 Ask the Wizard: |
Questions about Roulette |
I have a friend who was part of a casino staff who watched
over roulette tables, and he told me that when people start to win the casino changes
the croupier. I have also seen a member of the staff ask a croupier to spin the
roulette wheel at a different speed. Doesn't this mean that the casinos are certain
that the croupier can cause a non random series of numbers to appear?
Doesn't this mean that a gambler can look for a "lucky" table where the croupier
doing regular spins gives them a better chance of winning?
— Al from Melbourne Australia
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. July 11, 2008
In Roulette (0,00) board, what is the probability that any number will not have hit by the 200th spin? — J.F.W. from Marshall
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.
| 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.
September 30, 2007
I found these crappy roulette odds from the ferry between England and the Netherlands:
- 1 Number: 30 to 1
- 2 Numbers: 15 to 1
- 3 Numbers: 10 to 1
- 4 Numbers: 7 to 1
- 5 Numbers: 5 to 1
- 6 Numbers: 4 to 1
And it's American-style (double-zero), despite the fact that the ferry is going between two European countries. What are the odds? — Spanky McBluejay
Shame on that ferry operator. The house edge varies from 13.16% to 21.05%, as follows.
| Netherlands/England Ferry Roulette |
| Numbers |
Probability |
Pays |
House Edge |
| 1 | 2.63% | 30 | 18.42% |
| 2 | 5.26% | 15 | 15.79% |
| 3 | 7.89% | 10 | 13.16% |
| 4 | 10.53% | 7 | 15.79% |
| 5 | 13.16% | 5 | 21.05% |
| 6 | 15.79% | 4 | 21.05% |
September 30, 2007
Hello, for obvious reasons I would appreciate if you not share my
name with anyone. At the casino where I work, there is virtual unanimity amongst the roulette dealers that they can "control the spin" and hit sections of the wheel with ease or miss other sections on purpose. Given all the factors in the spin of a ball in roulette, including the canoes (bumps) on the side, direction of ball and speed of wheel, etc. -- do you think there is anything to this? Would it be enough that a crooked dealer could help a player overcome the huge house advantage?
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. September 11, 2007
I have been playing roulette at the X Internet casino and I feel there is some foul play involved in their random generation system. I wanted to report this to your website so other people are also aware of it. I have a screenshot of how the numbers appeared - 6, 6, 22, 22, 30, 22, 30, 9, 22. In 7 turns 22 appears 4 times. I would like to upload that screenshot to show this. Would this be considered a flaw in their random number generator, which by the way is audited by Price Waterhouse Cooper? – Ashwal from Waterloo
The probability of getting any number exactly four times in seven spins is 38 × combin(7,4) * (1/38) 4 × (37/38) 3 = 0.000589, or 1 in 1698. Assuming 200 spins per hour, you should see this about once every 8.5 hours. I’m sure you cherry picked this sequence out of lots of ordinary looking spins. So I’m afraid this evidence doesn’t nearly rise to the level required for a kind of case of foul play. May 21, 2007
I'm a long-time subscriber of your newsletter and still loving your web site. I came across a casino web site that offers roulette in which the wheel doesn't contain any zeros. It just has number 1-36, and all the standard roulette rules apply. Do you see any way to take advantage of this? I know you don't like betting systems but in this case there is no house edge. There must be a money management system that could work profitably with these table limits. Any advice is appreciated. – Mark from Gatineau, Quebec
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. March 29, 2007
If I show you a Roulette game with no zeros and all usual roulette rules apply, is it possible to win 100% of the time? – Jon P. from London
You say there's no winning system in roulette. Gonzalo Garcia-Pelayo
and his family won a lot of money in many casinos all over the world. They even published a book, and describe how they did it. What's your opinion? – Jose from Spain
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. February 14, 2007
Single 0 roulette. What is the probability of having any one number (0-36) come up more than once in three consecutive spins? Thanks for you time! – Jeff from Rochester
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%.
November 10, 2006
On a 38 number Roulette wheel....after hitting three reds, a green, and then another red....what are the odds, that the next consecutive three spins would come up red 23? – 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 54,872.
November 10, 2006
Hi Wiz. Great Site. Just wanted to know why many casinos impose bet limits on roulette tables. e.g. you can bet $100 on RED but only $10 on say the number 10. Is the casino avoiding some sort of risk or are they safe-guarding me? – Kevin from Johannesburg, South Africa
Thanks. I think they are trying to minimize risk by limiting the maximum win.
