Jim and Renee will play one game of Rock, Paper, Scissors. In this game, each will select and show a hand sign for one of the three items. Rock beats Scissors, Scissors beat Paper, and Paper beats Rock. Assuming that both Jim and Renee have an equal chance of choosing any one of the hand signs, what is the probability that Jim will win?
A. 5/6
B. 2/3
C. 1/2
D. 5/12
E. 1/3

Ans: E

The explaination to the above question is: No matter what sign Jim throws, there is one sign Renee could throw that would beat it, one that would tie, and one that would lose. Renee is equally likely to throw any one of the three signs. Therefore, the probability that Jim will win is 1/3.

However, this explaination seems incorrect. If we say the probability of Jim to win is 1/3 and that of Renee to win is 1/3, then what is the remaining 1/3 probability for?
I thought the answer is 1/2.

Tricky question, but I think the solution is:

Probability of choosing either a rock, paper, or scissor is 1/3 each.

R>S
S>P
P>R

The 3 scenarios where Jim will win this one game of rock, paper, scissors is:

Jim picks Rock, AND Renee picks Scissors. (1/3 * 1/3) = 1/9
Jim picks Scissors, AND Renee picks Paper. (1/3 * 1/3) = 1/9
Jim picks Paper, AND Renee picks Rock. (1/3 * 1/3) = 1/9

The 1st scenario OR the 2nd scenario OR the 3rd scenario will allow Jim to win his one game of rock, paper, scissors. So,

1/9 + 1/9 + 1/9 = 3/9 = 1/3

Answering ameya's question:

P(Jim wins) = 1/3
P(Renee wins) = 1/3
P(They tie) = 1/3

Rey