


08112010, 07:39 PM

#1

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What are the odds of a number coming up 5 out of 6 on roulette?
A few weeks ago, I saw the number 1 hit five out of six times. I've seen a lot in roulette, but I've never seen that. It hit once, went to another number, and then hit four in a row.
So I started wondering about the odds. You can calculate the odds of it hitting 4 in a row by multiplying 1/38 (double zero wheel) times itself four times. The odds of the same number hitting 4 in a row would be just over 2 million to one.
How would you calculate the odds of hitting 5 out of 6?
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08112010, 07:41 PM

#2

Lifer
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08112010, 07:43 PM

#3

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technically, the last number hit has no bearing on the next.
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08112010, 07:43 PM

#4

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Using combinations... I think.
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Last edited by Arcadio; 08112010 at 07:52 PM.



08112010, 07:47 PM

#5

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Quote:
Originally Posted by Gothgar
technically, the last number hit has no bearing on the next.

Not a relevant response.
I'm wondering what the odds are of a specific number coming up 5 times out of 6. I'm not asking anything about the next unknown event.
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08112010, 07:51 PM

#6

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It's just the sum of the individual combinations. In this case, it's sum of each place the failure would occur. There are C(6,1) = 6 ways for the failure to occur. 1st, 2nd, 3rd...6th.
So, 6 * (1/38)^5 * 37/38.
The general case is covered by bionomial distributions or something like that.
Last edited by a123456; 08112010 at 07:56 PM.



08112010, 07:51 PM

#7

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5 choose 6, multiplied by uhhh... the number of combinations times the uh... I'm not really sure how I ever passed statistics
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08112010, 07:57 PM

#8

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Quote:
Originally Posted by a123456
It's just the sum of the individual combinations. In this case, it's sum of each place the failure would occur. There are C(6,1) = 6 ways for the failure to occur. 1st, 2nd, 3rd...6th.
So, 6 * (1/38)^5 * 37/38.

Not completely clear to me. Are you saying "37/38" because the answer is dependent upon whether it's a zero or double zero wheel?
Or is the mathematical answer (6 times [1/38 to the fifth] times 37) divided by 38? That wouldn't make sense to me.
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08112010, 08:01 PM

#9

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Quote:
Originally Posted by Rio Rebel
Not completely clear to me. Are you saying "37/38" because the answer is dependent upon whether it's a zero or double zero wheel?
Or is the mathematical answer (6 times [1/38 to the fifth] times 37) divided by 38? That wouldn't make sense to me.

The latter.
Prob of failure at position 1 is (1/38)^5 * (37/38) since it's success*5 and failure for the latter. Prob of failure is (37/38) because there are 37 out of 38 ways to fail. Then, failure can be in 6 places so it's times 6.
Edit: Now that I think more, the whole thing might be multiplied by 38 since it's not a particular number but any number hitting 5 out of 6.
Last edited by a123456; 08112010 at 08:06 PM.



08112010, 08:04 PM

#10

Lifer
Join Date: Aug 2001
Posts: 19,916

You are talking about probability WITH replacement.
The odds are exponential, so if you are talking about roulette, it is the total number of slots on the wheel raised to the power of x number of draws.
Those are your odds, to 1, that the number will come x times.
38^6 is 3,010,936,384:1
err read the question wrong, 5 out of 6 is more complicated.
rationally you can think this through using the previous formula, however.
You would need 5/6 using the aforementioned formula, then to calculate the probability of FAILURE on the last draw and combine them to get the odds.
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Last edited by Acanthus; 08112010 at 08:07 PM.



