, 2021

Solution:

Let $n = 7$ and $p = \frac{1}{4}$.
Let $q_{k}$ be the probability that $n$ is the last number rolled, if $k$ numbers less than $n$ have already been rolled. We want $q_{0}$ and we know $q_{n-1} = 1$.
We have the relation
$$q_{k} = (1-p) \frac{k}{n-1} q_{k} + \left[1 - (1-p) \frac{k+1}{n-1}\right] q_{k+1}$$
This rearranges to
$$\left[1 - (1-p) \frac{k}{n-1}\right] q_{k} = \left[1 - (1-p) \frac{k+1}{n-1}\right] q_{k+1}$$
This means that the expression on the LHS does not depend on $k$, so
$$[1-0] \cdot q_{0} = [1-(1-p)] \cdot q_{n-1} = p$$

Solution 2:

For a given sequence of Teresa's rolls, let $x_{i}$ be the $i$th distinct number rolled. We want to compute the probability that $x_{7} = 7$.
For a given index $i$, we say that $x_{i}$ is correct if $x_{i}$ is the least positive integer not in $\{x_{1}, \ldots, x_{i-1}\}$. Note that the probability of a given sequence $x_{1}, \ldots, x_{7}$ depends only on the number of correct $x_{i}$, since the probability of rolling the correct number on a given roll is higher by a factor of $2$.
Now, suppose $x_{7} = 7$. Consider $x_{i}' = x_{i-1} + 1$ for $1 < i \leq 7$, and $x_{1}' = 1$. Note that this operation on sequences $x_{1}, \ldots, x_{7}$ pairs sequences ending in $7$ with sequences starting with $1$. Additionally, we have that $x_{7}$ and $x_{1}'$ are both correct, and that $x_{i}'$ is correct if and only if $x_{i-1}$ is correct. Thus $x_{1}, \ldots, x_{7}$ and $x_{1}', \ldots, x_{7}'$ have the same probability.
So, we conclude that the probability of $x_{7} = 7$ is the same as the probability of $x_{1} = 1$. But this is just $\frac{1}{4}$.