Ireland

(a) Denote the configuration of cards after round $k\ge 0$ by a binary sequence $a_1^{(k)},a_2^{(k)},a_3^{(k)},\dots ,a_n^{(k)}$, where the subscripts are interpreted modulo $n$ (here $k=0$ corresponds to the initial cards dealt to the players). The outcome of each round may then be computed recursively by $$a_i^{(k+1)}=a_i^{(k)}+a_{i-1}^{(k)}+a_{i-2}^{(k)}\forall i=1,2,\dots ,n(1)$$ for $k\ge 0$, where addition is modulo 2. Next, by induction on $r$ we show that for all $r\ge 0$ and all $k\ge 0$, $$a_i^{(k+2r)}=a_i^{(k)}+a_{i-2r}^{(k)}+a_{i-2^{r+1}}^{(k)}\forall i=1,2,\dots ,n.$$ This is obviously true for $r=0$, as it matches (1). Assuming this equation holds for a fixed $r\ge 0$, we have, for all $i=1,2,\dots ,n$ and all $m\ge 0$, $$a_i^{(m+2r)}=a_i^{(m)}+a_{i-2r}^{(m)}+a_{i-2^{r+1}}^{(m)}.$$ With $m=k+2^r$ it follows that for all $i=1,2,\dots ,n$ and all $k\ge 0$, $$\begin{array}{rl}{a}_i^{(k+2^{r+1})}& =a_i^{(k+2^r+2^r)}=a_i^{(k+2^r)}+a_{i-2^r}^{(k+2^r)}+a_{i-2^{r+1}}^{(k+2^r)}\\ & =(a_i^{(k)}+a_{i-2^r}^{(k)}+a_{i-2^{r+1}}^{(k)})\\ & +(a_{i-2^r}^{(k)}+a_{i-2^{r+1}}^{(k)}+a_{i-2^{r-2^{r+1}}}^{(k)})\\ & +(a_{i-2^{r+1}}^{(k)}+a_{i-2^{r-2^{r+1}}}^{(k)}+a_{i-2^{r+2}}^{(k)})\\ & =a_i^{(k)}+a_{i-2^{r+1}}^{(k)}+a_{i-2^{r+2}}^{(k)}\end{array}$$ and the results follows by the principle of induction. Therefore, if $n=2^r$ for any $r\ge 2$, after $n$ rounds we have $a_i^{(n)}=a_i^{(0)}+a_{i-n}^{(0)}$ and $a_{i-2n}^{(0)}=a_i^{(0)}$ for all $i$; we conclude that each such $n$ is playable.

(b) For any $n$ divisible by 3, we may set the initial configuration of cards to $$a_i^{(0)}=\{\begin{array}{ll}1& \text{if }i\equiv 0(\mathrm{mod}3)\\ 0& \text{otherwise.}\end{array}$$ Then it is easily shown that the next round yields the all-ones sequence, and this configuration will not change thereafter, i.e. $a_i^{(k)}=1$ for all $i=1,2,\dots ,n$ and for all $k\ge 1$. Therefore every $n$ divisible by 3 is not playable.