# Monsters’ gems

Once again, The Riddler does not disappoint! This puzzle is about slaying monsters and collecting gems.

A video game requires you to slay monsters to collect gems. Every time you slay a monster, it drops one of three types of gems: a common gem, an uncommon gem or a rare gem. The probabilities of these gems being dropped are in the ratio of 3:2:1 — three common gems for every two uncommon gems for every one rare gem, on average. If you slay monsters until you have at least one of each of the three types of gems, how many of the most common gems will you end up with, on average?

Here is my solution:
[Show Solution]

A more brute-force approach:
[Show Solution]

Yet another solution approach with very nice write-up can be found on Andrew Mascioli’s blog

## 4 thoughts on “Monsters’ gems”

1. Blazer says:

I used linearity of expectation so I said the expected value of common gems is just the expected value of number of gems divided by 2. Then I said the probability of having for example something like after n gems at least one of the two more common gems and none of the least common type is (5/6)^n-(1/2)^n-(1/3)^n and similar for other expressions so from there it is easy to find probability of needing exactly (n+1) gems which we can then sum from n=2 to infinity and compute using arithmetico-geometric sequence sum ideas where you write a sum as a sum of sums where each sum in the sum telescope and then that bigger sum telescopes as well but it is messy and you get the correct answer. These are computationally sort of isomorphic but the intuition/broad idea motivating the solutions is slightly different perhaps or not depending on how you view it.

1. Laurent says:

Good point about the linearity of expectation — I updated my solution to take this into account. It makes the recursions a bit simpler, and even more similar to the coin-flipping problem!