Question

I draw five cards from a randomly shuffled deck. What is the probability that those five cards are in either ascending or descending order

Answer 1

Answer: The probability of drawing 5 cards in either ascending or descending order out of the deck of 52 standard playing cards is 0.84%.

Step-by-step explanation: We have the following cards in a deck: 1(ace), 2, 3, 4, 5, 6, 7, 8, 9, 10, 12(Jack), 13(Queen), 14(King) (thirteen different in total). There are 4 copies of each of them yielding 52 cards in total. First to draw all of the cards in the ascending order all of the drawn cards need to be different. Imagine you have 13 piles of 4 identical cards. Let us calculate the number of ways you can select 5 different ones (the order matters). First you select 5 different piles (this secures that each card is different) and this is possible to do in $\frac{13!}{(13-5)!}$ (we use the formula for variations since the order matters). Now, each card in the pile can be selected in $4$ different ways so the total number of sequences with $5$ different cards is $4^5\frac{13!}{(13-5)!}$. Now we select out of these sequences the clases of those that contain exactly the same cards but in different order. Only 4 of the sequences within the same class will be in ascending order out of 4*5! which is the total number of the sequences within the class! This means that we have to multiply our total number of sequences of 5 different cards by $\frac{4}{4\cdot 5!}=\frac{1}{5!}$ and this yields the final answer of total number of ascending sequences to be

$4^5\frac{13!}{(13-5)!5!}.$  

The total number of possible ways to draw 5 out of 52 cards is just

$\frac{52!}{(52-5)!}.$

This yields for the probability

$\frac{4^5\frac{13!}{(13-5)!5!}}{\frac{52!}{(52-5)!}}=0.42\%$

Exactly the same calculation applies for the descending order. So the probabilty of the cards being in either ascending or descending order is just the sum of these two (the events are mutually exclusive, you cannot have both the ascending and descending order at the same time) yielding the final probability of $0.84\%$.