September 22, 2006
If I have a $200 bankroll that I don't mind losing, and keep playing $10 on one single number on European (single-zero) roulette, what are the probabilities of winning, $200, $500 or $1000? Assuming I'll stop after reaching the target. Thanks, great site, you wanted me to keep this short :) – 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.
September 13, 2006
www.ccc-casino.com has no zero roulette which they call Super Chance Roulette. Are there any systems that would be effective since there is no zero? Without the zero could one effectively play both black and red at the same time since there is no fear of the zero? – Jon from Danville, New Hampshire.
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.
July 31, 2006
I work in a casino and have a bet that says a roulette dealer cannot influence the outcome of a roll. There are definitely those who think it can be done. Not to a number of course, but perhaps a section of the wheel. What would you consider a good test to reasonably determine whether a dealer has influenced the results? Assuming the number of trials is reasonable for us to attempt I will gladly share the results. – Mark S. from Sault Ste. Marie
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.
Probability of at Least 50 to 70
Successful 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 |
July 11, 2006
Can you please explain to me how the table limits for roulette works and what is the difference between minimum limit for individual number and table. If possible, please give examples. – Nic
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.
June 9, 2006
Congratulations on a great site. I totally understand your anger over the spread of 6 to 5 Blackjack payouts but am very curious as to why Americans seem to accept 00 Roulette without any argument. This Roulette is almost criminal and should be ranked along side Keno & Slots. - Andrew from Sydney
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.
March 20, 2006What are the chances of a dealer hitting 5 of the same
number in 10 spins of the roulette wheel? -- Spence from
Red Deer, Alberta
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 359,275. Oct.
3, 2005
I was at Casino On Net. I was playing Roulette. I was
making safe bets, only betting on the 1st 12(L), 2nd 12(M)
& 3rd 12(H). I spun the wheel 5 times without betting,
waiting for a pattern of one of the sets to not come up so I
could bet on it, hoping this would shift probability of it
landing in my favor. 5 spins later L didn't show. I kept
betting in L, I figured the numbers 1, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11 OR 12 would land within 12 Spins so that I could
at least recoup my money...but they didn't. The table went
17 spins in a row without a LOW number & I went from
$258 to $0.00...it was Bonus Money anyway. This is a 3 part
question:
- 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. - Brad from
Sydney, Nova Scotia
- 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%. Sept.
25, 2005
I realize that decisions per hour in games like
blackjack and craps can depend heavily on factors like the
number of other players at the table, the hand shuffle vs.
machine shuffle, shooter and dealer speed. Still, I was
curious if you could give me a rough approximation of how
many decisions per hour an individual can expect at a
mostly-full craps table and a blackjack table with both a
hand shuffle and machine shuffle. This would help me
estimate my expected loss per hour and weigh it against the
comps I am being offered.
The following tables show the number of
hands/tosses per hour in blackjack, craps, and roulette.
The source of the tables is
Casino Operations Management by Jim Kilby. Aug.
28, 2005
|
Hands per Hour in Blackjack
|
|
Players
|
Hands per hour
|
|
1
|
209
|
|
2
|
139
|
|
3
|
105
|
|
4
|
84
|
|
5
|
70
|
|
6
|
60
|
|
7
|
52
|
|
Rolls per Hour in Craps
|
|
Players
|
Rolls per hour
|
|
1
|
249
|
|
3
|
216
|
|
5
|
144
|
|
7
|
135
|
|
9
|
123
|
|
11
|
102
|
In craps 29.6% of total rolls are come out rolls, on
average.
|
Spins per Hour in Roulette
|
|
Players
|
Spins per hour
|
|
1
|
112
|
|
2
|
76
|
|
3
|
60
|
|
4
|
55
|
|
5
|
48
|
|
6
|
35
|
What are the odds of playing 15 spins on European
roulette, covering at eight numbers and not getting any of
them?
The probability of losing any one spin is
1-(8/37) = 78.38%. So the probability of losing 15 spins
is .783815 = 2.59%.
Dec. 13, 2004
My question is based more from observations on your
part and rumors that I've heard on my part. If it's true
that Las Vegas dealers, when they are taught at dealer
school, learn how to spin and roll the ball the same way
then is it true after observing how a dealer spins, one can
determine the quadrant of the wheel the ball might land on?
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. Sept. 23,
2004
I live in NJ about two hours north of Atlantic City.
Do you have an idea as to where the closest European
Roulette Wheel to someone in my part of the country is?
