Mario, Yoshi, and Toadette 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

Step-by-step explanation: Hi!

well, the only case where someone wins is if two of them chose the same thing, and the other chose one of the two remaining possibilities.

an useful way of starting this is calculated al the possible combinations of rock, paper and scissors.

we have 3 persons and 3 options, ence the total number of permutations is 3*3*3 = 27

So, the cases where none wins are:

1) the 3 chose the same thing: there are 3 combinations here, one for each option

2) the 3 chose a different thing: here you think, the first one has 3 possibilities, then the second one must choose a different option, so has 2, and the third one only has 1 remaining option. so you have 6 combinations.

so there are 6 + 3 = 9 combinations where no one wins, if we want to se the probability, we compute 9/27 = 1/3 = 0.33%.

But the problem ask for it to happen 4 times in a row, so you will have 0.33*0.33*0.33*0.33 = 1/81 chances that this happens