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.
2 answers
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 $(1/3)^4 = \boxed{1/81}$
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 $(1/3)^4 = \boxed{1/81}$
1 answer