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March 4th, 2017 at 7:09:20 AM
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When I play roulette, I like to be on "repeaters" -- in other words, I bet on whatever number just came in. I am not a mathematician, and I get that the house edge always remains the same, but I am just wondering how one would calculate the odds of hitting a 'repeat' in a reasonable number of rolls.

For example, I know that if 17 just came up, and I bet it again, the odds of it repeating immediately are 1/38. But what are the odds say, of ANY number being immediately repeated in say 20 rolls? Whenever I walk up to a roulette table and look at the screen, it seems there are almost always one or more repeaters there.

Some casino boards show roulette numbers that are "hot" -- given the presumption that several numbers are coming up frequently (i.e., "hot"), it seems very likely that one OR MORE of these "hot" numbers would come up consecutively -- giving rise to the saying "nothing sweeter than a repeater".

So can anyone calculate the odds of some number repeating in any given 20 spins?

For example, I know that if 17 just came up, and I bet it again, the odds of it repeating immediately are 1/38. But what are the odds say, of ANY number being immediately repeated in say 20 rolls? Whenever I walk up to a roulette table and look at the screen, it seems there are almost always one or more repeaters there.

Some casino boards show roulette numbers that are "hot" -- given the presumption that several numbers are coming up frequently (i.e., "hot"), it seems very likely that one OR MORE of these "hot" numbers would come up consecutively -- giving rise to the saying "nothing sweeter than a repeater".

So can anyone calculate the odds of some number repeating in any given 20 spins?

March 4th, 2017 at 7:51:17 AM
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The odds of any immediate repeat happening in 20 spins (assuming you count the first number "repeating" if it is the same as the spin right before you started counting the 20 spins) is the same as the odds of any particular number hitting in 20 spins - on a double-zero wheel, this is 1 - (36/38)

^{20}, or about 2/3.
March 4th, 2017 at 9:07:02 AM
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What Don said is right.

Whether you pick a single number for all 20 spins, or pick a new number each time is irrelevant. The fact that you're letting the prior spin pick your new number is also irrelevant.

Whether you pick a single number for all 20 spins, or pick a new number each time is irrelevant. The fact that you're letting the prior spin pick your new number is also irrelevant.

I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁

March 4th, 2017 at 9:57:59 AM
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I come up with 0.397517521Quote:mbrother11So can anyone calculate the odds of some number repeating in any given 20 spins?

about a 40% chance of at least 1 repeater in 20 spins.

that starts at spin 1 (spin 1 can't be a repeat)

so equation is

=1-(37/38)^(20-1)

a quick sim shows this is right.

the distribution (just 100k groups of 20)

# of repeats: count

1: 30731

2: 7559

3: 1118

4: 123

5: 9

6: 1

total: 39541

here is a simple recursion in Excel

in Google

https://goo.gl/7zohHT

the median looks to be 27 spins

so one can do

or can don't

Sally

I agree, sure exciting when any number DOES repeat!

Last edited by: mustangsally on Mar 4, 2017

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March 7th, 2017 at 9:19:49 AM
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I'm not sure you understood my question...let me try wording it a different way... if you walk up to a table and the board shows the last 20 numbers that have come up, how likely is it that in that block of 20 there is some number that came up consecutively?

I know sometimes there will be no repeater ... but some nights I see 3-5 occurrences of SOME NUMBER repeating within the 20 or so that show up on the board.

So if I bet the previous number, I would have won quite a lot...Betting just $5 each time, if there were just 3 repeaters in 20 rolls, I would have spend $100 on those 20 rolls, and won $540. Granted, sometimes I'd lost...but on those times where 5 repeaters are in the group of 20, I'd win $900 for my $100 in bets.

I just am not good enough at math to determine the likelihood of profit betting on repeaters. I do know that my choice of 20 spins was simply due to how many previous spins showed up on the board.

To know whether a profit was likely using repeaters, I guess it matters how often a repeat happen in a typical run of 38 spins...and I'm thinking odds are pretty good that sometime in 38 spins one would see a number repeated. What I would be hoping for is that there would be more than one repeater in 38 spins.

Also note, I am not specifying that THE SAME number keep repeating -- just that ANY number does.

So does that change what you wrote me 3 days ago?

I know sometimes there will be no repeater ... but some nights I see 3-5 occurrences of SOME NUMBER repeating within the 20 or so that show up on the board.

So if I bet the previous number, I would have won quite a lot...Betting just $5 each time, if there were just 3 repeaters in 20 rolls, I would have spend $100 on those 20 rolls, and won $540. Granted, sometimes I'd lost...but on those times where 5 repeaters are in the group of 20, I'd win $900 for my $100 in bets.

I just am not good enough at math to determine the likelihood of profit betting on repeaters. I do know that my choice of 20 spins was simply due to how many previous spins showed up on the board.

To know whether a profit was likely using repeaters, I guess it matters how often a repeat happen in a typical run of 38 spins...and I'm thinking odds are pretty good that sometime in 38 spins one would see a number repeated. What I would be hoping for is that there would be more than one repeater in 38 spins.

Also note, I am not specifying that THE SAME number keep repeating -- just that ANY number does.

So does that change what you wrote me 3 days ago?

March 7th, 2017 at 9:30:11 AM
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Quote:mbrother11I'm not sure you understood my question...let me try wording it a different way... if you walk up to a table and the board shows the last 20 numbers that have come up, how likely is it that in that block of 20 there is some number that came up consecutively?

As I like to say, the probability of something happening = 1 minus the probability of it not happening.

On a double-zero roulette wheel, the probability of a spin being the same number as the previous one is 1/38, so the probability of the spin not being the same is 37/38. In order for there to be no repeats in 20 spins, this has to happen 19 times in a row; the probability that there are no repeaters is (37/38)

^{19}= 60.25%, so the probability of at least one repeater is 39.75%.

Note that this does not mean that, if there are no repeaters in the previous block of 19 numbers, then the probability of the next spin matching the previous one is 39.75%. The probability of a spin matching the previous spin is always 1/38.

And no, I don't know why I said 36/38 instead of 37/38 in my previous post, either - I was probably thinking of the 18/38 chance of any of the even-money bets (e.g. red) winning.

March 7th, 2017 at 11:19:16 AM
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Quote:mbrother11

So does that change what you wrote me 3 days ago?

Nope, if you do the math on what Sally simulated out, it comes out to ~5.3% house edge, which not coincidentally is the house edge on a double zero roulette wheel.

61% of the time you will walk away empty handed, the other 39% you win, sometimes even big, but in the long run, the losers overcome the winners.