Different Models of One Game

The ladder is $50,33,22,15,10,7,5,3,2,1$ and then $1$ for ever; it is the same under either tie rule, since no step lands on a half-integer. The fixed points of one rounded step are exactly $0$ and $1$. In the continuum game the unique outcome surviving iterated dominance is $0$ (classical, cited); the integer ladder never reaches it, because $\frac{2}{3}\cdot 1$ rounds back to $1$. In the finite discrete game itself (five players, the prize shared equally among those closest), everyone naming $1$ is a Nash equilibrium, and so is everyone naming $0$ (each payoff $\frac{1}{5}$; every unilateral deviation to any other of the $101$ integers earns $0$). The contrast between $1$ and $0$ is therefore between the rounded ladder and the continuum game, not between the ladder and the equilibria of the same discrete game.

The two players face different conditional structures. For player 2 with count $T_2=\tau \ge 1$, $P(T_1=\tau -1\mid T_2=\tau )=1/(2-\epsilon )=\frac{10}{19}$: the partner has received one message fewer. For player 1 in state $b$ with count $T_1=\tau \ge 0$, $P(T_2=\tau \mid T_1=\tau )=\frac{10}{19}$: the partner has received the same number, and never one fewer. In both cases $\frac{10}{19}$ is the probability that the partner did not receive the last message one sent. Since $B$ against an $A$-player loses $2$ and against a $B$-player gains $1$, $B$ is profitable only if that risk is below $\frac{1}{3}$; it never is. By induction alternating between the players — player 2 at count $0$, player 1 at count $0$, player 2 at count $1$, and so on — both players play $A$ at every count. A public announcement of state $b$ is a different protocol, not $\epsilon =0$ in the stopping rule: under it $(B,B)$, paying $(1,1)$, becomes an equilibrium, but so do $(A,A)$ and independent mixing with $P(B)=\frac{2}{3}$; playing $B$ requires selecting the efficient equilibrium.

With the uniform closure the rank recursion is literally the level recursion, so the two ladders coincide at every depth. With the equilibrium closure, in the weighted game (whose equilibrium mixture is not uniform): the level ladder is $S,R,P,S,R,P,\dots$ and the rank ladder is $R,P,S,R,P,S,\dots$: the same three-cycle shifted by one floor, ${\mathrm{rank}}_k={\mathrm{level}}_{k+1}$, so the two prescriptions disagree at every depth $k\ge 1$. (At the bottom the rank-$1$ agent faces the equilibrium mixture, is exactly indifferent, and the tie rule chooses R; level $1$ faces the uniform mixture and strictly prefers S.)

The defender catches the intruder exactly when $d+r$ is odd. For the defender, one level deeper than the intruder wins and two levels deeper loses again; for the intruder the same step is fatal — an intruder one level deeper than the defender is caught, and the intruder escapes exactly when the two ranks have the same parity. Depth confers no advantage of its own; only parity matters. The induced game over the parities of one's own rank is matching pennies, with the unique equilibrium $\frac{1}{2}$–$\frac{1}{2}$ for both and catch probability $\frac{1}{2}$. Over all integer ranks the equilibria are not unique: any distribution putting mass $\frac{1}{2}$ on each parity is one. The parity projection is unique, not the distribution of depth. No finite rank is safe.

The test tuned to a Bayesian (pass all good cases and $\frac{3}{7}$ of the bad) earns the sender $\frac{3}{5}$ against the Bayesian and against the credulous receiver, and leaves both with payoff $0$. The test tuned to credulity (pass everything) earns $1$ against the credulous receiver, who then loses $\frac{2}{5}$, and collapses to $0$ against a Bayesian, for whom an uninformative pass is no reason to act. No test earns anything against the test-blind receiver at this prior. Designing for credulity is thus profitable exactly when the model of the receiver is right, and maximally fragile when it is wrong ($1\to 0$); designing for Bayesian updating varies less across these models ($\frac{3}{5}$ or $0$). That is not minimax superiority — both tests have worst case $0$ — and under a prior $t$ on the Bayesian receiver (and $1-t$ on the credulous one) the Bayesian-tuned test earns $\frac{3}{5}$ against $1-t$ for the pass-everything test, so the preference flips at $t=\frac{2}{5}$.