Originally Posted by Steve922
I'm fine with the maths but can't see how those formulae are derived.
It's been a long time since my probability classes as well, but technically, I believed I used permutations (52P5) instead of combinations (52C5). Permutations consider all the different orderings and combinations do not. For instance, 52P5 = 52! / (52-5)! where 52C5 = 52P5 / 5!. Also, the great thing about probability is that there are usually many ways to work it all out. Here's how I did it:
First, consider the total number of ways to select exactly four cards of any single suit. For a single suit, there are 13P4 ways to select and order 4 cards of that 13 card suit. Since there are 4 suits, we multiply that by 4 to consider all suits. So there are 4*13P4 ways to select and order 4 cards of any single suit. For the remaining single card, it can be 52-13 = 39 different possible cards and that card can be placed into any of 5 positions. So there are 39*5 ways to select and order that remaining card. Therefore, there are 4*13P4*39*5 total ways to select and order the cards.
Second, to arrive at a probability, we just divide that total by the total number of ways to select and order 5 cards from a total of 52, which is 52P5.
So, the probability of a board with exactly 4 cards of a single suit is:
4*13P4*39*5 / 52P5 =
4*13*12*11*10*39*5 / 52*51*50*49*48 =
13384800 / 311875200 =
0.0429 = 4.29%
Note that this is the same answer that others have gotten using different methods, so that is definitely a plus!
BTW, I'm glad you asked the question because it forced me to refresh all that 52P5 and 52C5 notation for myself as well!