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inversehelix
inversehelix
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June 19th, 2010 at 5:55:19 PM permalink
Howdy,

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!
ahiromu
ahiromu
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June 19th, 2010 at 5:57:45 PM permalink
The chance to call a single number, let's say 6, then afterwards spin it twice in a row is 1 in 38^2 and thrice is 1 in 38^3. The chance for a number to repeat AFTER it has hit is only 1 in 38. You don't seem stubborn or set in your ways, so you probably understand that the ball has no memory and all that jazz. Basically, number repeating isn't out of the ordinary at all.

(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.
Its - Possessive; It's - "It is" / "It has"; There - Location; Their - Possessive; They're - "They are"
am19psu
am19psu
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June 19th, 2010 at 5:58:35 PM permalink
deleted. fail.
am19psu
am19psu
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June 19th, 2010 at 6:00:30 PM permalink
Quote: inversehelix

Howdy,

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.
inversehelix
inversehelix
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June 19th, 2010 at 6:06:51 PM permalink
thanks for the post, a little light went on in my head when I read this that the 38^2 is only if you're calling the number beforehand and that *that* number hits twice.

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.
ahiromu
ahiromu
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June 19th, 2010 at 6:15:32 PM permalink
Exact numbers might be a little beyond my paygrade, but some people here can definitely answer that question for you. Off the top of my head, I think a rough number might be 1/38*9999 (after the first spin) or ~263 times. I feel like I'm missing something, somewhere.
Its - Possessive; It's - "It is" / "It has"; There - Location; Their - Possessive; They're - "They are"
am19psu
am19psu
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June 19th, 2010 at 6:17:28 PM permalink
Since 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?

EDIT: ahiromu is correct, that it should be 9999, not 10000. It's not a significant difference either way
inversehelix
inversehelix
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June 19th, 2010 at 6:20:04 PM permalink
Quote: am19psu

Since 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.
DJTeddyBear
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June 20th, 2010 at 5:18:42 AM permalink
I 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.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
joshyeltman
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June 25th, 2010 at 2:14:20 PM permalink
On 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.
Wizard
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Wizard
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June 25th, 2010 at 3:04:23 PM permalink
Quote: joshyeltman

On 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?
摘xtraordinary claims require extraordinary evidence. -- Carl Sagan
Nareed
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June 25th, 2010 at 4:36:51 PM permalink
Quote: Wizard

I 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.
Donald Trump is a fucking criminal
DJTeddyBear
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June 25th, 2010 at 5:09:32 PM permalink
Quote: Nareed

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.

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.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
Wizard
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June 25th, 2010 at 10:39:18 PM permalink
The answer to the birthday problem is 23.

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.
摘xtraordinary claims require extraordinary evidence. -- Carl Sagan
Chuck
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June 26th, 2010 at 4:18:54 AM permalink
Quote: Wizard

As 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.
Wizard
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June 26th, 2010 at 11:46:16 AM permalink
Quote: Chuck

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.



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!
摘xtraordinary claims require extraordinary evidence. -- Carl Sagan
DJTeddyBear
DJTeddyBear
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June 26th, 2010 at 3:10:53 PM permalink
Um... Ew. Ick!

Can we please get this thread back on topic?
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
joshyeltman
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June 26th, 2010 at 11:40:47 PM permalink
What 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.
Wizard
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June 27th, 2010 at 7:21:34 AM permalink
Quote: joshyeltman

What 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.
摘xtraordinary claims require extraordinary evidence. -- Carl Sagan
LMyztik
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September 4th, 2014 at 4:26:42 PM permalink
Does anyone here know the name of the "roulette system" where people would bet most red numbers (no black numbers) and some other times most black numbers (no red numbers)?

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).
mustangsally
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September 4th, 2014 at 5:33:33 PM permalink
Quote: LMyztik

Is this just as bad as any other "method"?

I say it is just as good 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
I Heart Vi Hart
LMyztik
LMyztik
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September 6th, 2014 at 5:29:23 PM permalink
Does 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?
LMyztik
LMyztik
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September 6th, 2014 at 5:36:18 PM permalink
Quote: LMyztik

Does 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?
ThatDonGuy
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September 6th, 2014 at 6:21:16 PM permalink
Quote: LMyztik

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?


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.
jimstinehart
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September 12th, 2014 at 7:54:11 AM permalink
All 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.
LMyztik
LMyztik
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September 16th, 2014 at 7:58:02 PM permalink
Quote: jimstinehart

All 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.
onenickelmiracle
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September 16th, 2014 at 8:25:27 PM permalink
Triple numbers aren't very likely and wouldn't be even if major influence were possible.
I am a robot.
Buzzard
Buzzard
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September 17th, 2014 at 12:16:37 AM permalink
I KNEW I KNEW I bought John Patrick's winning Roulette system and lost. DAMN CROOKED CROUPIERS
Shed not for her the bitter tear Nor give the heart to vain regret Tis but the casket that lies here, The gem that filled it Sparkles yet
outlawslaw
outlawslaw
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September 17th, 2014 at 1:02:37 AM permalink
Quote: DJTeddyBear

I 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.
Not the first time ive heard that today....
LMyztik
LMyztik
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September 19th, 2014 at 5:44:38 PM permalink
Fun post, anyhow, does that "roulette betting system" I posted have a distinctive name? And anyone with numbers to compare the possible return of betting on almost all red numbers vs just betting red or vs betting any group of numbers (including black/red)?
MangoJ
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September 20th, 2014 at 9:10:04 AM permalink
Quote: jimstinehart

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.



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.
MSGambler
MSGambler
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April 17th, 2015 at 4:45:06 PM permalink
Please 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?


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!
AxelWolf
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April 17th, 2015 at 5:00:43 PM permalink
Quote: MSGambler

Please 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!

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.
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
mustangsally
mustangsally
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June 10th, 2018 at 8:22:29 PM permalink
Quote: MSGambler

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?

'you will see' would be
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%

Last edited by: mustangsally on Jun 10, 2018
I Heart Vi Hart
evs
evs
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August 15th, 2018 at 9:42:40 PM permalink
how does this apply to 50/50; 0.66 / 0.33 odds? (I'm catching on they must be equally likely)
Last edited by: evs on Aug 16, 2018
mustangsally
mustangsally
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August 16th, 2018 at 9:14:03 AM permalink
Quote: evs

how does this apply to 50/50; 0.66 / 0.33 odds? (I'm catching on they must be equally likely)

yes, the first part of this thread was about repeating numbers
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!

oh, the answer should be 3.5 draws on average
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
I Heart Vi Hart
evs
evs
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August 20th, 2018 at 3:48:42 AM permalink
Why do you need to wait for eight unique? Why can't you just make up the "eight unique" numbers at random?
Wizard
Administrator
Wizard
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August 20th, 2018 at 8:58:09 AM permalink
In eight spins, the probability of at least one repeat is 54.6485035%. This information will not help you at all at the wheel.
摘xtraordinary claims require extraordinary evidence. -- Carl Sagan
evs
evs
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August 20th, 2018 at 10:32:36 AM permalink
copied last 8 numbers from roulette=if I close(randomly selected) 8 numbers every time?
example: 1 throw: 13548072; 2 throw: 84913546...
unJon
unJon
Joined: Jul 1, 2018
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August 20th, 2018 at 11:13:57 AM permalink
Quote: Wizard

In eight spins, the probability of at least one repeat is 54.6485035%. This information will not help you at all at the wheel.

Sure it will! Side bet with other players. Just like the birthday bet. :-)
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

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