Probability of rolling a specific average (4, 5) with a six-sided die over n-rolls

Question

I'm wondering if someone could help me answer the following.

How could you estimate the probability of rolling a specific average value (e.g., rolling an average of 4 or of 5) over a set number of rolls (n).

For instance, how could you calculate the probability of:

1. Averaging exactly (or greater) than 4 over 15 rolls (n=15).

2. Averaging exactly (or greater) than 5 over 15 rolls (n-15).

I realize there is a different between exact and greater than claims, but I'd be interested in figuring out how estimate both.

Thanks.

Answer 1

Comment.

For the number of spots $D$ on a single roll of a fair die, $E(D) = 3.5$ and $SD(D) \approx 1.7.$

For the average $\bar D_{15}$ of 15 rolls, $\mu =E(\bar D_{15}) = 3.5$ and $\sigma =SD(\bar D_{15}) = SD(D)/\sqrt{15} \approx 0.44.$

Because $4$ is roughly $\mu + \sigma$, the probability to have the average for 15 dice above 4 will be around 15%. (And an average above 5 will be very rare, maybe about 0.0003.)

Is this the question you intended to ask? If so, the distribution of $\bar D_{15}$ is very nearly normal; you can use that to get more precise answers.