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Consider the following cooperative game played on the circle $S^1$, which we identify with $[0, 1]$ with its endpoints identified.

Let $0 < \alpha < 1$ be an irrational number. A total of $N \geq 2$ players each start at $0$. At every turn, the players each simultaneously bet heads or tails on a coin flip. If their choice matches, they move a distance of $\alpha$ clockwise, otherwise they move $\alpha$ counterclockwise.

A strategy is a sequence of heads/tails choices by the players that can depend on previous choices and coin flips, but not on current or future coin flips.

Question: Does there exist a strategy such that, almost surely the sequence of players’ positions is dense in $(S^1)^N$?

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  • $\begingroup$ I'm not quite seeing how this is cooperative. $\endgroup$ Commented Mar 15 at 10:02
  • $\begingroup$ Some ideas, assuming all of them bet on a single coin flip (I did not check them/can try to write them later if noone answers): I believe a set of measure $1$ of the set of all strategies $(\{0,1\}^N)^{\mathbb{N}}$ which do not depend on previous coin flips should work. One may also be able to take any finite sequence of vectors $v_1,\dots,v_k$ in $\{-\alpha,\alpha\}^N$ that generate $\mathbb{R}^N$ as an $\mathbb{R}^N$-vector space (so $k\geq N$), and choose them as bets in repeat $\endgroup$ Commented Mar 15 at 13:05
  • $\begingroup$ @user479223 It is cooperative in the sense that the players are trying to cooperate to make the sequence dense (and they can coordinate guesses). $\endgroup$ Commented Mar 16 at 2:33
  • $\begingroup$ I just mean I don't see how cooperation helps - how does one player's choices affect the other players? I am just dumb, please forgive me. $\endgroup$ Commented Mar 16 at 10:44
  • $\begingroup$ @user479223 no you aren’t :P. Maybe this is the part you were missing (and i should’ve made more clear) - they all bet on the same coin. $\endgroup$ Commented Mar 16 at 10:46

2 Answers 2

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Sure, let us take a sequence $v_{n, 1}, v_{n, 2},\ldots, v_{n, M(n)}$ of vectors in $\mathbb{R}^N$ such that each vector has coordinates $\pm \alpha$ and the fractional parts of partial sums of these vectors form $\frac{1}{n}$-net on $(S^1)^N$. Such a sequence exists for the same reason there is a robot which can solve every maze of size $n\times n$ -- just subdivide $(S^1)^N$ into finitely many parts, each of which has diameter smaller than $\frac{1}{2n}$, and then aim to visit each of them one by one (we can in two vectors shift any coordinate by $2\alpha$ while leaving each other fixed, and then each irrational rotation is dense. I think with this idea we can make what we want more explicit, by it is not the intuition I aim for).

Now just always bet this sequence of coinflips on repeat ($k$'th person bets tails if the corresponding coordinate is $+\alpha$ and heads if it is $-\alpha$). Sooner or later we will get all our guesses as we want, and regardless of the point where we start our points will form a $\frac{1}{n}$-net of the torus. Once this is achieved, go to the next $n$.

With a bit more suffering I think it is not hard to make the strategy independent of the previous coin flips (i.e. repeat $n$'th part of the strategy $K(n)$ many times where $K(n)$ is big enough, then we will get $\frac{1}{n}$-net for infinitely many $n$, hence we will have denseness).

