Solved – If I roll a die 25 times what is the probability that the mean outcome is greater than 4

Question

Consider a problem:

You take a fair die to a party and announce that you will roll it 25 times.

You will record each outcome and at the end average the 25 outcomes together to get their arithmetical mean.

You offer a bet: The player puts down \$1. If mean exceeds 4, you will give him $21 back, but otherwise, he loses his dollar. Is this a good bet for the player?

(Use Central limit theorem, CTL).

I tried to solve it. Please check whether it is true:

Answer 1

You are effectively taking the probability for a single dice roll being larger than 4 (which is indeed $\frac{1}{3}$). But that is a different thing than the mean of several dice rolls being larger than 4 (you can express this also as the sum of $x$ dice rolls being larger than $4x$).

Here I will provide an intuition behind the reason for your error by using the case of two dice rolls. Then I plot some further examples with more dice rolls to see what happens for even larger numbers. And eventually you will see that an approximation with the Normal distribution will be a good idea (although for 25 dice rolls you can also still calculate it exactly).

Two dice rolls example

The probabilities for the mean of dice rolls being above some number is not the same as the probability for a single dice roll being above some number. $P(\bar{x}_{\text{multiple rolls}}>4) \neq P(x_{\text{single roll}}>4)$

See for instance the possible outcomes of two dice rolls, where only $\frac{10}{36} < \frac{1}{3}$ have a mean above 4 (or total above 8)

$$\begin{array}{c|ccccc} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 3 & 4 & 5 & 6 & 7 & 8 & \color{red}9 \\ 4 & 5 & 6 & 7 & 8 & \color{red}9 & \color{red}{10} \\ 5 & 6 & 7 & 8 & \color{red}9 & \color{red}{10} & \color{red}{11}\\ 6 & 7 & 8 & \color{red}9 & \color{red}{10} & \color{red}{11} & \color{red}{12} \end{array}$$

More dice rolls

The image below shows how this continues for more dice rolls, by plotting the probability of the sum of dice rolls $X$.

25 dice rolls

• Explicit: You can calculate this explicitly by computing a table like above for the two dice rolls, but then for many instead. There has already been a question about this on this site (How to easily determine the results distribution for multiple dice?). Such calculation will give you a probability of: $$P(\bar{x}>4) = \frac{1823148354623298816}{6^{25}} \approx 0.0641 > \frac{1}{21} $$

• Approximation with normal distribution: In the image above you might note the bell shape curve of the normal distribution. The normal distribution is actually a quite good approximation for the mean of a dice roll (in fact the normal distribution was first described in relation to the approximation of coin flips, where deMoivre used a precursor of the normal distribution to approximated the binomial distribution, and you might see the dice roll as a multivariate generalization of the coin flip).

  This is 'what the question wants you to do': Use the normal distribution as an approximation for the mean of dice rolls. (and then use the resulting expression for the normal distribution to compute $P(\bar{x}>4)$). The question mentions CLT (the central limit theorem), and when you look that up you may find expressions for "approximating the mean of a sample based on the variance of the distribution" (if one would be pedantic then one could say that this approximation is not exactly the same as the 'central limit theorem', but many people mention/use this term when they employ this type of approximation).

  See the image below how the exact probability compares to the approximation with the normal distribution. The image on the right is the same function but plotted on a logarithmic scale to better show the difference. It seems only a slight difference but the true value and the estimated value for $P(\bar{x}>4)$ will be different by about 12 percent.