roulette question

what are the odds for a number to not come out for 190 consecutive spins at American roulette?thank you

For a specific number it would be (37/38)^190 = 0.0063. For any number it would be about 24%.

I think (37/38)^190, or about 0.63%. But others know the math better than I do.

EDIT: The above is for a particular number, but for any number, it would be, oh, about 24%.

[/sheepish grin]

Quote: uztucim

what are the odds for a number to not come out for 190 consecutive spins at American roulette?thank you

https://wizardofodds.com/ask-the-wizard/roulette/
question #2 at WoO. Almost Same question was asked but for 200 spins. The Wizard shows how it is calculated for 1 or more not to have hit.

thank you very much. i also have a question in the sportsbook category regarding your interpretation of Kelly's criterion i am looking for an answer.thank you so much!

Quote: Wizard

For a specific number it would be (37/38)^190 = 0.0063. For any number it would be about 24%.

I came up with an exact answer using The Wizards' formula
For any number it would be 0.216494 or 21.6494%

formula used from:https://wizardofodds.com/ask-the-wizard/roulette/
question #2 at WoO.

I added a handy table (from 43 to 400 spins, so it is big) for at least 1 number NOT hitting in my blog.
HERE
((37/38)^190)*38 = 0.239457152. This formula double counts as The Wizards' article explains.
From the article:
"The probability that any given number will not have hit is (37/38)^200 = 0.48%.

With 38 numbers, we could incorrectly say that the probability that any one of them would not be hit is 38 × (37/38)^200 = 18.34%."

I was doing exactly what The Wizard said not to do. using his formula in excel is very simple.
((37/38)^190)*38 = 0.239457152 says that "on average" .23 of a number will not have hit within 200 spins.
For higher spins the probability of both formulas converge to within <.5%, so I may just be splitting hairs to have an exact answer.
Thank you Wizard for the hard work on creating the formula at the WoO.

For 190 spins here are some numbers from a recent simulation.(only 67,000 trials, so my % is off by 1.5% or less)

not hit	unique #s	repeats		expected # of spins	at least 1 not hit
3	35	155	0.12%	159,179	20.22%
2	36	154	1.99%	9,565
1	37	153	18.11%	1,049
0	38	152	79.78%	238

The results show that there could be 1,2 3 or even more numbers in 190spins that have yet to hit.
With a larger simulation the relative frequencies would be as accurate as the table The Wizard showed in the same article.

My computer I used for simulations has crashed, so I have to get a new one, so if any one else wants to run a larger sample, then we could have even more accurate results.
good luck

@nope27 >> I left you another post on your blog but for some reason, it does not get posted at the bottom? Ken

Quote: nope27

I came up with an exact answer using The Wizards' formula
For any number it would be 0.216494 or 21.6494%

One can see using the ((37/38)^190)*38 = 0.239457152 formula it appears to be a correct answer.
But a few results below show why it is the worng formula to use and one must use the formula The Wizard of Odds made at his WoO site.


spins/prob
100/2.639953891
125/1.35534397
140/0.908492695
150/0.695829303
190/0.239457152
200/0.18340412

The result of 100 spins of 2.63 is an average of 100 spins of the numbers that have not hit.

Here is a shorter version table for at least 1 number not hitting in X spins.

spins	prob	1 in	expected # of spins
50	99.9999975%	.	.
60	99.9991%	.	.
70	99.97%	.	.
80	99.68%	.	.
90	98.45%	.	.
100	95.34%	.	.
110	89.73%	.	.
120	81.75%	.	.
130	72.13%	.	.
140	61.86%	.	.
150	51.77%	1.93	290
160	42.49%	2.35	377
170	34.33%	2.91	495
180	27.39%	3.65	657
190	21.65%	4.62	878
200	16.98%	5.89	1,178
210	13.25%	7.55	1,585
220	10.29%	9.72	2,138
230	7.97%	12.55	2,887
240	6.15%	16.26	3,902
250	4.74%	21.10	5,274
260	3.65%	27.41	7,128
270	2.80%	35.67	9,630
280	2.15%	46.44	13,003
290	1.65%	60.51	17,547
300	1.27%	78.88	23,663
310	0.97%	102.86	31,886
320	0.75%	134.17	42,935
330	0.57%	175.06	57,770
340	0.44%	228.44	77,670
350	0.34%	298.14	104,349
360	0.26%	389.14	140,091
370	0.20%	507.96	187,944
380	0.15%	663.09	251,973
390	0.12%	865.63	337,594
400	0.09%	1130.07	452,028

This is nice information to see but how useful is it?

Quote:

nope27: I added a handy table (from 43 to 400 spins, so it is big) for at least 1 number NOT hitting in my blog. HERE

Great job on the table, nope27. That's the probability of at least one number not hitting at all. Can you do the probability of at least one number hitting once, twice, three times, etc.?

That table that's posted isn't going to help you beat the game. You simply can't step outside of probability by watching on the side lines.

Quote: Keyser

You simply can't step outside of probability

I heard that somewhere. Urban legend?

I'm sure this has been asked before, but I am trying to explain this to others and need some clarification.

