Rock-paper-scissors

Question

Sheldon, Leonard and Wolowitz play a sizzling game of rock-paper-scissors for three players. In every round, each player simultaneously shows one of the three shapes. Rock beats scissors, scissors beats paper, and paper beats rock. If in a round exactly two distinct shapes are shown then 1 point is added to the score of the one player or to the two players who showed the winning shape, otherwise no point is added.

After many rounds of playing it occurred that each of the shapes had been shown exactly the same number of times. Is it possible that at this moment the total sum of points of Sheldon and Leonard and Howard was 2015?

Answer 1

This is not possible.

After each round, let $n_R$, $n_P$, and $n_S$ denote the total number of rocks, papers, and scissors' that have appeared, and let $S$, $L$, and $W$ be the scores of Sheldon, Leonard, and Wolowitz. Define a quantity $$ Q:=S+L+W+2\cdot n_R+ n_P. $$ It can be checked that on each round, the value of $Q$ either stays the same, increases by $3$, or increases by $6$. Since the value of $Q$ begins at $0$, $Q$ will always be divisible by $3$. However, if each shape has appeared the same number of times and the sum of the scores is $2015$, then $$ Q=S+L+W+n_R+2 n_P=2015+3 n_R=3(671+n_R) +2 $$ is not divisible by $3$.

Answer 2

Let R, P, and S be the number of rocks, papers, and scissors that have been shown so far, and let Q be the total sum of points.

I claim that if 3 does not divide Q, then R/P/S are distinct. By symmetry, if we prove that R/P must be distinct, then also P/S and S/R must be distinct. We will consider the effect that one round can have on the following two values: R-P mod 3 and Q mod 3.

• Q does not change. This happens when there is a 3 of a kind of a 1 1 1. In either case clearly R-P mod 3 does not change.
• Q increases by 1. This happens in 3 cases: 2R P, 2P S, 2S R. Clearly in all three cases R-P increases by 1 (mod 3).
• Q increases by 2. This happens in 3 cases: R 2P, P 2S, S 2R. Clearly in all three cases R-P increases by 2 (mod 3).

Therefore it is always true that $R-P \equiv Q \pmod{3}$, so if $Q \not\equiv 0 \pmod{3}$, we have $R-P\not\equiv0\pmod{3}$ or $R\not\equiv P\pmod{3}$ which implies that $R \ne P$, as claimed.

By the contrapositive of my claim, $R=P=Q \implies Q\equiv 0\pmod{3}\implies Q \ne 2015$. In fact, just two of R/P/Q being equal would imply the same thing.

Answer 3

It is not possible

If we use variables instead of rock, paper, and scissors there are only 4 possible transactions.(let $A<B<C<A$)

As for number of times played we will only take note of the change of the difference to the average.

1: ABC ->  0A  0B  0C -> 0 points
2: AAA -> +2A -1B -1C -> 0 points
3: AAB -> +1A  0B -1C -> 1 point
4: ABB ->  0A +1B -1C -> 2 points

If we get a "1" there is no change whatsoever.

If we get a "2" AAA it can be solved(to a state where every difference to the average is 0) by:

• two "2" BBB and CCC -> result 0 points total
• one "2" BBB one "3" CCA and one "4" BCC -> result 3 points total
• one "3" BBC and one "4" BCC -> result 3 points total

If we get a "3" AAB (and without using anymore "2" since we already found all solutions with "2") it can be solved by:

• two "3" BBC and CCA -> result 3 points total
• one "4" BCC -> result 3 points total

If we get a "4" ABB (without using anymore "2" or "3") it can be solved by:

• two "4" BCC and CAA -> result 6 points total

We can now see that whatever we do to get each shape the same number of times our result must be dividable by 3, which 2015 is clearly not.