From Brett Humphreys comes a card-counting conundrum:
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 — that is, each participant is dealt five cards that they can’t trade.
As the number of suits increases, the probability of each hand changes. With four suits, a straight is more likely than a full house (a three-of-a-kind and a different two-of-a-kind in the same hand). 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. You are dealt seven cards, among which you pick the best five-card hand. How many suits would the deck need so that a straight (not including a straight flush) is less likely than a full house?
I calculated the following hands for 3-30 suits.
Choose five cards from the deck.
$${13i \choose 5}$$One rank can be repeated with two different suits. Choose three ranks from the remaining twelve, and they can be of any suit.
$$\dfrac{13 \cdot {i \choose 2} \cdot {12 \choose 3} \cdot i^3}{13i \choose 5}$$Two ranks can be repeated with two different suits. Choose one ranks from the remaining eleven, and it can be of any suit.
$$\dfrac{{13 \choose 2} \cdot {i \choose 2}^2 \cdot 11i}{13i \choose 5}$$One rank can be repeated with three different suits. Choose two ranks from the remaining twelve, and they can be of any suit.
$$\dfrac{13 \cdot {i \choose 3} \cdot {12 \choose 2} \cdot i^2}{13i \choose 5}$$There are ten straights, starting with A-10. Any of the cards can be any of the suits. There are nine straight flushes of each suit, and a royal flush of each suit.
$$\dfrac{10i^5 - 10i}{13i \choose 5}$$Choose 5 cards of the same suit. Subtract the nine straight flushes of each suit, and a royal flush of each suit.
$$\dfrac{i\cdot{13 \choose 5} - 10i}{13i \choose 5}$$Choose 3 cards of a rank of any suit. Choose 2 cards of another rank of any suit.
$$\dfrac{13{i \choose 3} \cdot 12{i \choose 2}}{13i \choose 5}$$One rank can be repeated with four different suits. Choose one rank from the remaining twelve, and it can be of any suit.
$$\dfrac{13{i \choose 4} \cdot 12{i \choose 1}}{13i \choose 5}$$There are nine straight flushes of each suit, starting with A-9.
$$\dfrac{9i}{13i \choose 5}$$There is a royal flushes of each suit.
$$\dfrac{i}{13i \choose 5}$$One rank can be repeated with five different suits.
$$\dfrac{13{i \choose 5}}{13i \choose 5}$$Anything left over.