The chart shows that at 400 spins it is very unlikely that all the numbers will not have appeared. So after 400 spins does that mean the chances of seeing that number appear at sometime in the future has increased because the chances of not seeing it continue to decrease?

Quote: beenitty

I'm sure this has been asked before, but I am trying to explain this to others and need some clarification.

The chart shows that at 400 spins it is very unlikely that all the numbers will not have appeared. So after 400 spins does that mean the chances of seeing that number appear at sometime in the future has increased because the chances of not seeing it continue to decrease?

your chances of seeing a number in the future does not change ever. Every spin has the same odds of a number appearing.

I agree.

So could you say the following:

After 400 spins you will have a distribution of occurrences. Lets say the last number to not appear is 00 on American Roulette Wheel.

Number/Appearances

00/0
0/15
1/20
2/17
3/11
.
.
.
36/16

So the odds of any number is the same 2.6315%. Over time you will see the distribution become even, lets say at spin # infinity.
So if we believe that all the numbers will have an even distribution at infinity is it not true that between spin 400 and infinity we will see 00 appear relatively more than its counterparts?

No.

Think about the fraction 400 / infinity
Now think about what was discussed in this thread:
What is .99 repeating as a fraction?

Bottom line, 400 is nothing compared to infinity.

Besides - even if something like this was the case, how do you know that 00 did not occur, say, 35 times in the previous 400 spins?

I'm not disagreeing I just want to make sure that I'm explaining this correctly.

So you're saying that 400/infinity is really small. Lets call it zero.

So am I correct in saying that it will happen more often but the probability does not increase?

Also consider this.

If you do a simulation of a random number generator and track the occurrences of each number, again you will see a distribution.

Lets say you take a snapshot of the distribution at 10000, 50000, 100000, etc etc up to 1M spins.

Now lets say that at 100000 spins you stopped and looked at the distribution. In theory you should see every number (American Roulette, 100000/38 = 2631) about 2631 times. You wont because the distribution wont be exactly even but I'm sure you get my point. Some numbers will have appeared more than this and some less. So say you have a number that has only appeared 500 times.

It is wrong to assume that at 1M spins (or more) you will see that numbers should converge towards the mean of 1/38 or 2.63%?

Or is the issue that 1M is also very small compared to infinity so the numbers will eventually appear at the frequency of the mean but it may take until the end of time for this to be achieved.

Quote: beenitty

every number (American Roulette, 100000/38 = 2631) about 2631 times. You wont because the distribution wont be exactly even but I'm sure you get my point. Some numbers will have appeared more than this and some less. So say you have a number that has only appeared 500 times.

It is wrong to assume that at 1M spins (or more) you will see that numbers should converge towards the mean of 1/38 or 2.63%?

If a number has appeared 2131 times less than chance would have dictated after 100,000 spins, then it is on target to be still 2131 short of expectation after 1,000,000 spins, or any other huge number of future spins. Fate doesn't dictate that it will regress to the norm in terms of absolute count, only in percentage terms.

eg, in your example, instead of appearing 1/38th of the time, or 2.631578947368421%, it instead came out 500/100,000 = 0.5%: It's 2.131578947368421% short, which is way out.

But if after 1,000,000 spins it is still 2131 appearances short, then it will have appeared (1,000,000/38 - 2131) = 24184 times: It will have appeared 2.418478947368421% of the time. At 2.418478947368421% It will be closer to the expected 2.63158% and so can be said to be regressing to the mean.

Now lets say we spin 1,000,000,000,000 times starting out 2131 appearances short and ending at 2131 short. It would then have appeared (1,000,000,000,000/38 -2131) times = 26315787342 times or 2.6315787342% : It has pretty much regressed almost exactly to the expected appearance frequency, but it is still numerically just as far away.

Purely and simply, regressing to the mean does not mean any short term variation gets cancelled out, only that it becomes statistically insignificant in percentage terms.


Oh, and just to be controversial 400/infinity = 0 EXACTLY. Not close to, but exactly: That's the nature of infinity.

Very nice example.

Thanks

Also as your sample increases the Result in % terms gets closer to the Expecation (Mean).
BUt also as your sample increases in absolute terms the difference from the Mean increases.

For example you toss a coin a 100 times. Results will say vary from 40-60 Heads +-10 in absolute terms and +-10%.
If you toss a coin 1000 times the Results will vary from say 450-550 Heads +-50 in absolute terms and +-5%.
So in % terms the results get closer to the mean (down from 10% to 5%) but in absolute terms they go up from 10 to 50.

So regression to the Mean does not mean that the future results will cover the difference in absolute terms and not even that the difference in absolute terms will decrease. In absolute terms It is as likely for the difference to increase as it is to decrease. Only in % terms the difference will decrease and regress to the mean.

Quote: AceTwo

...also as your sample increases in absolute terms the difference from the Mean increases.

Quite so. After a billion billion spins, the numerical distance from the expected value can be millions too many or millions too few and still be within a TINY fraction of expected value in percentage terms. And there's no knowing whether it will be millions too many or millions too few. The only (near) certainty is that the observed 2100 or so initial shortfall (in the example) would pale into insignificance.