1. Four people are playing rock-paper-scissors. They each play one of the three options (rock, paper, or scissors) independently at random, with equal probability of each choice. Compute the probability that someone beats everyone else.
(In rock-paper-scissors, a player that plays rock beats a player that plays scissors, a player that plays paper beats a player that plays rock, and a player that plays scissors beats a player that plays paper.)
Answer:
$\frac{4}{27}$
Solution: As the four players and three events are symmetric, the probability a particular player makes a particular move and beats everyone else is the same regardless of the choice of player or move. So, focusing on one such scenario, the desired probability is 12 times the probability that player 1 plays rock and beats everyone else.

In this case, player 1 plays rock and all other players must play scissors. All four of these events have probability $\frac{1}{3}$, so this scenario has probability $\frac{1}{3^{4}}=\frac{1}{81}$. Thus,
$$
\mathbb{P}(\text { one beats all })=12 \cdot \frac{1}{81}=\frac{4}{27} .
$$