Posted by PLEASE HELP on Sunday, November 25, 2012 at 3:45pm.
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?!

PROBABILITY QUESTION PLEASE HELP!!  bobpursley, Sunday, November 25, 2012 at 3:52pm
Pr=1prwinning=1= 1*1/3*2/3=12/9=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

PROBABILITY QUESTION PLEASE HELP!!  Answer!!, Sunday, December 30, 2012 at 7:01pm
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 .

PROBABILITY QUESTION PLEASE HELP!!  HAHAHA, Friday, November 8, 2013 at 5:58pm
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

PROBABILITY QUESTION PLEASE HELP!!  Alecsun, Monday, January 6, 2014 at 2:25am
1/81

PROBABILITY QUESTION PLEASE HELP!!  LOLOL, Tuesday, August 5, 2014 at 1:45pm
1/81
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