Confused with standard deviation on roulette

Hi guys. I’m new to the forum - just signed up today. I did a bit of hunting on the internet as to where I’d get the best response to my roulette question and ended up here. I look forward to joining your community :)

I’m from the UK and am interested in the version of the game with the single zero. I have a reasonable understanding of the maths involved and calculated the standard deviation to be 5.8378 for a single number bet at 35/1. I know this is correct as I’ve googled it! It’s the next bit of my query I’m confused with...

Say I wanted to place £10 on a number for 100 consecutive spins. Regardless of any wins/losses my total wagering at this point would be £1,000. I want to use the 68-95-99.7 rule to determine what range my results should fall into. I’m assuming the following things are correct:
After 100 spins my EV would be -£27.03. The standard deviation for 1 spin would be £58.38 and therefore the standard deviation for 100 spins would be £583.78 (sqr root of 100 spins = 10).

If I’m right with my calculations then 68% of the time my results should fall in the range £610.81 loss to £556.76 win (one standard deviation from the mean). If I use the 95% rule and use 2 SDs then my results would range from £1,194.59 loss to £1,140.54 win. Obviously this is impossible as the most I could have staked at this point is £1,000. If I use 3 SDs then it gets even more unlikely as the loss figure goes to £1,778.38 and the win to £1,724.32.

Should I instead (in the case of the 95% rule) say that I could potentially win £1,140 but would have X% chance of going broke?

I’m confused. I’d love to know where I’m going wrong here. I’ve used Excel and the SDEVPA calculation to get to this point. Any comments would be greatly appreciated.

Regards
Kev

Hi Kev,
Welcome to the forum with your good question.
Your calculations seem fine for the range of outcomes. The problem we have is that obviously you cannot lose more than you stake.

If you draw the chart of a normal distribution, the leftmost tail is chopped off at the zero point on the x axis. But that tail is still there and the area it encloses is still there. You have to imagine that you are continuing to wager from your reserve bankroll.

The 68-95-99.7 rule relates to the area enclosed by the bell curve and you have to consider the area that would have existed if you'd continued to play into negative bankroll.

Be aware, too, that for some games or betting patterns, the distribution is absolutely not a symmetrical bell shaped normal distribution.

You might not get too much help with mathematical analysis of roulette, because the math guys don't try to wrestle with the Expected Loss=(1/37) x (Total Wagered) truth, which you just cannot overturn.

The distribution will not be normal in 100 games. You would be better off using the binomial distribution for this and just calculate the probability of winning 0, 1, 2, 3 ... times.

Kev,

The probability of you losing all 100 spins and going El Busto is as follows:

(36/37)^100 = 0.06457697093 or 6.457697093%

That's why you are getting the wonky result for the 95% Confidence Interval, so you were on the right track with that theory. The reason why is simply that you have greater than a 5% probability of busting out, so the "Expected," 2SD left side value looks too high (low). The Standard Deviation doesn't know or care that your bankroll is only $1,000.

Quote: kevdev

I know this is correct as I’ve googled it!

very poor assumption
not everything shown on Google is 100% true and correct
simple elementary math skills can verify the value, done correctly of course.

Quote: kevdev

I’m assuming the following things are correct:
After 100 spins my EV would be -£27.03. The standard deviation for 1 spin would be £58.38 and therefore the standard deviation for 100 spins would be £583.78 (sqr root of 100 spins = 10).

If I’m right with my calculations then 68% of the time my results should fall in the range £610.81 loss to £556.76 win (one standard deviation from the mean).

you have to start over. you have wrong assumptions and one cant , as you did, simply use results for risk of ruin. this is not the way to do this

error #1
after 100 spins at 0 roulette
the distribution or results is NOT even close to being normal.
in other words, you can NOT use the 68-95 etc etc for your calculations and expect to get even close to a correct answer.
might think you are correct, start here
https://www.mathsisfun.com/data/standard-normal-distribution.html



this is the same problem the video poker people have using the normal distribution (bell curve) for even 4,000 hands played.

this has been gone over before here on Wizard of Vegas and I suggest you use search to discover the truth about what you are after.

error #2
WHY use the bell curve for something that is so easily calculated using a computer?
why?

error #3
see who can find it

good luck
back to basics for you is my suggestion

all those that follow that you are 100% correct using your math for what you want are also 100% wrong.
too bad

Quote: CrystalMath

The distribution will not be normal in 100 games.

you are in a very small minority, you know. most believe that using the bell curve here are right in their doing.
what do you have to show that what you say is more correct.

