There are three jars. Each of three coins is placed in one of the three jars, chosen at random and independently of the placements of the other coins. What is the expected number of coins in a jar with the most coins?
$\textbf{(A) } \frac{4}{3} \qquad\textbf{(B) } \frac{13}{9} \qquad\textbf{(C) } \frac{5}{3} \qquad\textbf{(D) } \frac{17}{9} \qquad\textbf{(E) } 2$
Difficulty: 40/100 — A little hard to get into, but once the cases are established, it is not exceptionally difficult to compute.
Core Concepts: casework, expected value
Challenges: If the contestant is not familiar with expected value, it can be hard to begin the casework. Working with the coins particularly when they are indistinguishable can be confusing to comprehend.
Techniques: Using casework to find an expected value.
Error-prone Steps: Incorrectly determining the number of possibilities for a case.
Ideal Time:
Experienced: ≤ 1 min
Intermediate: ≤ 2-3 min
Beginner: ≤ 4-5 min