Theoretical Poker Math/Probability Question

This is purely a theoretical question. Please excuse the long-windedness of this post.


In poker, the more improbable hand is ranked higher than those more probable. For example, it is less likely to be dealt a four of a kind than a full house in 5 card stud. So, full houses lose to a four of a kind.


Now, my question. In N-card stud, how are the probabilities of the hands affected, and at what values of N are the probability crossovers and for what hand ranks? I am assuming that straights and flushes are still based on 5 cards, and you must play your best 5 card hand.


For example, at N = 2, a pair is less likely than nothing, and so the hand ranks still hold. A Royal Flush is better than a four of a kind... a pair is better than a high card. (Note that even though Royal Flush all the way down to three of a kind have equal likelihood at N = 2 (Zero probability), we are consistent, and no crossover has taken place).


At N = 11, a high card is MUCH less likely than a pair. The only way this can happen is if you have A234 6789 JQK. Any other 11 cards give you at least a pair or higher.


So the question is at which values of N are the crossovers, and how does the probability-based hand rankings change as N approaches 52. At N = 52, we are once again consistent with our 5-card stud probability-based rankings system with Royal Flush having probability = 1, and everything else equal to zero (recall that we are ALWAYS going to choose the best 5 card hand according to the traditional poker rankings). This is consistent, since we are only concerned with crossovers, and there is no crossover at N = 52 compared to our usual N = 5 case.


Any ideas?