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Use coordinates $1,2,3$ for paper, scissors and stone. Let $X^N$ take values in the lattice simplex $S_N=\{(n_1,n_2,n_3)/N:n_i\ge 0,{\sum }_i{n}_i=N\}$. A uniformly selected unordered pair consists of types $i$ and $j$, $i\ne j$, with probability $2n_i{n}_j/(N(N-1))$. Losing copies the winner, so the possible jumps of the continuous-time Markov chain areTies make no change. For arbitrary event rate $\lambda$, the respective transition rates are $2\lambda N{x}_1{x}_3/(N-1)$, $2\lambda N{x}_2{x}_1/(N-1)$ and $2\lambda N{x}_3{x}_2/(N-1)$. ChooseThen the rates are exactly $N{x}_1{x}_3$, $N{x}_2{x}_1$, $N{x}_3{x}_2$. The infinitesimal generator of this cyclic imitation chain iswhere $\mathcal{J}=\{e_1-e_3,e_2-e_1,e_3-e_2\}$ and the three $\beta$ functions are, in that order, $x_1{x}_3,x_2{x}_1,x_3{x}_2$. Its drift is precisely

Here is a precise fluid limit with its error control. Suppose $X_0^N\to x_0$ in probability, and let $x$ solve $\dot{x}=F(x)$, $x(0)=x_0$. For every finite $T$,Indeed, the martingale decomposition of a density-dependent jump process givesThis follows by compensating each transition count by its integrated rate; the compensated counts are square-integrable martingales, with variance equal to the expected compensator. Since $\Vert \ell {\Vert }^2=2$ and $x_1{x}_2+x_2{x}_3+x_3{x}_1\le 1/3$ on the simplex, the Doob L2 maximal inequality, applied coordinatewise, givesThe polynomial drift is Lipschitz on the compact simplex, with some finite constant $L$. It preserves the sum of the coordinates, and each coordinate solves an equation of the form ${\dot{x}}_i=x_i{h}_i(x)$, so nonnegativity is preserved. Thus the ordinary differential equation stays in the simplex and exists for all time. The Gronwall inequality giveswhich proves the claim by Markov's inequality. This proves the required version directly, rather than invoking an unspecified approximation theorem.

For an interior trajectory, differentiating the logarithm of the product of numbers givesThe same identity for the product itself holds on the boundary, so $x_1(t)x_2(t)x_3(t)$ is conserved. A strictly positive initial product keeps the deterministic trajectory away from the boundary. Except at $(1/3,1/3,1/3)$ these interior trajectories are periodic: the strictly concave function ${\sum }_i\log x_i$ has smooth compact level curves surrounding its unique maximum, and the vector field is nonzero and tangent to each such curve.

The finite Markov chain behaves differently at long times. For $P(x)=x_1{x}_2{x}_3$, the change in $P$ during the first jump isThe other two terms are cyclic permutations. Their first-order contributions cancel, leaving the exact product decay in a cyclic imitation chain identityFor $N\ge 3$, while all three counts are positive, their product is at least $N-2$, so $P(X_t^N)\ge (N-2)/N^3$. A missing type cannot reappear. If $\tau$ is the first loss of a type, thenOnce only two types remain, one always wins, so its count increases until an absorbing monochromatic vertex is reached, almost surely in finite time. The cases of one or two initially present types follow the same argument without the first stage. Every finite chain eventually becomes monochromatic, whereas an interior deterministic solution retains its positive product. The fluid limit holds on each fixed finite interval; it does not justify exchanging the limits $N\to \infty$ and $t\to \infty$.

For the cyclic imitation chain, its infinitesimal generator satisfies ${\mathcal{L}}_N(x_1{x}_2{x}_3)=-3x_1{x}_2{x}_3/N$. Therefore $\mathbb{E}[X_1^N(t)X_2^N(t)X_3^N(t)]=e^{-3t/N}\mathbb{E}[X_1^N(0)X_2^N(0)X_3^N(0)]$. While all three types remain, the product of numbers is at least $(N-2)/N^3$ for $N\ge 3$, so this identity proves that one type is lost almost surely. The remaining pair has a fixed winner, which eventually occupies the whole population. The conserved product in the deterministic fluid limit therefore describes fixed finite horizons, not the chain's infinite-time behavior.