to single imprisonment.

let p_{2} = ZR + p_{1} = 0.013323464.

The player can repeat returning to the first level from zero to an infinite number of times. So the probability of a push is:

ZR + Z p_{2}R + Z p_{2} p_{2}R + Z p_{2} p_{2} p_{2}R + ... =

Z × (1/(1- p_{2})) × R = 0.013325830.

pr(push) = 0.013325830.

pr(win) = 18/37 = 0.4864864865.

pr(loss) = 1-pr(push)-pr(win) = 0.5001876839.

Expected value = pr(win)-pr(loss) = -0.0137011974.

**Infinite imprisonment**

Not that you asked, Maff, but in Spain I hear they allow infinite imprisonment. Let p = probability of a push. This is also the probability of starting at level x of imprisonment, falling deeper in levels, but eventually climbing back up to x.

p = ZR + ZpR + ZppR + ZpppR + ...

p^{2} - p + ZR = 0

By the Quadratic Formula p = (1-(1-4*RZ)^{1/2})/2 = 0.0133258620.

pr(push) = 0.0133258620.

pr(win) = 18/37 = 0.4864864865.

pr(loss) = 1-pr(push)-pr(win) = 0.5001876515.

Expected value = pr(win)-pr(loss) = -0.0137011650.

**Zeros ignored imprisonment**

Finally, at some casinos zeros are simply ignored after the first zero which imprisons the bet. There the probability of a push is simply (1/37)×(1/2) = 0.0135135135.

The following table summarizes all four kinds of rules.

### Imprisonment Odds

Imprisonment | Win | Push | Loss | Expected Value |

Single | 0.48648649 | 0.01314828 | 0.50036523 | -0.01387874 |

Double | 0.48648649 | 0.01332346 | 0.50019005 | -0.01370356 |

Triple | 0.48648649 | 0.01332583 | 0.50018768 | -0.01370120 |

Infinite | 0.48648649 | 0.01332586 | 0.50018765 | -0.01370117 |

Zeros ignored | 0.48648649 | 0.01351351 | 0.50000000 | -0.01351351 |

I’d like to thank ChesterDog and weaselman for their mathematical help. I’d also like to give thanks and a plug to professor G. Artico and polarprof.it for the HTML of the summation formula above.

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

Is there any pattern to the order of the numbers on the roulette wheel?

"Anonymous" .

The idea is that the wheel should be balanced. In other words, if you took the average of a section of consecutive numbers on the wheel, they would be close to the average number 18.5. I thought of a way to explain how the numbers are distributed on a double-zero wheel, as follows.

- Position the 00 at the 12:00 position and the 0 at the 6:00 position.
- The numbers at the top of the wheel are 13, 1, 00, 27, 10*, 25. This must simply be memorized.
- All low odd numbers (from 1 to 17) are on the left side and the high odds (19 to 35) on the right side.
- Starting with the 1 and 13, move four positions counter-clockwise to get the next odd, but do not pass the zero.
- As an exception to rule 4, do not place the 19 four positions counter-clockwise from the 17, because the 19 is a high odd and belongs on the right side. Instead, put the 11, the only low odd not otherwise covered in rule 4.
- All even numbers are directly across the wheel, 180 degrees, from the preceding odd number.

I'm at a loss to explain the order on a single-zero wheel. For now, I can say that if you put the zero in the 12:00 position, then:

- On the left side will be all red low numbers and black high numbers.
- On the right side will be all red high numbers and black low numbers.

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