Note that your approach only works for target N>3 (that is q>2), as p2+(1−p)2=1/3 does not have a real solution. Open question: How about 1-coin solution emulating a 3-sided dice?
See my solution (one long comment for all 3 puzzles) for some thoughts on rational coins—there are some constructions, but more importantly I suspect that no finite set of rational coins without the 1/2-coin can generate any fair dice. (My lemma does not cover all rational coins yet, though; see by the end of the comment. I would be curious to see it turn out either way!)
Update: I believe the solution by @UnexpectedValues can be adapted to work for all natural numbers.
Proof: By Dirichlet prime number theorem, for every N, there is a prime q of the form q=(N−1)+kN as long as N-1 and N are co-prime, which is true whenever N>2. Then N|q+1and the solution by UnexpectedValues can be used with appropriate p. Such p exists whenever q>2which is always the case for N>3. For N=3, one can take e.g. q=5 (since 3|5+1). And N≤2 is trivial.
A beautiful solution! Some remarks and pointers
Note that your approach only works for target N>3 (that is q>2), as p2+(1−p)2=1/3 does not have a real solution. Open question: How about 1-coin solution emulating a 3-sided dice?
See my solution (one long comment for all 3 puzzles) for some thoughts on rational coins—there are some constructions, but more importantly I suspect that no finite set of rational coins without the 1/2-coin can generate any fair dice. (My lemma does not cover all rational coins yet, though; see by the end of the comment. I would be curious to see it turn out either way!)
Update: I believe the solution by @UnexpectedValues can be adapted to work for all natural numbers.
Proof: By Dirichlet prime number theorem, for every N, there is a prime q of the form q=(N−1)+kN as long as N-1 and N are co-prime, which is true whenever N>2. Then N|q+1and the solution by UnexpectedValues can be used with appropriate p. Such p exists whenever q>2which is always the case for N>3. For N=3, one can take e.g. q=5 (since 3|5+1). And N≤2 is trivial.