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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.

S Wiley Wakeman
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  • An exact solution is setting $S:=n\cdot A$, where $A$ is your desired average. After follow this answer and the probability that you want, for some fixed $A$ in $n$ rolls, is $[x^S]g(x)/6^n$ (for fair dice of six sides). To get probabilities "more or equal than $A$" you only need add all the probabilities of all of these averages. For an estimation you can use the CDF of a normal distribution. – Masacroso Mar 09 '17 at 16:03

1 Answers1

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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.

BruceET
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