Roulette Repeaters
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I'm a roulette dealer and am getting more and more into learning the math behind all of these games. My understanding is that the odds of a number repeating on a 38 number wheel is 1 in 38^2, and a threepeat being 1 in 38^3
Yes? No?
I'm curious especially about the repeat, because 38^2 is 1,444 as I recall, which would mean I'd probably only see one every few days, but I see them a few times per shift, which makes me question my math.
So, any input? Thanks!
(Might be bullshit):
Seeing three numbers in a row, then four, will be 1 in 38^2 and 1 in 38^3. Seeing 3-4 numbers in a row is maybe once a shift/day kind of a thing.
Quote: inversehelixHowdy,
I'm a roulette dealer and am getting more and more into learning the math behind all of these games. My understanding is that the odds of a number repeating on a 38 number wheel is 1 in 38^2, and a threepeat being 1 in 38^3
Yes? No?
I'm curious especially about the repeat, because 38^2 is 1,444 as I recall, which would mean I'd probably only see one every few days, but I see them a few times per shift, which makes me question my math.
So, any input? Thanks!
The odds of any repeated number, given a fair wheel, is 1/38. So you should be seeing it a few times per shift.
Now with that in mind, I should probably alter my question a bit.
What is the chance of *any* repeater? And how do I calculate it? Let's say we're not calling a number first, but simply some random number coming up twice? To put some numbers out there, On a 38 number wheel across 10k spins, how many repeating numbers can I expect to see?
As for the ball not having a memory. . . heh yeah. I get it. Players come to my table all the time and tell me about their new system and how I have to read _____ because it's going to put me out of a job. I just smile and politely collect their money.
On a slight tangent, since some of you may enjoy this, I e-mailed James Randi (randi.org for those not familiar), and he was willing to accept challenges from people who claim to have betting systems that can beat the house edge, especially if they're charging for access to those systems.
EDIT: ahiromu is correct, that it should be 9999, not 10000. It's not a significant difference either way
Quote: am19psuSince the probability of a repeating event is 1/38 = .0263157, then the expected number of repeaters over 10k spins is 10000*.0263157 = 263.1. Does this make sense?
Perfect sense, Thanks!
Actually I feel kind of stupid for not figuring that one out myself. I'm new to calculating these sorts of things, but the answers always make sense.
Seriously, thank you.
How often do you see a three-peat?
Have you ever seen, or hear about, a four-peat? How about a five-peat?
The reason I ask is because I've noticed that it's very unusual for a table's spin history display to not have at least one repeat on it. And that display only shows, what?, about 20 spins? Quite often I've seen more than one repeat. So it got me wondering about the three-peat and more.
Quote: joshyeltmanOn average, how many spins does it take to see a previously rolled number? Not necessarily back to back, but to repeat any number that has previously been rolled. A couple of us have made guesses, but we can't figure out an accurate way of calculating the average.
I get a mean of 8.408797212, but invite somebody else to confirm. This is similar to the "birthday problem," which asks, how many people do you need for there to be a 50% or greater chance of a common birthday?
Quote: WizardI get a mean of 7.408797212, but invite somebody else to confirm. This is similar to the "birthday problem," which asks, how many people do you need for there to be a 50% or greater chance of a common birthday?
Funny thing, once I was discussing the birthday problem with someone who turned out to have the same birthday as me, albeit in a different year.
Different year is the key. Make it the same year and the number skyrockets.Quote: NareedFunny thing, once I was discussing the birthday problem with someone who turned out to have the same birthday as me, albeit in a different year.
I remember this being discussed one night on Carson's Tonight Show. I believe the number was in the low 20's.
Johnny "tested" it by proposing to ask the people in one section of the audience. The rows only had 6 or 8 people in them. He was on the third or fourth person in the first row when someone in the second row reacted that they had the same day as the person he had just asked.
Back at my last real job there were about 30 people in the office, and we had two sets of common birthdays, as I recall. There were also a whole bunch of birthdays in June, about 25% of the entire office. If I were still there I'd be enjoying longer breaks and eating lots of cake about this time. As long as I'm on the topic, my two brothers have the same birthday, but four years apart.
Quote: WizardAs long as I'm on the topic, my two brothers have the same birthday, but four years apart.