There are lots of single zero wheels in Atlantic
City. Most of the casinos there have them, but at a $25
minimum. Sept. 14,
2004
What is my risk of ruin with the following betting
pattern in roulette?
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.
Aug. 12, 2004
Brilliant Site. My question is whether or not free
online demo games follow the same system as their money
playing counterparts. I ask because I was playing
Netgaming.com's
free roulette and it appeared as though the system was
purposely letting you win in order to bait you into playing
for money. Do online sites do this? I was basically trying
the Martingale, which I know is false, but I turned the 1000
very quickly and easily into 10000.
To answer your question I gave Netgaming a try.
In their single zero roulette game I placed 200 bets on
red. My results were 133 wins and 67 losses. The
probability of 133 or more wins in 200 spins is 1 in
3,788,515. So obviously they were letting me win. Let the
record show I do not approve of manipulating the odds for
any reason. So to Netgaming.com,
and any other casinos that do this, I say shame on you.
However to fiddle with the odds with real money would be a
much greater sin and worthy of inclusion on my blacklist.
May 13, 2004
There is a
story today about a British man who will bet his
life savings on one roulette roll. My friend and I have been
debating about what the best casino bet is for this type of
wager. If you can only place one bet, and you wish to
maximize your odds, what is the best game to play and what
is the best bet?
First, let me say this guy was a fool. He bet
$138,000 on a normal American roulette wheel with two
zeros and a house edge of 5.26%. This amounted to an
expected loss of $7263. 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%.
May 5, 2004
Hi Mike, Just to lighten the mood a little, I remember
watching you play roulette on a special Vegas Challenge show
a while ago. Obviously you knew your expected value wouldn't
change regardless of strategy. I remember you playing only
one number, but I forgot if you changed after a win. Not
sure what I'd do in your situation, but out of curiosity,
did you play the single number as opposed to an outside bet
like red/black to maximize the variance? That way if you get
"lucky" you'll be ahead by the most? Anyway, I enjoyed the
fact that you played a really conservative and boring
strategy.
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. April
11, 2004
Betting all 38 numbers on roulette would make it
impossible to beat the odds even for a short time, and with
a $1 bet per number there would be a loss of $2 per roll of
the wheel.(0, 00 wheel, without advantageous rules for even
money bets) Would it seem reasonable that there should be an
optimum range of numbers to bet based on statistics?
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. Oct. 15,
2003
I played the same number 1000 spins in a row on a 0,00
wheel and hit 6 times. What are the chances of hitting 6 or
fewer times in this scenario? Thanks, 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.
June 27, 2003
|
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.
June 27, 2003
Dear Sir, In a single zero roulette game, the
PROBABILTY of winning increases if you place a portion of
your money on fewer numbers for more spins versus covering
more numbers per spin, an example: If you are willing to
risk 500$ in order to win 250$ then you could: Option (A):
place 250$ on any of two dozen and should you be a winner
you will win 250$. The probabilty of that happening is
24/37=(.648648). Option (B): Place 125 on any one dozen and
should you be a winner you will win 250$ and walk away.
However, should you lose you can now bet187.5$ on the same
dozen and should you be a winner you will win 375$ which
will get you the 250$ and the 125$ you lost on the previous
spin. Now should you lose on both spins you still have
187.5$ to play with and you can place 20.833333$ on any nine
numbers and should you be a winner you will get 750$ which
is equal to your 500 original capital plus 250$ in winning
which was your goal. The probability of that happening
meaning either hitting a dozen OR nine numbers at LEAST once
in three spin is equal
to[1-(25/37)x(25/37)x(28/37)]=0.65451. Hence, for
the SAME capital and for the SAME payoff you are able to
increase your PROBABILITY of success as in option (B) if you
play fewer numbers with less money but for MAYBE more
spins.(As you might win on the first spin) You can even
improve your probability further if you play only six
numberes at a time and try to win 250$. Any
Explanation??!!!! Assuring you of my highest regards and
awaiting the favor of your reply I remain.
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.
April 17, 2003
Hello. I¥ve been an avid roulette gambler for
some years now and for the first time ever I¥m thinking
of trying out a roulette system...Now I know how you feel
about these so called "systems" and the scammers behind
them, and belive me, I feel the same way, but I¥ve come
across two systems which can¥t be ignored...
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 sceptic 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 -
Johan
7500 spins? Is that it? Anyone can win 7.94%
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.
