Of course, there's a sort of dual way to generalize the $n=2$ case: we take hyperplanes passing through $v$ and tangent to the ball, and ask if as few as $n$ such hyperplanes can contain the ball in the intersection of their halfspaces. In other words, the "front-on" view of the two approaches looks like en.wikipedia.org/wiki/Method_of_exhaustion

Yes, it turns out the statement is trivial, but the "nontrivial intuition" the question is trying to capture is interesting, and addressed nicely in Sam Hopkins' comment. Namely, if we had an oracle for BB(n), we could automate the proof/disproof of any well-defined claim.

Can one show that the $y_1,\dots,y_n$ are negatively associated random variables? Then since the marginals are the same, it would imply that any Chernoff-type bound that applies to the x's also applies to the y's.

Another question is if we can prove monotonicity - if it is optimal to bet with $h$ heads and $t$ tails, is it optimal to bet with $h' \geq h$ heads and $t' \leq t$ tails?

Can we get anything by writing down the optimal strategy via backward induction? E.g. we can define $f(h,t)$ to be the EV of betting at the current round $n = h+t$ given there have been $h$ heads and $t$ tails, followed by playing optimally thereafter; and $g(h,t)$ to be the EV of not betting and playing optimally thereafter; and $V(h,t) = \max\{f(h,t), g(h,t)\}$, and we can write $f(h,t)$ and $g(h,t)$ in terms of $V(h+1,t)$ and $V(h,t+1)$. Etc.

For the record, I agree with Will that this proof is not correct. And the reason is roughly that one should consider betting even when this round's EV is negative, if the EV of the next round conditioned on a heads this round is positive. For example, if $p$ were uniformly either $0$ or $0.9$, this answer would say to wait to bet until you see a head, but betting immediately is much better, because in the world where $p=0.9$, you usually have a 2x or 4x head start on an approaching-infinitely-large EV, which is worth giving up an additive $1$ in the world where $p=0$.

Anyway, there is the voting button.

I think we should be slow to confidently dismiss questions outside our areas of familiarity, even when there are clear problems with the question.

@TimothyChow It may also help to mention that to prior-me, and presumably other untrained people, intuitively there can appear to be a difference between "BB(n) has a well-defined value in ZFC" and "ZFC proves BB(n) = m for some m".

@TimothyChow to me, that appears to confuse two notions of determinable, i.e. of course BB(n) is not computable by reduction from halting, but it's much less obvious that BB(n) could be independent of ZFC.

What is an example of a point in $S \setminus T$?

@WillSawin speaking for myself, my intuition that BB(n) is determinable would be that every Turing Machine either halts or does not, and there is a finite set of Turing Machines of size n. (See my previous comment for how I learned the flaw in this reasoning.)

"any BB number seems determinable" - Noah Schweber gave me a an illuminating comment: there are Turing Machines that "do not halt", except that in some models of ZFC, they "do halt" in a technically finite but nonstandard number of steps. So BB(n) can be independent of ZFC.

Though I wondered if the way antes or blinds work in poker would change the conclusion.

@Eilon I now see the question is well-posed and agree with the answer. "Chip EV" just refers to each round as resulting in a number of chips for each player, which is zero-sum and in [-k,k] where k is the smaller of the two players' starting chips in the round. Since by the last paragraph, each round is a symmetric game (with the "number of chips" utility function), each player has a strategy guaranteeing E[chips] >= 0.

I don't immediately see how to apply the data processing inequality here, but it sure feels similar. jkds, can you say if something goes wrong with a direct proof (plugging in for $q_1,q_2$)?