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inversehelix
inversehelix
Joined: Jun 19, 2010
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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
Joined: Jan 15, 2010
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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
Joined: May 12, 2010
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June 19th, 2010 at 5:58:35 PM permalink
deleted. fail.
am19psu
am19psu
Joined: May 12, 2010
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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
Joined: Jun 19, 2010
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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
Joined: Jan 15, 2010
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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
Joined: May 12, 2010
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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
Joined: Jun 19, 2010
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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
DJTeddyBear
Joined: Nov 2, 2009
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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.
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁 Note that the same could be said for Religion. I.E. Religion is nothing more than organized superstition. 🤗
joshyeltman
joshyeltman
Joined: Jun 25, 2010
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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.

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