03-17-2023 , 01:53 PM
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?
03-17-2023 , 06:55 PM
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.

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