Brett plays poker with a large group of friends. With so many friends playing at the same time, Brett needs more than the 52 cards in a standard deck. This got Brett and his friends wondering about a deck with more than four suits.

Suppose you have a deck with more than four suits, but still 13 cards per suit. And further suppose that you’re playing a game of five-card stud.

As the number of suits increases, the probability of each hand changes. With four suits, a straight is more likely than a full house. How many suits would the deck need so that a straight (not including a straight flush) is less likely than a full house?

Extra credit: Instead of five-card stud, suppose you’re playing seven-card stud. How many suits would the deck need so that a straight (not including a straight flush) is less likely than a full house?

I stole this from FiveThirtyEight.

I know we (and by "we" I mean someone other than me) could just brute-force it and count up the combos or run a simulation, but are there better ways to do this?

We can express each hand frequency in terms of s, the number of suits. Their probabilities have the same denominator, so we can just compare the number of combos.

N(straights) = 10(s^5 - s)
N(boats) = 13*12*C(s,2)*C(s,3) = 13*12[s(s-1)/2][s(s-1)(s-2)/6] = 13(s-2)(s^2)(s-1)^2
So we need s such that 13(s^5 - 4s^4 + 5s^3 - 2s^2) > 10(s^5 - s)
which means 3s^5 - 52s^4 + 65s^3 - 26s^2 + 10s > 0
Wolfram Alpha says s>16.0133, so we need at least 17 suits in 5-card stud.

Repeating the procedure for 7-card wouldn't be much harder.