On the average, how many times must a die be thrown until one gets a 6?
Approach A: Analytical
Let $p =~^1/_6$ as the probability of getting a 6, and $q = 1 - p$ of not getting one. Then:
| case | throws | probability |
|---|---|---|
| 6 | 1 | $p$ |
| x6 | 2 | $pq$ |
| xx6 | 3 | $pq^2$ |
| xxx6 | 4 | $pq^3$ |
| … | $n$ | $pq^{n-1}$ |
The mean (or expected value) is, by definition:
Hence:
As a check, if we sum all probabilities we have:
Approach B: Distributions
We use a negative binomial to model the variable. The definition of a negative binomial is:
(…) the number of successes in a sequence of iid Bernoulli trials before a specified (non-random) number of failures (denoted r) occurs is given by NBin(r, p).
Let $X \sim NBin(1,~^1/_6)$, that is we define a single failure as the chance of throwing a 6. The expected value of a negative binomial is given by:
The solution to our problem is given by $\mathbb{E}[X] + 1$ since we want to include the last throw, hence:
Notes. Wikipedia presents the mean using the probability of a success instead of a failure:
Where $q = (1 - p)$.