
Last time we solved some probability puzzles involving coin flips. This time we'll look at puzzles involving cards.
Example 1. How many ways are there to order 3 cards: a jack (J), a queen (Q), and a king (K)?
By order them I mean put one on top, then one in the middle, then one on the bottom. There are three choices for the first card: it can be A, Q, or K. That leaves two choices for what the second card can be, and just one for the third. So, there are
ways to order the cards.
Example 2. How many ways are there to order all 52 cards in an ordinary deck?
By the same reasoning, the answer is
This is a huge number. We call it 52 factorial, or 52! for short. I guess the exclamation mark emphasizes how huge this number is. In fact
This is smaller than the number of atoms in the observable universe, which is about \( 10^{80}.\) But it's much bigger than the number of galaxies in the observable universe, which is about \( 10^{11}\), or even the number of stars in the observable universe, which is roughly \( 10^{22}.\) It's impressive that we can hold such a big number in our hand... in the form of possible ways to order a deck of cards!
Definition 1. We say a deck is wellshuffled if each of the possible ways of ordering the cards in the deck has the same probability.
Example 3. If a deck of cards is wellshuffled, what's the probability that it's in this order?
Since all orders have the same probability, and there are \( 52!\) of them, the probability that they're in any particular order is
So, the answer is
Suppose you take the top \( k\) cards from a wellshuffled deck of \( n\) cards. You'll get a subset of cards—though card players call this a hand of cards instead of a subset. And, there are \( n\) choose \( k\) possible hands you could get! Remember from last time:
Definition 2. The binomial coefficient
called \( n\) choose \( k\) is the number of ways of choosing a subset of \( k\) things from a set of \( n\) things.
I guess cardplayers call a set a 'deck', and a subset a 'hand'. But now we can write a cool new formula for \( n\) choose \( k.\) Just multiply the top and bottom of that big fraction by
We get
I won't do it here, but here's something you can prove using stuff I've told you. Suppose you have a wellshuffled deck of \( n\) cards and you draw a hand of \( k\) cards. Then each of these hands is equally probable!
Using this we can solve lots of puzzles.
Example 4. If you draw a hand of 5 cards from a wellshuffled standard deck, what's the probability that you get the 10, jack, queen, king and ace of spades?
Since I'm claiming that all hands are equally probable, we just need to count the number of hands, and take the reciprocal of that.
There are
5card hands drawn from a 52card deck. So, the probability of getting any particular hand is
We can simplify this a bit since 50 is 5 × 10 and 48 is twice 4 × 3 × 2 × 1. So, the probability is
The hand we just saw:
is an example of a 'royal flush'... the best kind of hand in poker!
Definition 3. A straight is a hand of five cards that can be arranged in a consecutive sequence, for example:
Definition 4. A straight flush is a straight whose cards are all of the same suit, for example:
Definition 5. A royal flush is a straight flush where the cards go from 10 to ace, for example:
Example 5. If you draw a 5card hand from a standard deck, what is the probability that it is a royal flush?
We have seen that each 5card hand has probability
There are just 4 royal flushes, one for each suit. So, the probability of getting a royal flush is
Suppose you have a wellshuffled standard deck of 52 cards, and you draw a hand of 5 cards.
Puzzle 1. What is the probability that the hand is a straight flush?
Puzzle 2. What is the probability that the hand is a straight flush but not a royal flush?
Puzzle 3. What is the probability that the hand is a straight?
Puzzle 4. What is the probability that the hand is a straight but not a straight flush?
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