Cards

Mathematics related to cards.

Combinatorics and Counting

Probability of 5-Card Hands

Here, we give two examples that illustrate the care that is needed when setting up the problem and deciding how to model the process. In the first procedure, we are looking for five-card hands with at least one particular face value - which we model as a single event. In the second procedure, we are looking for five-card hands with exactly one occurrence of a particular face value, which requires us to think about each card being dealt as separate events.

Note that the total number of 5-card hands is given by $ \binom{52}{5} = 2,598,960 $.

Hand with at least one X

Suppose we wish to know the probability that a given 5-card hand will contain at least once ace.

In this case, it is slightly easier to enumerate the number of hands that do not contain any aces. We re-pose the problem as counting the number of ways we can form 5-card hands from 48 cards (the 52 total minus the 4 aces). This probability is given by:

$ \binom{48}{5} = 1,712,304 $

This gives the total number of 5-card hands that contain at least one ace as the total number of possible hands minus the number of hands without aces:

$ \binom{52}{5} - \binom{48}{5} = 886, 656 $

Now, you may be wondering why we use the binomial distribution, if we are trying to model a process with replacement? Isn't the hypergeometric distribution more appropriate? You would almost be correct - except that we are analyzing a single event.

In other words, in this process, success is not defined as "the next card being dealt comes up an ace", with many trials and many outcomes. Success is defined as "one of five cards, pulled at random from the deck, will be an ace." That means that pulling five cards is a single event, a single independent trial, with a single outcome.

Another way to think about it is, our success criteria applies to all of the cards at once, and is only met or not when all five cards have been dealt.

We will see in the next section that if we are concerned with the actual process of dealing the hand (i.e., we check criteria as the deal is happening), then we must treat each card as a separate event - and then we no longer have independent sampling with replacement, and we cannot use the binomial distribution. Instead, we have sampling without replacement, and we must use the hypergeometric distribution.

Hand with exactly one X

Now suppose we wish to know the probability that a given 5-card hand contains exactly one ace.

Combinatoric Solitaire

Knuth mentions an anecdote about runs/sequences of increasing integers - there was a mathematician/scientist who would play a form of solitaire in which he would place cards on an existing pile as their face value increased, and start a new pile as their face values decreased. The question is, how many piles will result, and playing this form of solitaire is a form of "empirical" sampling.

Related

• Shuffling