posted by on .

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.

PLEASE HELP this is so hard I do not have any idea because I fail at probability. How do we do this?!

Pr=1-prwinning=1=- 1*1/3*2/3=1-2/9=7/9 is probablility of no one winning each game

Pr this happening four games= that to the 4th power, or (7/9)^4 = about .366

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 .

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 ( ways), so there are outcomes in which no one wins.

Therefore, the probability that no one wins a given round is . This means the probability that no one wins any of four rounds is .

You are all wrong, sorry to say. There are 3 ways they are all the same, and 6 ways they are all the same. added up, that's 9. Now the number of possibilities in total are 27 for each round. That's 9/27=1/3 for each round. (1/3)^4=1/81

1/81