[popsicesHoupe]

1. excel
2. stdev([cell range])

[teentodiefows]

find the mean
then find the variance
then take the square root of the variance.

[averkif]

The population standard deviation can be simply calculated as follows:

1) sum all values
2) sum the squares of all values
3) square the result from (1)
4) Divide (3) by N (the number of values)
5) Subtract (4) from (2)
6) Divide (5) by N
7) Take the square root of (6)

So, if we have 3, 6, and 7 as values:

1) = 3+6+7 = 16
2) = 9 + 36 + 49 = 94
3) = 16*16 = 256
4) = 256/3 = 85.33
5) = 94 - 85.33 = 8.67
6) = 8.67/3 = 2.89
7) = SQRT(2.89) = 1.7

[Obgrfbke]

uuuuh yeah.

SE = Stdev/sqrt(n)

[eXC3Kvnn]

Originally posted by Lawrence of Arabia
uuuuh yeah.

SE = Stdev/sqrt(n) You must be an idiot...

[JewJoleSole]

oh, one more thing. you proved my point in your previous post

6) Divide (5) by N
7) Take the square root of (6)

what happens when n=100. when n= 1000? when n= 10000

[chipkluchi]

Originally posted by Lawrence of Arabia
oh, one more thing. you proved my point in your previous post

6) Divide (5) by N
7) Take the square root of (6)

what happens when n=100. when n= 1000? when n= 10000

What happens is that the difference in the sums grows linearly, you idiot.

[NanoGordeno]

Let's make this absolutely clear: you are claiming that the standard deviation of any set of independent dice rolls approaches 0 as the size of the set approaches infinity?

thats correct. its called the Law of Large Numbers. The variance of X_bar which is equal to sigma^2/n approaches zero as n increases.

[mtvlover571]

In fact, if you were to roll a die many times (or to roll many dice at once) then the set of values you receive would have a standard deviation which approaches sqrt(35/12)

duh

[VewCoorkPow]

swiss math?

[opdirorg]

Originally posted by Lawrence of Arabia
In fact, if you were to roll a die many times (or to roll many dice at once) then the set of values you receive would have a standard deviation which approaches sqrt(35/12)

duh

no. your sample average X_bar would have a standard deviation approaching zero when you roll a dice many times. You're a ****ing idiot. The set of values is different from the mean of the set.

A set of independently determined means has a variance which approaches 0 as the number of trials in each determination of the mean increases. Any given set of actual trials has a variance which approaches 35/12

You're like a ****ing child.

[Foucceedo]

Lawrence reminds me of the undergrads I TA. They know a bunch of formulas, but don't have a ****ing clue what any of them mean, or where they're applicable.

[Kghyutgykim]

standard deviation of any set of dice rolls iid approaches zero as the number of elements in the set (n) approaches infinity: FACT. as you increase the sample size of each set, your variance approaches zero.

the variance of x_bar decreases as the number of trials in each determination of x_bar increases: FACT

[HawksBurnDown]

Originally posted by self biased
unfortunately i don't have access to a laptop when i'm playing a game. i need to be able to figure out if a set of rolls is within one or not more or less on the fly. What you want to do is the following (assuming normal 6-sided dice):

1) Take the total number of dice rolled (if you're interested in two dice rolled 5 times, then use 2*5 = 10). Multiply by 3.5
2) Sum the values of the rolls (you should be summing 10 numbers between 1 and 6 in above example)
3) subtract (1) from (2)
4) Multiply 1.7 * sqrt(total number of dice rolled)
5) Divide (3) by (4)

(5) is the number of standard deviations high (+ve) or low (-ve) the series of rolls was

Please note that this is only a good estimator of probabilities for large total numbers of dice.

If I rolled 100 dice and ended up with a score of 380 then:

(1) is 350
(2) is 380
(3) is 30
(4) is 17
(5) is 1.8 or so (which is not that unusual)

[Stoniaanapy]

Originally posted by Lawrence of Arabia
which means that your little example here

The population standard deviation can be simply calculated as follows:

1) sum all values
2) sum the squares of all values
3) square the result from (1)
4) Divide (3) by N (the number of values)
5) Subtract (4) from (2)
6) Divide (5) by N
7) Take the square root of (6)

So, if we have 3, 6, and 7 as values:

1) = 3+6+7 = 16
2) = 9 + 36 + 49 = 94
3) = 16*16 = 256
4) = 256/3 = 85.33
5) = 94 - 85.33 = 8.67
6) = 8.67/3 = 2.89
7) = SQRT(2.89) = 1.7

approaches zero as n increases. Could you please provide me with an estimate of the standard deviation of a set of 100 dice rolls then?

[zttrftwsq]

Just because I'm a nice guy, I'll provide you with an excel spreadsheet which does it for 1000 rolls

The standard deviation (both pop and sample) for those 1000 pseudo-random rolls is provided and is suspiciously close to sqrt(35/12). Rolls are generated by taking a uniform deviate from 0 to 1, multiplying by 6 and taking the ceiling.

[Overlord]

everyone is right but too stubborn to admit it.

[itepearce]

This thread makes me cry. Is LoA really about to get a degree in economics?

[Efksqhyu]

Originally posted by KrazyHorse
No, lawrence doesn't understand the difference between the set and the mean of the set.

hes like a kid with all the legos but no schematic to build something with.

[LomodiorCon]

I don't know any of my TA's.