... Generalize this to $n$ players. Compute the probabilities $p_1, ..., p_n$ that each player wins.
Proposed Solution:
Let $p_k$ be the probability of player $k$ winning in the set $\{p_1, ..., p_n\}$
Then:
$$p_k = (\frac{5}{6})^{k - 1} * p_1$$
The tree shows a recursive definition for $p_1$
$$p_1 = (\frac{5}{6}) * (\frac{5}{6})^{n - 1} * p_1 + \frac{1}{6}$$ Combining $p_1$. $$p_1 = \frac{\frac{1}{6}}{1 - (\frac{5}{6})^n}$$
Substituting $p_1$ into $p_k$
$$p_k = \frac{(\frac{5}{6})^{k - 1} * \frac{1}{6}}{1 - (\frac{5}{6})^n}$$
This is the general case. Alice, Bob, and Claire have probabilities to win $p_1, p_2, p_3$ respectively.
We can substitute $n = 3$ and $k = \{1, 2, 3\}$ to get the results: $\{\frac{36}{91}, \frac{30}{91}, \frac{25}{91}\}$ respectively. And these add up to 1 since the probability for no one to win approaches zero.
Can somebody verify this solution?.