What’s the expected number of throws to see all of the faces of a die?
This is called the coupon collector problem. For a fair n-sided die, the expected number of attempts needed to get all n values is \begin{equation}
n \Sigma_{k=1}^n \frac{1}{k}
\end{equation} which, for large n is approximately n log n.
The time until the first result appears is 1. After that, the random time until a second (different) result appears is geometrically distributed with parameter of success 5/6, hence with mean 6/5 (recall that the mean of a geometrically distributed random variable is the inverse of its parameter). After that, the random time until a third (different) result appears is geometrically distributed with parameter of success 4/6, hence with mean 6/4. And so on, until the random time of appearance of the last and sixth result, which is geometrically distributed with parameter of success 1/6, hence with mean 6/1. This shows that the mean total time to get all six results is \begin{equation}
6 \Sigma_{k=1}^6 \frac{1}{k} = \frac{147}{10} = 14.7
\end{equation}
n \Sigma_{k=1}^n \frac{1}{k}
\end{equation} which, for large n is approximately n log n.
The time until the first result appears is 1. After that, the random time until a second (different) result appears is geometrically distributed with parameter of success 5/6, hence with mean 6/5 (recall that the mean of a geometrically distributed random variable is the inverse of its parameter). After that, the random time until a third (different) result appears is geometrically distributed with parameter of success 4/6, hence with mean 6/4. And so on, until the random time of appearance of the last and sixth result, which is geometrically distributed with parameter of success 1/6, hence with mean 6/1. This shows that the mean total time to get all six results is \begin{equation}
6 \Sigma_{k=1}^6 \frac{1}{k} = \frac{147}{10} = 14.7
\end{equation}
Correct Answer: 14.7
My Notes
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