I also have two brothers with the same birthday, two years apart. I counted back and could never figure out what was so special about the conception date.
Quote: ChuckI also have two brothers with the same birthday, two years apart. I counted back and could never figure out what was so special about the conception date.
I never thought of that before, but both my younger brothers would have been conceived right around my birthday. Hey, I'm the one who is supposed to have fun that day!
Can we please get this thread back on topic?
Quote: joshyeltmanWhat I really want to know is what formula you used to come up with that number. I arbitrarily guessed around 8, and I'm glad I was close, but I'd like to know how to calculate this.
First you need to know the formula for the probability of a repeat repeat number with exactly n spins. That probability is 1-(37/38)*(36/38)*...*((38-n+1)/38). Here are some values by n:
2 0.026316
3 0.077562
4 0.150386
5 0.239819
6 0.339843
7 0.444078
8 0.546485
9 0.641962
10 0.726760
11 0.798666
12 0.856947
13 0.902121
14 0.935606
15 0.959330
16 0.975384
17 0.985749
18 0.992124
19 0.995855
20 0.997927
21 0.999018
22 0.999561
23 0.999815
24 0.999927
25 0.999973
26 0.999991
27 0.999997
28 0.999999
Let f(n) be the probability of a repeat within n spins. Let g(n) be the proability that the first repeat is on exactly the nth spin. Then g(n)=f(n)-f(n-1). Here are some values for g(n)
2 0.026316
3 0.051247
4 0.072824
5 0.089433
6 0.100024
7 0.104235
8 0.102407
9 0.095477
10 0.084799
11 0.071905
12 0.058281
13 0.045175
14 0.033485
15 0.023724
16 0.016054
17 0.010365
18 0.006376
19 0.003731
20 0.002073
21 0.001091
22 0.000543
23 0.000254
24 0.000112
25 0.000046
26 0.000018
27 0.000006
28 0.000002
29 0.000001
Then you take the dot product of the above table to get the expected number of spins, which is...
2*0.026316 + 3*0.051247 + 4*0.072824 + ... = 8.408795574
p.s. In writing this up I realize my original answer was off by exactly 1, I think because I wasn't counting the first spin.
I've seen only one person doing it and it is amusing since most red numbers are covered and the chances of red/black are higher than the straight bets.
Does this increase the chances or reduces the house edge in any way? Is this just as bad as any other "method"?
i.e.: on a single zero roulette, making $5 bets, straight bets on 5-7-9-12-14-16-18-21-23-25 + Split bets on 27/30 and 0/1. (All red numbers and covering 0).
I say it is just as good as any other "method"Quote: LMyztikIs this just as bad as any other "method"?
and way better than betting one unit on every single number.
I really have seen this played a few times and it was amusing (on a Roulette machine that is)
I tried it once too
Sally
Quote: LMyztikDoes anyone know what is this called? And also is there someone who could put on some numbers here? What is the math involved which will show that such a method diminishes in any way the house advantage or not at all? Anyone up for the challenge?
I mean, it sounds definitely better than betting one unit on every single number, but what about the statistics? Would it be the same probability? Intuitively it sounds that straight betting 10 red numbers and 2 red splits would be the same probability as to bet any other 10 mixed color numbers and 2 mixed splits. Would the math be correct or is there an increased percentage since we are covering?
Quote: LMyztikI mean, it sounds definitely better than betting one unit on every single number, but what about the statistics? Would it be the same probability? Intuitively it sounds that straight betting 10 red numbers and 2 red splits would be the same probability as to bet any other 10 mixed color numbers and 2 mixed splits. Would the math be correct or is there an increased percentage since we are covering?
The math is correct - as long as you don't bet the 0-00-1-2-3 combination, each bet in roulette has the same expected result; you lose 1/19 of it. When you leave out numbers, you increase the amount you win if one of them comes up, but you also increase the chance that none of them will win.