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    $\begingroup$ Isn't what you mention in the first two lines impossible for big $n$, due to the set $\{-\alpha,\alpha\}^N$ being finite? $\endgroup$ Commented Mar 15 at 13:06
  • $\begingroup$ Doesn't sound like much suffering, the expected time to find a net is $2^{M(n)}$ repeats, it's just a Bernoulli process. $\endgroup$ Commented Mar 15 at 14:19
  • $\begingroup$ @VilleSalo it is indeed not much, but using even a single brain cell more than is required by the question is pain:) $\endgroup$ Commented Mar 15 at 14:29
  • $\begingroup$ @SaúlRM I have no idea what you mean, sure, $\{-\alpha, \alpha\}^N$ is finite, but we have here $\{-\alpha, \alpha\}^{NM(n)}$, which can be arbitrarily large. The only way I see is if for some reason you interpreted partial sums as partial sums of entries of each vector, and not of the vectors themselves ($v_{n, 1}$, $v_{n, 1}+v_{n, 2}$, $v_{n, 1}+v_{n,2}+v_{n,3},\ldots$) but then you would have had a compilation error couple of words later, as the partial sums of entries of a vector are elements of $S^1$ and not of $(S^1)^N$. $\endgroup$ Commented Mar 15 at 14:31
  • $\begingroup$ @AlekseiKulikov thanks. I somehow missed the "partial sums" part. Then my answer is similar to this, although the implementation is a bit different (which also makes it periodic and independent of previous flips), so I will leave it uploaded $\endgroup$ Commented Mar 15 at 14:40
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Here is a strategy which is periodic and independent of previous flips.

As in Aleksei Kulikov's answer, we identify heads with $\alpha$ and tails with $-\alpha$, so that a bet is an element of $\{-\alpha,\alpha\}^N\subseteq\mathbb{R}^N$. Let $(e_i)_{i=1}^N$ be the usual basis of $\mathbb{R}^N$.

Consider the choices $v_1,\dots,v_n$ where $v_i=\alpha\sum_{j=1}^Ne_j-2\alpha e_i$. We will repeat the bets $v_1,v_2,\dots,v_n$ periodically, with period $n$. As $(\mathbb{S}^1)^N$ is second countable, we just need to check that, given some nonempty open set $O\subseteq(\mathbb{S}^1)^N$, $O$ is visited by the players with probability $1$.

To see why, note that the position of our players in $(\mathbb{S}^1)^N$ at time $n$ is given by $\pi\left(\sum_{i=1}^Na_i(n)v_i\right)$, where $\pi:\mathbb{R}^N\to(\mathbb{S}^1)^N$ is taking modulo $1$ in all coordinates, and $a_i(n)$ are integer values such that, from step $n-1$ to step $n$, we add $\pm 1$ to $a_{n\text{ mod }N}$ and keep the other coefficients the same.

So we can see the problem as a sequence of vectors (a random walk) $u(n)=(a_1(n),\dots,a_N(n))_{n\in\mathbb{N}}$ in $\mathbb{Z}^N$, where at each step we add $\pm e_{n\text{ mod }N}$. Let $$P:\mathbb{Z}^N\to(\mathbb{S}^1)^N;(a_1,\dots,a_N)\mapsto\pi\left(\sum_{i=1}^Na_iv_i\right)$$

Claim: Let $A=\{u\in\mathbb{Z}^N;P(u)\in O\}\subseteq\mathbb{Z}^N$. Then there is a finite set $F$ such that $A$ intersects $u+F$ for all $u\in\mathbb{Z}^N$.

Proof: Let $\varepsilon>0$ be such that $O$ contains an $\varepsilon$-ball, and choose the set $F$ such that $P(F)$ is $\varepsilon$-dense in $(\mathbb{S}^1)^N$. We are using here that $P(\mathbb{Z}^N)$ is dense in $(\mathbb{S}^1)^N$. $\square$

The claim implies that there is $\varepsilon>0$ and $L\in\mathbb{N}$ such that, for any $n$ and any initial position $u(n)$, the probability that at least one of the vectors $u(n+1),\dots,u(n+L)$ is inside $A$ is $>\varepsilon$ (e.g. let $L=n\cdot M$, where $M$ is the maximum distance between $0$ and any point of $F$).

Thus, for each $k$, the probability that no vector $u(n)$, for $n=1,\dots,kL$, is in $A$, is at most $(1-\varepsilon)^k\to0$. So almost surely $u(n)$ is in $A$ for some $n$, implying that our players will almost surely land in $O$.

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