I really want to know this
enjoy

Quote: Mission146

Kev,

The probability of you losing all 100 spins and going El Busto is as follows:

(36/37)^100 = 0.06457697093 or 6.457697093%

That's why you are getting the wonky result for the 95% Confidence Interval, so you were on the right track with that theory. The reason why is simply that you have greater than a 5% probability of busting out, so the "Expected," 2SD left side value looks too high (low). The Standard Deviation doesn't know or care that your bankroll is only $1,000.

looks like about everything you said here in your conclusions is wrong.
why is that so?
I realize we all have opinions
OP is after way more than that
right?

Quote: kevdev

Say I wanted to place £10 on a number for 100 consecutive spins.
Should I instead (in the case of the 95% rule) say that I could potentially win £1,140 but would have X% chance of going broke?

what is your starting stake here? I missed it
thanks

this is very easy to calculate NOT using the normal distribution
it really is

but more information is required
not just opinions

Hi everyone. Thanks very much for your prompt replies.

@OnceDear - useful comments, thank you. I appreciate the welcome!

@CrystalMath @Mission146 - that makes perfect sense. Since seeing your posts I used the following parameters for a binomial distribution: 100 trials, 0 wins and a 1/37 chance of success. I think that pretty much gives me the information I need.

Quote: 7craps

very poor assumption
not everything shown on Google is 100% true and correct

Sorry, that was said with my tongue firmly in my cheek. I do have the math skills to check that one.

When I mentioned X% chance of going broke what I really should have said was my chance of having zero successes. I was sort of implying that I couldn’t possibly lose more than £1000 so that would have been my bankroll in this case. I apologise for not making that clearer.

Quote: 7craps

this has been gone over before here on Wizard of Vegas and I suggest you use search to discover the truth about what you are after.

error #2
WHY use the bell curve for something that is so easily calculated using a computer?
why?

error #3
see who can find it

good luck
back to basics for you is my suggestion

all those that follow that you are 100% correct using your math for what you want are also 100% wrong.
too bad

I’m really grateful to you for taking the time out to answer my question but I have searched this forum, the net and spoken to various people before posting here. I haven’t found the answer I was looking for.

Regarding error #2 - I don’t know why! I’m asking a question! I said I had a “reasonable” understanding of standard deviation and wanted to know why the empirical rule wouldn’t work. The binomial distribution clears things up for me but I didn’t understand why I couldn’t get sensible results. I’ve since repeated the £100 wager for 10,000 spins using the 68-95-99.7 rule and can see that I could expect to lose a max of £20,216 based on 3 SD. Much more sensible

Thanks all
Kev

The main reason for using a normal distribution approximation is because an exact calculation using the binomial would be very tedious if not effectively impossible. As pointed out above, your initial question can be solved with an easy equation.

As a general rule of thumb, a standard distribution will only start becoming somewhat accurate (defined here as within 10% relative to binomial within 2 standard deviations of the mean) when your expected value is at least 5 times the standard deviation. So in this case that would take (1/p -1) * 5^2 = 900 spins where p = 1/37. That would give a mean of 24.32 with SD of 4.86.

Accuracy of the standard distribution increases with N. So if you were to spin 20,000 times, the standard distribution should be very accurate ...probably well within 0.1% within 2 SDs of the mean , and within 10% all the way out to 5 SDs.

All of the above may be wrong since I have poor reading skills and no formal training in statistics/probability. It’s just based on my experience so caveat emptor.