March 21, 2003
I was fascinated to read your comment regarding little
steel balls not having a memory. I have a roulette wheel at
home and the little balls jump up and down with glee when I
come home from hospital. Clearly that's because they
remember me, and what's more I think they would be offended
by your remarks. On my planet roulette is not allowed
because the little balls can be bribed so easily. I think
your planet stinks and now I can't afford to repair my
spacecraft to get back home.
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.
Nov. 28, 2002
Hi. You say all betting systems will fail. If you play
roulette and bet one unit on number 1-12 and 2 units on
number 13-24 wouldn't you then have 66.66% chance to break
even or win. - Atle from Porsgrunn, Norway
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.
Nov. 3, 2002
My co-worker D. insists that he has perfected a way to
consistently win at Roulette. I'm not convinced. Is he just
lucky or is there any system that works? - Tom Davey from
Merritt Island, USA
He is just lucky. As I have said thousands of
times, no betting system can pass the test of time.
May 8, 2002
Hi Wizard, you've got a great site. In double 0
roulette, I realize all the bets have the same high house
edge, but I'm just finishing a stats course and it seems to
me that not all the bets are quite the same due to their
standard deviations. A $1 bet on Red, for instance, has an
S.D. of 1.012019 while a $1 bet on a single number has an
S.D. of 5.839971, according to my calculations. Thus, the
expected probability of coming out ahead over 1, 100, and
10000 trials, respectively is 0.4793, 0.3015, and 0.0000 for
an even-money bet, and 0.4964, 0.4641, and 0.1837 for a
single number bet. Is my analysis correct? (I assumed
normality) Thanks! - Mike from Toronto, Canada
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.
April 22, 2002
I have read everything you have to say about roulette,
but see nothing about roulette spinners. Under the watchful
eye of the pit boss the roulette spinner throws "sections"
to improve the odds of the house. The spinner that can throw
greens, not every time but with good percentages of time,
maybe one out of 7 or 8 when he wants to wipe out a large
progressive better.
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. Jan. 2,
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
What would be the odds of having a red number turn up
18 times in a row on Roulette. - Doug Jeffrey from Eugene,
USA
(18/38)18 =~ 1 in 693745.
Nov. 11, 2001
Everybody says that roulette cannot be beaten in the
long run with mathematical systems. But how can you explain
the fact that there are professional gamblers who make their
living at roulette? I don't think it's just boast. They
actually win more than they lose with everyday playing. -
Denis
I have never seen any evidence that such a
player exists. Anyone who is claiming this either is just
lucky or is lying for the sake of selling worthless
betting systems. May 13,
2001
On average, in single 0 roulette, how often will a
number repeat (ex. two 8's in a row) over the course of 36
spins? - Jon Moriarty from Danville, New Hampshire
The expected number of repeats is 36/37.
May 1, 2001
Hey, Wiz. I am curious to know what the expected
number of spins is for an American Roulette wheel before all
38 numbers have been chosen at least once. Is this
proportional to the number of selections (38) or is it
exponentially related to this number? I tried figuring this
out for a 6-sided die but got stuck fast. - Scott R. Walshon
from Elmhurst, Illionois
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 product of the
expected number of spins at each step:
(38/38)*(38/37)*(38/36)*Ö*(38/1)=160.66.
March 11, 2001
While Roulette clearly cannot be beaten by chance, I
have heard that it can be beaten by physics 2 ways (in
theory). Way one: a high tech device, which measures the
velocity of the ball against the velocity of the wheel and
predicts the outcome sector of the wheel with like a 40%
accuracy. Way two: Wheel bias. Obviously a wheel would have
to have a bias of at least 5.26% to get the player to an
even keel. The question is, how many spins would you say,
wizard does it take to determine wheel bias, if there is
any? - JF from Providence, USA
Any device to measure ball and wheel velocity
would not be very welcome around a roulette table.
However some people claim to be able to judge in their
head roughly where the ball will land, with enough
accuracy to overcome the house edge. I have yet to be
convinced that anybody is winning long term with this
method. Roulette wheels today are very high quality and
the bias should be negligible. Iím even more
skeptical anyone is winning tracking wheel bias. Some
casinos track all the numbers to check for bias
themselves. Feb. 10,
2001
Say, wizard I was wondering about roulette. With the
boss media software, you can spin the wheel without actually
placing a bet. Isn't this in the player's favor? With some
of the roulette betting systems I have seen, like the
Martingale you double up when you lose and such. Can't you
just watch the wheel without placing a bet and based on the
previous spins place your bet. An example would be to spin
the wheel 5 times before placing a bet. Suppose that all the
numbers were odd. Wouldn't it make sense to begin playing
the wheel by betting on even? I know this kinda buys in to
the gamblerís fallacy and the wheel doesn't care what
the last spin was, but also the probability diminishes with
each spin that an odd number will continue to come up. Am I
on to something or just going over an old theory? - Craig
from Detroit, USA
Youíre just rehashing the
gamblerís 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.
