# Expectation of Multiple Dice Rolls(Central Limit Theorem)

You take $n = 50$ dice rolls and calculate the average value of these dice rolls. What is the probability (approximately -i.e using CLT) that this average is greater than or equal to $4$?

My attempt:

We know that each die has expectation and $3.5$ and variance $35/12$. So the sum has variance $3.5n = 105$ and variance $\frac{35}{12}n = 145.8\overline{3}$. So, I was thinking to evaluate the cumulative probability of a normal distribution with parameters $\mu = 105, \sigma^2 = 145.8\overline{3}$, however I'm not sure if this is correct.

• You need to compute the expectation and variance of the average rather than the sum of 50 outcomes. – Math Lover Nov 13 '17 at 23:20
• Ok, that's what I thought. what would I be computing the expectation and variance with respect to? I.e it is clear to me during the initial calculation for a single die that i take the weighted sum of each outcome 1 through 6, however it does not seem obvious to me what the calculation would look like in this case. – rubikscube09 Nov 13 '17 at 23:22
• Let $Y = \sum X_i$. Then $m = Y/50$. You have found the expectation and variance of $Y$. Can you compute those of $m$? – Math Lover Nov 13 '17 at 23:27
• So clearly $E[m] = 3.5$ and for the variance, we would have $Var [m] = \frac{1}{n^2} Var[Y]$ which would be $(35/12)/50$. – rubikscube09 Nov 13 '17 at 23:35
• Yes, use them to solve your problem. – Math Lover Nov 13 '17 at 23:37

For completeness, your normal approximation might suggest a probability of about $$1-\Phi\left(\dfrac{4-3.5}{\sqrt{(35/12)/50}}\right) \approx 0.019217$$ where $\Phi$ is the cumulative distribution function of a standard normal distribution
This is not bad as the actual probability is about $0.021086$, but you can do slightly better with a continuity correction. A mean of at least $4$ corresponds to a sum of at least $200$ or a sum strictly more than $199$ on this discrete distribution, so checking $4$ or $\frac{199}{50}=3.98$ are both arguable on the continuous normal distribution, and a compromise would be to check $3.99$ so about the closer $1-\Phi\left(\tfrac{4-3.5}{\sqrt{(35/12)/50}}\right) \approx 0.021240$
As a warning, a normal approximation like this can be relatively poor in the tails of a distribution. If for example you had been looking for the probability of a mean of at least $5$, the actual probability would have been about $5.8 \times 10^{-11}$ but a straight normal approximation might have been about $2.6\times 10^{-10}$ and with continuity correction about $3.4\times 10^{-10}$, both several times too high