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Roll 8 fair dice. $X$ is the number of dice that land on $6$. How would you calculate the variance of $X$?

Can we treat this problem as a binomial such that $p$ = probability of landing on a $6$, $p = \frac{1}{6}$ and $q$ = probability of not landing on a $6$, $q=\frac{5}{6}$. Then the variance $\operatorname{Var}(X) = n p q = 8 \times \frac{1}{6} \times \frac{5}{6}$.

Would this be correct?

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Yes, it is correct. –  Sasha Oct 23 '11 at 4:58
    
And this simple observation is the beginning of a long and fascinating story since the variance, being quadratic, should be of the order of the square of the mean, thus of the order of $n^2$ when $n$ grows large, while it is only of order $n$. A first consequence is called the law of large numbers but there are many others... –  Did Oct 23 '11 at 6:39

2 Answers 2

up vote 0 down vote accepted

The calculation is absolutely correct. Suppose that you repeat an experiment independently $n$ times, and the probability of "success" each time is $p$. If the random variable $X$ is the number of successes, then $X$ has binomial distribution, with mean $np$ and variance $npq=np(1-p)$.

The above dice problem fits the requirements perfectly.

There is a problem with the title of the post. You are finding the variance of a random variable, not of an event. The concept of variance does not even apply apply to events. An event is just a subset of the sample space. So getting an odd number of $6$'s, for example, is an event.

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As I said in the comment, your solution is correct. Here is a simulation confirmation using Mathematica. The function RandomChoice is used to generate a million repetitions of 8 rolls. Command Tally shows the number of different outcomes generated:

In[56]:= sample = 
  Map[Count[#, 6] &, RandomChoice[Range[6], {10^6, 8}]];

In[57]:= Tally[sample]

Out[57]= {{0, 232113}, {3, 104193}, {2, 260399}, {1, 372385}, {4, 
  26287}, {5, 4152}, {6, 441}, {7, 30}}

In[58]:= {Variance[sample], 8 1/6  5/6} // N

Out[58]= {1.1125, 1.11111}
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