Can the chunks be empty? If not, what do you intend for $t=1$ or $2$? I think $\beta$ should be less than $1/3$, otherwise you won't find three disjoint sets of size at least $\beta\ell$ in a set of size $\ell$. Do you know of such a construction for 2 subsets and 2 chunks?

If $N=4$ had a solution, then both pairs of (positive) integers would have the same sum and product. Any two numbers are determined by those values, up to ordering.

Then the minimum is $N=0$.

The "fundamental" condition is not necessary. A minimal solution won't include both $x$ and $-x$.

@Per: It's my understanding that Young tableaux are diagrams filled with anything, and standard Young tableaux are filled with $[n]$ in increasing order along rows and down columns. That may not be universal, and if it's confusing I can change it. The problem doesn't seem to me to be related to the usual representation topics that standard Young tableaux are used for, but if it is that would be good too.

@MaskedAvenger: No problem, thanks for reading it. It turned out to be more complicated to write down than I thought it would be. I'll make some clarifying edits.

@MaskedAvenger: questions asked are distinct, no two students will be asked the same one. So the $n-a$ available questions will decrease, or not, when one is asked, depending on whether a student worked on that question.

The sum of two algebraic numbers a root of the resultant of their minimal polynomials. For more, see math.stackexchange.com/questions/155122/…

Which rotations are the obvious ones? All of them? Or only those $4$ leaving the base on the bottom, as you would pack actual boxes?

Why are you asking for $p$ to be prime?

The sine values for $c=1/10$ and $d=3/10$ differ by $1/2$, so the answer to question 2 is no.

If $n=2m$, then $a_m =m$, since the summands of $\sum^m_1a_jn^{(m-j)}$ are all multiples of $m$ except $a_m$. By the same reasoning, every even position must be an even digit, and hence odd positions are odd, so they are not independent.

You're right, I included the bet in the winnings. How are the cards distributed?

I don't know that this has been studied mathematically by game theorists, but an equivalent scenario comes up often in actual poker, especially limit games. After all cards are dealt, a player can estimate their own chances of winning at showdown against a range of opponents cards, with the perceived range based on previous actions (both actions in the hand, and general playing tendencies).

Are all players equally likely to have A or B? If so, the example game won't be much fun. If the third player faces two callers, then the worst scenario is to have card B while one other player has B and one has A, giving a $40%$ chance to win half the pot by calling, and a $60%$ chance to lose the bet. $(0.2)(33)-(.6)(10)=0.6>0$, so the third player should always call. The second player knows this, so facing one call, she should always call. The first player knows both will call if he does, so he should call. The probability of A winning needs to be $>\frac{3x+3}{5x+3}$ for bet $x$ and ante 1