Quote: jimstinehartAll of the posts on this thread misakenly assume that actual human roulette dealers in Vegas spin totally random numbers at all times. That simply is not true. If one plays both a lot of real roulette and also the free roulette game on your laptop, you can easily see that the two sources generate different types of spins. My laptop game frequently spins the same number 3 times in a row (the number 36 on the very last game I played), whereas in thousands of real roulette games at Vegas and elsewhere, I have hardly ever seen a human dealer do that. Why? Because a human dealer would be silently accused by some customers of rigging the game: either spinning 36 to help a secret friend who is betting 36, or more likely to hurt most of the players, who are not betting 36. That could in turn lead to people leaving the game and going to another casino to bet at a roulette table that does not seem to be "rigged", which could get the human dealer in trouble with his pit boss and the casino. No, a human dealer who spun 36 three times in a row would not in fact be "rigging" the game, but the casino wants to avoid the impression that its roulette tables are rigged. The casino wants its dealers to generate spins that have the a-p-p-e-a-r-a-n-c-e of being random, rather than wanting to generate truly random spins. So 3-peats are discouraged, because many players would misinterpret a 3-peat as a "rigged" set of spins that the dealer has done for some nefarious purpose. To avoid that, and keep the customers happy (rather than trying to help or hurt a particular bettor), a human dealer will try hard to spin the ball in a different part of the wheel than where the ball has just landed twice on the same number, to avoid a 3-peater and possible problems with his customers and his pit boss. It's just human nature. Las Vegas roulette dealers are not machines. The spins at real roulette tables are not truly random.
What a bunch of nonesense.
Quote: DJTeddyBearI see you got the math figured out. But it piqued my curiosity.
How often do you see a three-peat?
Have you ever seen, or hear about, a four-peat? How about a five-peat?
The reason I ask is because I've noticed that it's very unusual for a table's spin history display to not have at least one repeat on it. And that display only shows, what?, about 20 spins? Quite often I've seen more than one repeat. So it got me wondering about the three-peat and more.
At an Indian casino in San Diego the number 1 came up 4 times in a row to which i got up and walked muttering BS under my breath. It happens just feels like BS when it does.
Quote: jimstinehartMy laptop game frequently spins the same number 3 times in a row (the number 36 on the very last game I played), whereas in thousands of real roulette games at Vegas and elsewhere, I have hardly ever seen a human dealer do that.
If you really believe that, then go to any human dealer roulette game, wait for a double number to come, and simply bet anything *but* that number.
Oh well, I forgot: even if a human dealer could do that feat consistently, it would be zero EV on a single-zero wheel and still -EV on a double-zero wheel.
1. Please help me understand this information. Are you saying that any number has a roughly 54% chance of a repeat in the first 8 spins? Or, that 54% of the time, you will see a repeat in 8 spins?
2. Have you ever tested the Bodog American Roulette PRACTICE game to verify it is a fair game?
I seem to win VERY FREQUENTLY on that game so I have either beaten Roulette, or the practice game software is not random. Please advise since you link to them. IIRC, you have blacklisted sites for having unfair practice games. Is there any VERIFIED fair practice roulette site that you are familiar with?
Thank you for a great site!
I believe the Wizard has full confidence in the Bodog/bovada brands including the practice modes. However you should ask him directly there's a possibility bodogs in not the same as bovada.Quote: MSGamblerPlease help me with two related questions to the above chart.
1. Please help me understand this information. Are you saying that any number has a roughly 54% chance of a repeat in the first 8 spins? Or, that 54% of the time, you will see a repeat in 8 spins?
2. Have you ever tested the Bodog American Roulette PRACTICE game to verify it is a fair game?
http://casino.bodog.eu/lobby/american-roulette-fun
I seem to win VERY FREQUENTLY on that game so I have either beaten Roulette, or the practice game software is not random. Please advise since you link to them. IIRC, you have blacklisted sites for having unfair practice games. Is there any VERIFIED fair practice roulette site that you are familiar with?
Thank you for a great site!
'you will see' would beQuote: MSGambler1. Please help me understand this information.
Are you saying that any number has a roughly 54% chance of a repeat in the first 8 spins?
