Play rock, scissor and paper. If you win by rock, you get 2 points

Question

Play rock, scissor and paper. If you win by rock, you get 2 points, if you win by paper or scissor, you get one point. If you are tie, you get zero point. Provide a strategy that you can get more points than your rival.

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First of all, what are the probabilities like in this game?

Because you have $1/3$ chance to win on every move, but you also have $1/3$ chance of selecting a move(R, S, or P). Should we account for that in probability or not?

You cannot always pick rock because your opponent knows that you have an incentive to do that and thus he would choose paper and beat you.

Answer 1

Perhaps we should calculate the Nash equilibrium for this game. The payoff matrix is as below. Rows/columns are indexed as R, P, S and the payoffs in the matrix are for the player whose options are on the left, call it player 1.

$$ \begin{pmatrix} 0 & -1 & 2 \\ 1 & 0 & -1 \\ -2 & 1 & 0 \end{pmatrix} $$

Let the player 1 play rock, paper or scissors with probabilities $p_1, p_2$ and $1-p_1-p_2$ respectively. Similarly define $q_1, q_2$ for player 2. Now we can write the expected value for the payoff of player 1 when playing R, P, S are

$$ \begin{aligned} &E_R = p_1(-q_2 + 2(1-q_1-q_2)) \\ &E_P = p_2(q_1 - (1-q_1-q_2)) \\ &E_S = (1-p_1-p_2)(-2q_1+q_2) \end{aligned}$$

Now, we know that neither player will have the upperhand, so all these expected values should be equal to $0$. The bottom two say that we should have $q_1 = \frac{1}{4}, q_2=\frac{1}{2}$, (it will be similarly for $p$'s by symmetry). So you play paper with probability half and the others with one fourth. But I don't know why the first equation doesn't work. Making some of the probabilities zero doesn't work either, or I can't see it.

EDIT: Oh, but yes the first equation does also give $\frac{1}{2}+2(1-\frac{1}{4}-\frac{1}{2}) = 0$. Well anyway, I also did this simulation and it backs up that indeed $[\frac{1}{4},\frac{1}{2},\frac{1}{4}]$ is correct.

def getRand(ps):
    u = random()
    v = 0
    ret = -1
    while v<u and ret<len(ps):
        ret += 1
        v += ps[ret]
    return ret
        
#0=R, 1=P, 2=S
def simuGame(ps, pick1=None):
    v1 = pick1 if pick1!=None else getRand(ps)
    v2 = getRand(ps)
    if v1==v2: return 0
    if v1==0 and v2==2:
        return 2
    if v1==0 and v2==1:
        return -1
    if v1==1 and v2==0:
        return 1
    if v1==1 and v2==2:
        return -1
    if v1==2 and v2==0:
        return -2
    if v1==2 and v2==1:
        return 1
        
simuN = 10000
ptsRPS = [0]*3
pickProbs = [0.25, 0.5, 0.25]
for i in range(simuN):
    for pick1 in range(3):
        ptsRPS[pick1] += simuGame(pickProbs, 0)
    
print ([float(x/simuN) for x in ptsRPS])

Answer 2

This is impossible. Suppose such a strategy existed; one where you would end up with more points than your opponent. Then your opponent can use the exact same strategy and beat you, because neither player has an advantage (like the first player (White) in chess is supposed to have a slight edge in general). Therefore your opponent can get more points than you, contradicting you having got more points than them. Therefore our assumption that such a strategy exists is wrong: no such strategy exists.

In order for us to have a winning strategy, we would need to know in advance any strategies our opponent is likely to use. Kind of like, you can't reliably win in poker in the long run, unless you know/figure out the tendencies of opponents. Sure you can play GTO (game theory optimal), but this doesn't guarantee you winning, because your opponent might also be playing GTO, in which case both you and opponent's EV's will be $0$...