How Likely Are You to Be Dealt a Royal Flush?

In the game of poker, you have a royal flush when you have a 10, J, Q, K, and A all of the same suit. Assuming a uniform random distribution of cards in the deck, what is the probability of being dealt a royal flush?

A royal flush consists of a certain five cards of the same suit. Assume we are dealt one of these cards. Then the suit is determined, and we now need the remaining four cards of that suit. So, let's take the probability of being dealt one of the certain five cards (10, J, Q, K, A) of any suit. Since there are five such cards per suit, and there are four suits (the set of suits is \( \{ \diamondsuit, \heartsuit, \clubsuit, \spadesuit \} \)), then we can get any one of those cards to start building a royal flush. Therefore, the probability of getting a starting card is \[ P(\text{starting card}) = \frac{20}{52}. \] Once we have a starting card of a certain suit, our suit is restricted. That is, we may only get the remaining four cards from the same suit as our starting card. Then the probability of getting one of the four cards from the same suit is \[ P(\text{one of the four from the same suit}) = \frac{4}{51}. \] Likewise, after we get the second card, there are three left to get, so the probabilities are \( 3/50 \), then \( 2/49 \), and then \( 1/48 \). What, then, is the probability that we will get all of the necessary cards? It is the product of all these probabilities: \[ P(\text{all}) = \frac{20}{52} \cdot \frac{4}{51} \cdot \frac{3}{50} \cdot \frac{2}{49} \cdot \frac{1}{48} = \frac{1}{649\,740} = 1.539\ldots \times 10^{-6} = 0.000\,153\,9\ldots \%. \] Not very good odds.

Calculating the Number of Trailing Zeros in a Factorial

The factorial function, denoted \( n! \), where \( n \in \mathbb{Z}^* \), is defined as follows: \[ n! = \begin{cases} 1 & \text{if $n = 0$,} \\ n(n - 1)! & \text{otherwise}. \end{cases} \] In the case of \( n = 10 \), \( 10! = 3\,628\,800 \). Note that that value has two trailing zeros. When \( n = 20 \), \( 20! = 2\,432\,902\,008\,176\,640\,000 \) and has four trailing zeros. Can we devise an algorithm to determine how many trailing zeroes there are in \( n! \) without calculating \( n! \)?

Let us define a function \( z : \mathbb{N} \to \mathbb{N} \) that accepts an argument \( n \) that yields the number of trailing zeros in \( n! \). So, using our two examples above, we know that \( z(10) = 2 \) and \( z(20) = 4 \). Now, what is \( z(100) \)?

An integer \( a \) has \( k \) trailing zeros iff \( a = b \cdot 10^k \) for some integer \( b \). Now, the prime factors of \( 10 \) are \( 2 \) and \( 5 \). So, what if we consider the prime factors of each factor in a factorial product? For example, consider the first six terms (after \( 1 \)) of \( 100! \) and their prime factors. We omit \( 1 \) because of its idempotence: \[ \begin{align} 2 &= 2 \\ 3 &= 3 \\ 4 &= 2^2 \\ 5 &= 5 \\ 6 &= 2 \cdot 3 \\ 7 &= 7 \end{align} \] We see that \( 7! \) contains one \( 5 \) and (more than) one \( 2 \). Therefore, \( 7! \) contains a factor of \( 10 \), so we should expect there to be one trailing zero in \( 7! \). And we see that \( 7! = 5040 \), which does indeed have one trailing zero. This suggests that we could, for any integer \( n \) in \( n! \), cycle through the integers from \( 2 \) to \( n \), finding the prime factorization of each integer, and counting pairs of \( 2 \) and \( 5 \) that we find.

You might have noticed from our example of \( 7! \) that there are four \( 2 \)s and only one \( 5 \). Consider that there are \( n/2 \) even factors in \( n! \)—that is, every other integer factor of \( n! \), starting at \(2\), is even. So there are \( n/2 \) even factors. Each of these has at least one \( 2 \) in its prime factorization, and many of them have more than one. Now consider how many factors of \( 5 \) there are in \( n! \): there are, in fact, \( n / 5 \). So, we can see that, for every \( 5 \) we find in \( n! \), there is a \( 2 \) we can match with it. This suggests that we only need count the \( 5 \)s and not the pairs of \( 2 \) and \( 5 \).

This now suggests an algorithm, presented here in Java:

Using this algorithm, we find that \( z(100) = 24 \).

The Beauty of Scheme

Bertrand Russell once noted that

Mathematics, rightly viewed, possesses not only truth, but supreme beauty—a beauty cold and austere, like that of sculpture, without appeal to any part of our weaker nature, without the gorgeous trappings of painting or music, yet sublimely pure, and capable of a stern perfection such as only the greatest art can show. The true spirit of delight, the exaltation, the sense of being more than Man, which is the touchstone of the highest excellence, is to be found in mathematics as surely as poetry.

I think much the same can be said of code, especially code written in certain languages. There is something so elegantly simple and beautiful about Scheme (and Racket), for example, that shares the same aesthetics of mathematics.