Or, that 54% of the time, you will see a repeat in 8 spins?
on average of course.
so say you look at any set of 8 spins. on average 54% (about 5 out of 9)
will show at least one repeat, maybe more.
of course that is still the probability too.
the trick is betting on it
here is the distribution I got from some R code here
https://sites.google.com/view/krapstuff/birthday-paradox
> bday.tMax.dist.cum(38)#38 Roulette numbers
Time difference of 0.009552002 secs
[1] "for 38 items, mean:8.4088 "
Draw X Draw X Prob cumulative: (X or less)
[1,] 1 0 0
[2,] 2 0.026315789474 0.026315789474
[3,] 3 0.051246537396 0.07756232687
[4,] 4 0.072824026826 0.1503863537
[5,] 5 0.0894330154 0.2398193691
[6,] 6 0.10002376722 0.33984313632
[7,] 7 0.10423529427 0.44407843059
[8,] 8 0.10240660489 0.54648503548
[9,] 9 0.095476834636 0.64196187011
[10,] 10 0.084798504447 0.72676037456
[11,] 11 0.071905164589 0.79866553915
[12,] 12 0.058281028141 0.85694656729
[13,] 13 0.045174768224 0.90212133551
[14,] 14 0.033484806271 0.93560614179
[15,] 15 0.023724053026 0.95933019481
[16,] 16 0.016053870469 0.97538406528
[17,] 17 0.010364604092 0.98574866937
[18,] 18 0.0063755952804 0.99212426465
[19,] 19 0.0037306114799 0.99585487613
[20,] 20 0.0020725619333 0.99792743807
[21,] 21 0.0010908220701 0.99901826014
[22,] 22 0.00054254045067 0.99956080059
[23,] 23 0.00025427334405 0.99981507393
[24,] 24 0.00011192893614 0.99992700287
[25,] 25 4.6103451955e-05 0.99997310632
[26,] 26 1.7693210729e-05 0.99999079953
[27,] 27 6.2950581329e-06 0.99999709459
[28,] 28 2.0643712905e-06 0.99999915896
[29,] 29 6.1971379872e-07 0.99999977867
[30,] 30 1.6890695642e-07 0.99999994758
[31,] 31 4.1383737053e-08 0.99999998896
[32,] 32 9.0027778853e-09 0.99999999797
[33,] 33 1.7119034349e-09 0.99999999968
[34,] 34 2.787474343e-10 0.99999999996
[35,] 35 3.7788727138e-11 1 <<<<too lazy to make this more accurate!
[36,] 36 4.0947537147e-12 1
[37,] 37 3.3250631668e-13 1
[38,] 38 1.7986452803e-14 1
[39,] 39 4.8612034602e-16 1
this is asked a lot
as is the 37 wheel and 1d6 (one standard die) looking for distribution of the first repeat (some call repetition)
Sally
almost forgot sim data
1 million sets of 8 spins
results from 0 repeats to 1 to 6
group middle freq freq/100
--------------------------------------------
-0.5 <= x < 0.50 0.00 453851 45.39%
0.50 <= x < 1.50 1.00 409193 40.92%
1.50 <= x < 2.50 2.00 121785 12.18%
2.50 <= x < 3.50 3.00 14465 1.45%
3.50 <= x < 4.50 4.00 693 0.07%
4.50 <= x < 5.50 5.00 12 0.00%
5.50 <= x < 6.50 6.00 1 0.00%
yes, the first part of this thread was about repeating numbersQuote: evshow does this apply to 50/50; 0.66 / 0.33 odds? (I'm catching on they must be equally likely)
the second part about the first repeat
where there is an equal probability to get any number at any draw
I like the skittles version
buy a pack and no looking
take out 1 and place on the table (or something)
do not eat it!
now take another (without replacement of the first draw)
wait!
this is different from the roulette example
ok
after the each draw, replace the one back into the pack, shake up
draw another.
how many draws on average until a color repeats.
I eat that one!
the distribution
[1] "for 5 items, mean:3.5104 "
Draw X Draw X Prob cumulative: (X or less)
[1,] 1 0 0
[2,] 2 0.2 0.2
[3,] 3 0.32 0.52
[4,] 4 0.288 0.808
[5,] 5 0.1536 0.9616
[6,] 6 0.0384 1
Sally
example: 1 throw: 13548072; 2 throw: 84913546...
Sure it will! Side bet with other players. Just like the birthday bet. :-)Quote: WizardIn eight spins, the probability of at least one repeat is 54.6485035%. This information will not help you at all at the wheel.
