Three people play a game of "nonconformity": They each choose rock, paper, or scissors. If two of the three people choose the same symbol, and the third person chooses a different symbol, then the one who chose the different symbol wins. Otherwise, no one wins.

If they play 4 rounds of this game, all choosing their symbols at random, what's the probability that nobody wins any of the 4 games? Express your answer as a common fraction.

1/81

First, we compute the total number of possible outcomes for each round. Each person can choose any of three different symbols, so the total number of possible outcomes is $3^3 = 27$.

Now, we compute the number of outcomes in which no one wins. No one wins if all three symbols are the same (3 ways, one for each symbol), or each person presents a different symbol ($3! = 6$ ways), so there are $3 + 6 = 9$ outcomes in which no one wins.

Therefore, the probability that no one wins a given round is $9/27 = 1/3$. This means the probability that no one wins any of four rounds is $(1/3)^4 = \boxed{1/81}$

To find the probability that nobody wins any of the 4 games, we need to find the probability for each round of the game and then multiply them together.

In each round, there are 3 possible outcomes for each person (rock, paper, or scissors). Since they choose their symbols at random, the probability of each person choosing any particular symbol is 1/3.

Since there are 3 people playing, the probability that all 3 people choose the same symbol is (1/3) * (1/3) * (1/3) = 1/27.

The probability that nobody wins in each round is the complement of all 3 people choosing the same symbol. So, it is 1 - 1/27 = 26/27.

Since we have 4 rounds of the game, we need to multiply the probabilities of nobody winning in each round, which gives us (26/27)^4.

Therefore, the probability that nobody wins any of the 4 games is 26/27 raised to the power of 4, or (26/27)^4.