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I know that for 50%-50% situations like coin toss the standard deviation for N number of toss is given by $$\sigma = \sqrt{.25N} = \sqrt{N}/2$$

What is the type though that I can calculate the standard deviation in situations that 3 results are equal to come out? (33,33% - 33,33% - 33,33%)

Or what is the type to calculate standard deviation for 4-way results (25% - 25% - 25% - 25%),5-way results, 6-way results, 12-way results etc

Is it any more generalised rule?

If not i would only like to know the type for 3,6 and 12-way results

Thank you!

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  • $\begingroup$ The standard deviation is a measure of dispersion for a random variable taking (one dimensional) real values (or values from a subset of the reals such as the non-negative integers used to count the number of successes or heads of coin tosses). What do you mean by standard deviations when there are three or more different outcomes? $\endgroup$
    – Henry
    Apr 14, 2016 at 19:01
  • $\begingroup$ what i mean as 2-way result is a trial that has 2 possible equal to come outcomes (such as heads and tails in coin flips) What I mean as 3-way 4-way etc result is a trial that can have 3 or 4 outcomes. For example a 6-way would be the roll of a die. It has 6 possible outcomes in every trial (1,2,3,4,5 or 6) Calculating the σ in coin flip (a 2-way result) is σ=sqrt(N) /2. for example if i have 10000 trials the σ would be σ=sqrt(N) /2 = sqrt(10000) /2 = 100/2 =50. so heads would be 5000 +/-50 and tails would be 5000 +/-50 . is there a similar type to calculate for 3,6 and 12-way results? $\endgroup$
    – chester
    Apr 14, 2016 at 19:48
  • $\begingroup$ For throwing several dice, you can meaningfully calculate the variance and standard deviation of the total number of pips as this is one dimensional. For multiple outcomes you might consider something like a covariance matrix or a chi-squared statistic $\endgroup$
    – Henry
    Apr 14, 2016 at 21:08

1 Answer 1

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First you have a sequence of i.i.d. Bernoulli trials $X_i \sim \text{Bernoulli}(p)$, then the sum of them (the total number of "success") follows a Binomial distribution: $$ \sum_{i=1}^n X_i \sim \text{Binomial}(n, p)$$ and thus the variance is given by $np(1 - p)$. In particular when $p = 0.5$, you have your stated result.

When you generalized to more than two outcomes per trial, say $k$ outcome per trial, then the frequencies of each outcome jointly follows a multinomial distribution, and each individual frequency will marginally follow a Binomial distribution.

So it depends on what you want - and if you really want to find out the standard deviation. The standard deviation is the same, except that the $p$ is now changed to $\frac {1} {k}$.

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  • $\begingroup$ i'm not that advanced into statistics and bionomials. I understand most parts of your explanation but i'm not sure that I can implement them into what I need. What i actually need is the type to calculate the standard deviation for 3,6 and 12 outcomes per trial. For example I know for 2 outcomes per trial that if i have 10000 trials the σ would be σ=sqrt(N) /2 = sqrt(10000) /2 = 100/2 =50 Is there any specific type like this that I could calculate for 3,6 and 12 outcomes per trial? $\endgroup$
    – chester
    Apr 14, 2016 at 18:00
  • $\begingroup$ In the previous answer I personally left out the formula you want (at least not stated explicitly) because I am not completely sure if that is what you want and hope you at least have some what more clear idea before moving on. If each trial has $k$ outcomes and each outcome is equally likely with probability $\frac{1}{k}$, then the number of occurrence of a particular outcome out of $n$ iid trial will follow $\text{Binomial}(n,\frac{1}{k})$ as stated, and the standard deviation will be $\sqrt{n(\frac{1}{k})(1-\frac{1}{k})}$ $\endgroup$
    – BGM
    Apr 15, 2016 at 2:36
  • $\begingroup$ Yes. That is the type I was looking for. The general one so that it can be used for all. thanks :) $\endgroup$
    – chester
    Apr 18, 2016 at 15:10

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