Problem:
Francisco rolls a fair die and records the value he rolls. Afterwards, he continues rolling the die until he obtains a value at least as large as the first roll.
Let
$N$be the number of rolls after the first roll.Find
$E[N]$.
Let $X_i$ be the value of the $i$th roll where $X_i \in \{1, 2, 3, 4, 5, 6\}$ and each have an equal probability of occurring.
The expected number of rolls required after the first roll depends entirely on $X_1$. To model a single roll, we can use the geometric distribution where the expected value is given by:
$$E[N] = \frac{1}{p}$$
where $p$ is the probability of success.
For example, if $X_1 = 1$, the next roll is guaranteed to be greater than or equal to $1$ so the expected number of rolls is $1$ but how about when $X_1 = 2$?
To solve this case, we count the number of sides which are greater than or equal to 2 which is $\{2, 3, 4, 5, 6\}$ which is of length of 5. This means the probability of rolling a value greater than or equal to 2 is going to be 5 over 6.
$$E[N] = \frac{1}{(\frac{5}{6})} = 1.2$$
and for $X_1 = 3$ we have $\{3, 4, 5, 6\}$ which is of length 4 giving us a probability of $\frac{4}{6}$.
$$E[N] = \frac{1}{(\frac{4}{6})} = 1.5$$
Now to solve the entire problem, we can use this exact same process but for each possible outcome of $X_1$ and then sum the results:
E = 0
for i in range(1, 7):
E += (1/(7 - i))
print(E)
2.45
To understand why we can sum the expected values of the random variables is due to the law of total expectation which you can learn more about here.