Jan. 20, 2001
I read your page on systems and I have been telling
people this for years! I deal roulette in a casino and I
have seen all of the systems at one time or another. I have
seen one system that, even though on a computer simulation
might not work (probably won't), "Seems" to work in real
life. That means that I have seen it win more than lose. 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 doesnt work when they are walking off
my table 2 grand richer:-) - ? from ?
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. Jan. 14,
2001
Isn't it an even worse roulette betting strategy to
bet multiple numbers on the inside during one bet (as most
players do) vs. making a sequence of independent bets on one
number? For example, if one had $100, betting 10 bets of $10
dollars on the number "8" would lose less than betting $10
on 10 numbers on one spin? It seems to me that "hedging"
just guarantees that certain (in the above case 9 bets) will
ALWAYS lose? You don't address "hedging" on your page? -
Kevin Albright of Dallas, USA
See my Ten
Commandments of Gambling, number six is to never
hedge your 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%. Jan. 14, 2001
I was playing roulette last nigh using the
"Martingale" method of doubling down twice after the 1st
loss. Dumb I know but I usually donít lose much and I
gamble a long time. Anyway, what ended the game for me was I
was betting even, and in 4 rolls the number 9 came up in 3
of the spins. What are the odds of that? Does that sound
suspicious? For that matter, have casinos ever been caught
cheating? - Jim West of St. Peters., USA
The probability of getting any number 3 times
out of 4 is 38*4*(1/38)3*(37/38) = 1/5932.
However if you play long enough you almost canít
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. Jan.
14, 2001
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 1st column has more
blacks. Does betting like this lower the house edge?
ñ Brian from Pennsylvania, USA
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 8 reds and 4 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 ñ 0*10
ñ 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 ñ
0*14 ñ 2*12)/38 = -2/38. Both combinations weight
the various outcomes differently but they average to the
same number. Dec. 24,
2000
Is there a way of combining bets in roulette to
maximize one's odds? For example, a dozen bet pays out 2 to
1. If I place two dozen bets, say the first and second set
of 12, I have a 63.16% chance of having it pay off. These
are better odds then a simple red/black, even/odd, or
high/low bet. Although I really only gain 1 to 1 rather than
2 to 1(if I win, since part of my bet has to lose since the
winning number cannot be in both the first and second set of
twelves), the odds have been slightly shifted in my favor by
combining two bets. Have the odds on these sorts of
combinations been determined? If they have been, where might
I be able to find them? K from USA
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.
Nov. 19, 2000
Suppose I were to stroll into a casino and place a bet
on 2 of the 2:1 payoffs in roulette, such as $100 on 1-12
and $100 on 13-24. In a one time deal, isn't my odds of
taking home $100 a generous 63%? I'm not talking about the
long run; just a one-time bet. ñ Andrew Chen from San
Diego, USA
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%. Oct. 5,
2000
www.ccc-casino.com has no zero roulette which they
call Super Chance Roulette. Are there any systems that would
be effective since there is no zero? Without the zero could
one effectively play both black and red at the same time
since there is no fear of the zero? - Jon Moriarty 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. If any
viewer knows if there is some hidden catch to this game
please let me know. Sept. 10,
2000
Q: What is the weakness in playing roulette #'s to
repeat, within about 50 or so spins, using the law of
unequal distribution? I'm sure there is one, just not sure
of what exactly it is. Thank you. - Colin Andrews from
Staten Island, New York
A: There is no advantage or disadvantage in
playing any number for any reason on a fair roulette
wheel. I checked my statistics books and could find no
mention of the "law of unequal distribution."
July 18, 2000
Q: If one bet on two columns in roulette the
probability of winning would be 24/38, or 63%. This seems
like a winning strategy to me, what is your opinion?
-Anonymous
A: 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%.
Feb. 5, 2000
Q: How would you fare if you played roulette like
this: Bet $5 on both 0 and 00, bet $15 on two of the
columns. Wouldn't you have like a 70% of winning? - Matthew
Moran of Kansas City, USA
A: You would have a 5.3% chance of winning $140,
a 63.2% of winning $5, and a 31.6% chance of losing $40.
As with any method of playing roulette you will lose over
the long run. Feb. 19,
2000
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