Standard deviation of 7's per 100 rolls?

i read that on avg, a 7 is rolled once every 6 times.
thus a 7 will come up ~17times every 100 rolls.

do you normally use 1 sigma (+/- 34%) as the standard deviation? why not 1/2 sigma? or some other multiple of sigma?

if 1 sigma, then the range for 7s per 100 rolls is between 11 and 22?

Quote: 100xOdds

i read that on avg, a 7 is rolled once every 6 times.
thus a 7 will come up ~17times every 100 rolls.

do you normally use 1 sigma (+/- 34%) as the standard deviation? why not 1/2 sigma? or some other multiple of sigma?

if 1 sigma, then the range for 7s per 100 rolls is between 11 and 22?

1 sigma *is* one standard deviation. But that does not mean you multiply 16.66666 * 1.34.

Back to the textbook!! [g]

Yeah one sigma is equal to one standard deviation. You can look at multiples of standard deviations if you wish to change the amount of "confidence" you want something to fall under. One common use is to look within 2 standard deviations because about 95% of results will fall within 2 SDs. This is only accurate with a lot of trials though...100 is so, so. The Binomial Distribution would probably be more accurate

The SD for rolling a 7 though is:

sqrt[.1666 (1-0.16666)^2 + .83333(0 - 0.166666)^2] = 0.372678 per roll

Standard Deviation scales with the square root of the number of trials, so:

SD of 1 roll*sqrt(total rolls) = SD of total rolls

So for 100 rolls:

0.372678*sqrt(100) = 3.72678 rolls

The expected number of sevens is 100/6 = 16.6666

To be within 1 SD (68.7% of all results): 16.6666 +/- 3.72678 = 12.940 to 20.393 rolls

To be within 2 SD (95.5% of all results): 16.6666 +/- 2*3.72678 = 9.2131 to 24.120 rolls

Edited for brainfart.

Here is what I have

Quote: 100xOdds

i read that on avg, a 7 is rolled once every 6 times.

P = 6/36 (1/6) or 1 in 6 (1/P)

Quote: 100xOdds

thus a 7 will come up ~17times every 100 rolls.

N*P = expected number of 7s (successes)aka(mean or average or expected value - ev)
N = # or trials or 100 in your example
so
100 * 1/6 = 100/6 or 16 and 4/6 times
(16 and 2/3) OK ~17times
"thus a 7 will come up"
"will" IMO should not be used

Quote: 100xOdds

do you normally use 1 sigma (+/- 34%) as the standard deviation? why not 1/2 sigma? or some other multiple of sigma?

if 1 sigma, then the range for 7s per 100 rolls is between 11 and 22?

Now you want to know about the central limit theorem or the famous bell curve.
"The central limit theorem says that the distribution of an average of many independent, identically distributed random variables tends toward the famous bell-shaped normal distribution"

The standard deviation (binomial standard deviation or BSD - not BFD) is the square root of the variance.
variance = N*P*(1-P)
One can find the proof of this in many probability books.
standard deviation is the square root of variance
SD = SQRT(N*P*(1-P))
It (BSD) is in the same "unit" as the expected number of successes in N trials is.
Right?
What is is??
equals.

We know the EV = 16 2/3 (N*P)
SD = 3.726779962
SQRT(N*P*(1-P))
(left to reader to plug in the values and solve)

So, 1 SD (sigma) is the 68% value you can find here
http://en.wikipedia.org/wiki/68-95-99.7_rule
http://en.wikipedia.org/wiki/Standard_deviation

16.67 + and -
3.726779962
1 SD range
1 20.39344663
ev 16.66666667
-1 12.9398867

SD range expanded out

Same chance of ending up at 1 or -1 or 2 or -2 and so on.
Nothing can be done about it as long as the dice rolls are independent and fair six-sided dice are used.

advanced concept in spoiler

One can also use the binomial distribution formula for N<= 1000 or calculators for this.
http://stattrek.com/probability-distributions/binomial.aspx?Tutorial=Stat

The ev and sd just does a very good job at approximating the binomial without a lot of math
(as long as n*p and n*1-p >=5 in most cases)
Most times we can calculate the SD in our head when P is close to 0.50
Just divide N by 4 and find your square root of that number.
this was not meant to be an exhaustive study of the central limit theorem.
Some other items were left out on purpose, like the .5 correction to name just one.

nice to have this now in one place
pool is now open!
Good Luck

Quote: tringlomane

...The SD for rolling a 7 though is:

sqrt[.1666 (1)^2 - .83333(0)^2] = 0.408248290463863 per roll...

I would revise your calculation to: sqrt[.1666 (1-.1666)^2 + .83333(0 - .1666)^2] = 0.37267799625 per roll.

Quote: 7craps

1 20.39344663
ev 16.66666667
-1 12.9398867

Quote: 100xOdds

if 1 sigma, then the range for 7s per 100 rolls is between 11 and 22?

well, you came pretty close doing it the wrong way, 100xodds [g]

Some things to for sure take away from this if you are like me and don't easily start doing the math:

* what the 68–95–99.7 rule means - you were off base with your thinking. 68.27%,95.45%,99.73% that is. For example, the dark blue region means a session's results are expected to fall into the dark blue 68.27% of the time. It doesnt mean you add or subtract 34% to the mean, though, like I think you were trying.



* standard procedure is to have a set of results and try to get some meaning from it. A mean is established, etc. With gambling it seems this is often changed to projecting results knowing what they *should* average out to and trying to get meaning from that. One cannot get certainty, but can get to a range representing 99.73%, before even, say, picking up the dice. [somebody can confirm I have that right]

* standard deviation does not bow to simple calculations like you were attempting. Then again, just about anybody not intimidated can use some formulas - there's no advanced algebra or anything.

Quote: ChesterDog

Quote: tringlomane

...The SD for rolling a 7 though is:

sqrt[.1666 (1)^2 - .83333(0)^2] = 0.408248290463863 per roll...

I would revise your calculation to: sqrt[.1666 (1-.1666)^2 + .83333(0 - .1666)^2] = 0.37267799625 per roll.

Oops, that's right, that's what I get for trying to do things too fast.

odiousgambit,

yes, i was multiplying by 34%.. whoops. it seemed logical at the time.
looks like i need to learn ALOT about statistics calcs