08112010, 08:13 PM

#11

Lifer
Join Date: Jul 2001
Location: London, UK
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a123456 is right.
I'll show an example:
Let's say you want the probability of the exact series you saw. Namely:
1,(something other than 1),1,1,1,1
The odds of getting a 1 are 1/38. The odds of getting anything other than a 1 are 37/38. Thus, you just multiply the probabilities:
P = (1/38)*(37/38)*(1/38)*(1/38)*(1/38)*(1/38) = (37/38)*(1/38)^5
However, your question was what are the chances of a number coming up 5 times out of 6, and you didn't specify an order. In this case, there are 6 different ways for that to happen. The non1 spin could be the first one, the second, the third... etc. So because there are 6 ways for this to happen, it's just the probability calculated above, multiplied by 6. So:
P(5 out of 6 on roulette) = 6*(37/38)*(1/38)^5 ~ 1 in 13.5 million



08112010, 08:15 PM

#12

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Quote:
Originally Posted by a123456
The latter.
Prob of failure at position 1 is (1/38)^5 * (37/38) since it's success*5 and failure for the latter. Prob of failure is (37/38) because there are 37 out of 38 ways to fail. Then, failure can be in 6 places so it's times 6.
Edit: Now that I think more, the whole thing might be multiplied by 38 since it's not a particular number but any number hitting 5 out of 6.

Thanks. Great explanation.
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08112010, 08:40 PM

#13

Lifer
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Wikipedia is your friend.
http://en.wikipedia.org/wiki/Roulett...an_roulette.29
It's 35 to 1 for hitting any single number.
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08112010, 08:44 PM

#14

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08112010, 08:54 PM

#15

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Quote:
Originally Posted by a123456
(1/38)^5 * (37/38)

This!
Edit: Oops, this!
Quote:
Originally Posted by silverpig
However, your question was what are the chances of a number coming up 5 times out of 6, and you didn't specify an order. In this case, there are 6 different ways for that to happen. The non1 spin could be the first one, the second, the third... etc. So because there are 6 ways for this to happen, it's just the probability calculated above, multiplied by 6. So:
P(5 out of 6 on roulette) = 6*(37/38)*(1/38)^5 ~ 1 in 13.5 million

Last edited by Gigantopithecus; 08112010 at 08:59 PM.



08112010, 09:33 PM

#16

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I saw this when I was playing once
33
11
22
33
11
22
4
Never going to forget the $10 bucks I lost on 33



08112010, 11:22 PM

#17

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Quote:
Originally Posted by silverpig
a123456 is right.

I think I'm going to make an argument that while he may be technically correct, in the spirit and context of the question, he's not quite correct.
If someone says "I saw the number 1 hit five out of six times" then the intended meaning is generally not that the first or sixth spins were the failure. The failure would have occurred on the 2nd, 3rd, 4th, or 5th spin. Thus, the multiplier is 4, not 6. If he had observed 1, 1, 1, 1, 1, 17, then he would have said "I saw the number 1 hit five times in a row" not "five times out of six."
But, even this isn't quite correct, for it assumes that there were only 6 spins, of which the first and sixth spin were 1s and only one of the other spins wasn't a 1. To understand this argument better, let's reduce it to something easier. What if I said "what are the odds of me flipping a coin and getting heads twice in a row?" Well, if there are ONLY 2 flips, then the odds of getting heads twice in a row is 1/4. Pretty simple. But, if I were working at a casino flipping the same coin over and over and over and over, then what are the odds of me getting heads twice in a row (at some point)? Pretty damn likely. And, don't forget, in 6 consecutive flips, "2 flips in a row" occurs 5 times, not 3 times.
Also, to correct Rio, the probability of getting the same thing 4 times in a row is not (1/38)^4. It's actually (1/38)^3. The probability of getting a particular thing 4 times in a row is to the 4th power, but if we spin it 3 times, it's going to come up with *some* number for the first spin 100 percent; of the time. 1/38 of the time, the 2nd spin matches the first spin. i.e. 2 of the same in a row is 1/38. But the probability of getting two 7s in a row is (1/38)^2
Last edited by DrPizza; 08112010 at 11:52 PM.



