Question

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?!

Answer 1

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

Answer 2

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 .

Answer 3

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

Answer 4

1/81

Answer 5

1/81

Answer 6

It's 1/81

Answer 7

it is 1/81

Answer 8

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

Answer 9

It is 1/81

Answer 10

1/81