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In general, multiplication of games doesn't respect equality (although it does on the surreals). Indeed there can't be an associative and distributive multiplication on games with $1$ as the unit, because then we would have $0 =0\times\frac{1}{2} = (*+*)\times\frac{1}{2} = *\times(1+1)\times\frac{1}{2} = *$.
The simplest example of a two-sided nonorientable surface is probably the embedding of the Klein bottle $K$ into $K\times\mathbb R$ by $k\mapsto (k,0)$. Then its impossible for the point $(k,r)$ to go from $r>0$ to $r<0$ without passing through $r=0$.
Thanks! I only convinced myself you were right when I thought of a simpler example: the circle is orientable, but if you embed it in a Möbius strip in the obvious way then it's one-sided.
I think you're right. The tesseravore could use the strategy of always eating one of the squares in the angel's movement range diametrically opposed to the origin. My intuition is that this would be enough bias to cause the angel's random walk to always return to the origin. (My intuition isn't strong here though, high dimensional spaces are weird.)
Why did you restrict the devil's strategy to not know the angel's position? I can't see a winning strategy for the devil even if it knows where the angel is.
According to wikipedia, $\Gamma(\frac{1}{2}-m) = \frac{(-4)^mm!}{(2m)!}\sqrt{\pi}$. Substituting this in to the above formula gives $|B^{-(2m+1)}| = \frac{(2m)!}{(-\frac{\pi}{4})^mm!}$.
@JoelDavidHamkins When I say that $\sf{ETCS+R}$ is biinterpretable with $\sf ZFC$, I mean exactly that. (You have to get the language right for it to work out perfectly; morphisms have equality but objects don't.) In the questions at the end I ask for biinterpretability, then mutual interpretability, then equiconsistency, which I believe are in decreasing order of strength. Proving any of those would be interesting, but biinterpretability would be best. (Conversely disproving any of those would be interesting but disproving equiconsistency would be best.)
@PaulTaylor The idea would be to demonstrate that the choice of sets as foundations was arbitrary (or at least motivated only be ease-of-use rather than necessity). Alternatively, if you want to show that the choice of sets wasn't arbitrary then you could prove that $\sf{ETCG+R}$ isn't biinterpretable with $\sf ZFC$.