08112010, 11:27 PM

#18

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This is why statistics can suck my left nut.
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08122010, 06:04 AM

#19

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Quote:
Originally Posted by mmntech

The payout is 35 to 1, not the odds. The odds are 37 or 38 to 1, depending on whether the wheel has a double zero along with a zero.
Actually reading is your friend.
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08122010, 06:09 AM

#20

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Quote:
Originally Posted by DrPizza
I think I'm going to make an argument that while he may be technically correct, in the spirit and context of the question, he's not quite correct.
If someone says "I saw the number 1 hit five out of six times" then the intended meaning is generally not that the first or sixth spins were the failure. The failure would have occurred on the 2nd, 3rd, 4th, or 5th spin. Thus, the multiplier is 4, not 6. If he had observed 1, 1, 1, 1, 1, 17, then he would have said "I saw the number 1 hit five times in a row" not "five times out of six."
But, even this isn't quite correct, for it assumes that there were only 6 spins, of which the first and sixth spin were 1s and only one of the other spins wasn't a 1. To understand this argument better, let's reduce it to something easier. What if I said "what are the odds of me flipping a coin and getting heads twice in a row?" Well, if there are ONLY 2 flips, then the odds of getting heads twice in a row is 1/4. Pretty simple. But, if I were working at a casino flipping the same coin over and over and over and over, then what are the odds of me getting heads twice in a row (at some point)? Pretty damn likely. And, don't forget, in 6 consecutive flips, "2 flips in a row" occurs 5 times, not 3 times.
Also, to correct Rio, the probability of getting the same thing 4 times in a row is not (1/38)^4. It's actually (1/38)^3. The probability of getting a particular thing 4 times in a row is to the 4th power, but if we spin it 3 times, it's going to come up with *some* number for the first spin 100 percent; of the time. 1/38 of the time, the 2nd spin matches the first spin. i.e. 2 of the same in a row is 1/38. But the probability of getting two 7s in a row is (1/38)^2

While you're semantically correct, I was looking for the technical response. I could alter the formula above to remove the first and last spins, and multiply by four instead of six.
But I was really curious about the odds of starting at a specific point in time, and a specific number (1) coming up 5 times out of six.
Your comments are true and interesting, but the formula above is what I was looking for.
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08122010, 06:13 AM

#21

Lifer
Join Date: Apr 2002
Location: Columbus of Ohio
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I saw 36 come up 4 times in a row in Vegas.
It was 6am and everyone was dead tired.
There were 2 balls in the wheel! One stayed on 36 while the dealer kept spinning the second one.
The sensor was reading 36 every time.
By the time someone noticed it, it was too late...



08122010, 07:27 AM

#22

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Quote:
Originally Posted by Rio Rebel
While you're semantically correct, I was looking for the technical response. I could alter the formula above to remove the first and last spins, and multiply by four instead of six.
But I was really curious about the odds of starting at a specific point in time, and a specific number (1) coming up 5 times out of six.
Your comments are true and interesting, but the formula above is what I was looking for.

Exactly 5 out of 6? Or "at least 5 out of 6?" I was hoping to make the problem at least slightly more interesting than a 10th grade math problem.



08122010, 07:51 AM

#23

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Quote:
Originally Posted by edro
I saw 36 come up 4 times in a row in Vegas.
It was 6am and everyone was dead tired.
There were 2 balls in the wheel! One stayed on 36 while the dealer kept spinning the second one.
The sensor was reading 36 every time.
By the time someone noticed it, it was too late...

How do they screw that up? The dealer is only given one ball. When they change it out, he puts in on the table and the other ball gets put next to it. The pit boss takes the old ball and the dealer takes the new ball. They're pretty careful about that shit.
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08122010, 09:31 AM

#24

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I thought everyone on ATOT was an INTJ, but it's like none of you ever passed first year statistics, which isn't even statistics; it's probability.



08122010, 10:52 AM

#25

Member
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Posts: 147

Quote:
Originally Posted by Rio Rebel
The payout is 35 to 1, not the odds. The odds are 37 or 38 to 1, depending on whether the wheel has a double zero along with a zero.
Actually reading is your friend.

Roulette isn't a random game though  the croupier has an influence on the outcome.
My inlaws win at Roulette all the time and I'm convinced it's because they tip well. I doubt the croupier can place the ball in a specific slot, but I have no doubt they can place it in a relatively small section of the wheel at any given time, making certain numbers more likely to win.



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