“No two rainbows are the same. Neither are two packs of Skittles. Enjoy an odd mix.” – Skittles label

Analyzing packs of Skittles (or sometimes M&Ms) seems to be a very common exercise in introductory statistics. How many candies are in each pack? How many candies of each individual color? Are the colors uniformly distributed?

The motivation for this post is to ask some questions raised by the claim in the above quote:

*How many*different possible packs of Skittles are there? Here we consider two packs of Skittles to be distinguishable only by the*number*of candies of each color.- What is the
*probability*that two randomly purchased packs of Skittles are the same? - What is the
*expected number*of packs that must be purchased until first encountering a duplicate?

**Number of possible packs**

If there are candies in a 2.17-ounce pack, and each candy is one of colors, then the number of possible distinguishable packs is

Interestingly, a Skittles commercial from the 1990s suggests that there are 371,292 different possible packs (about 27 seconds into the video). It’s not clear whether this number is based on any mathematics or just marketing… but it’s actually reasonably close to the value 367,290 that we get with the above formula assuming that there are candies in a pack.

However, the total number of candies varies from pack to pack. Most studies suggest an average of about 60 candies per pack– with 635,376 possible packs of *exactly* that size– and this study includes a couple of outlier packs with as few as 42 and as many as 85 candies. We can sum the binomial coefficients over this range of possible pack sizes, to get 42,578,514 possible distinguishable packs… but fortunately, we can more conservatively allow for all pack sizes from *empty* up to, say, 100 candies, and still remain within an order of magnitude, and confidently assert that there are at most 100 million different possible packs of Skittles.

Mars Wrigley Confectionery advertises 200 million individual candies produced every day; if we make the extremely conservative assumption that these are distributed among equal *numbers* of packs of various sizes (i.e., for very 2.17-ounce pack, there is, for example, a corresponding 3.4-*pound* party pack), so that the 2.17-ounce packs make up only about 1.6% of total production, this works out to approximately 50,000 packs per day. By the pigeonhole principle, after only about 1800 days of production, there *must* be two identical packs of Skittles out there somewhere… but identical packs are likely much more common than this worst-case analysis suggests, as we’ll see shortly.

**Probability of identical packs**

The second question requires a bit more work. Let’s temporarily assume that there are always exactly candies in every pack, so that there are , or 635,376 possible packs. We might naively assume that the probability that two randomly purchased packs are identical is 1/635376. But this is only true if every possible pack is equally likely, e.g., a pack of 60 all-red candies is just as likely as a pack with 12 candies of each color. I can’t envision a method of production and packaging that would yield this uniform distribution.

Instead, let’s assume that each *individual* candy’s color is *independently* and identically distributed, with each of the colors equally likely. For example, imagine an equally large number of candies of each color mixed together in one giant urn (I’m a mathematician, so I have to call it an urn), and dispensed roughly 60 at a time into each pack.

This tracks very well with the actual observed distribution of colors of candies within and across packs. Some packs have more yellows, some have more reds, etc., but there is almost never a color missing, and almost never exactly the same number of every color… but with more and more observed packs, the overall distribution of colors is indeed very *nearly* uniform. In other words, although it can be unintuitive, giving the impression of “designed” non-uniformity of the distribution of colors, we should *expect* variability using this uniform model, even for a seemingly “large” sample from, say, a party-size bag of over a thousand candies.

So, given two randomly purchased packs each with candies of equally likely colors, what is the probability that they are identical? We can compute this probability exactly as the coefficient of an appropriate generating function:

For completeness, following is example C++ source code for computing these probabilities using arbitrary-precision rational arithmetic:

#include "math_Rational.h" #include <map> #include <iostream> using namespace math; // Map powers of x to coefficients. typedef std::map<int, Rational> Poly; // Return f(x)*g(x) mod x^m. Poly product_mod(Poly f, Poly g, int m) { Poly h; for (int k = 0; k < m; ++k) { for (int j = 0; j <= k; ++j) { h[k] += f[j] * g[k - j]; } } return h; } Rational p(int n, int d) { // Compute g(x) = 1 + ... + (x^n/n!)^2. Poly g; g[0] = 1; Rational factorial = 1, d_2n = 1; for (int k = 1; k <= n; ++k) { factorial *= k; d_2n *= (d * d); g[2 * k] = 1 / (factorial * factorial); } // Compute f(x) = g(x)^d mod x^(2n+1). Poly f; f[0] = 1; for (int k = 0; k < d; ++k) { f = product_mod(f, g, 2 * n + 1); } // Return [x^(2n)]f(x) * (n!)^2 / d^(2n). return f[2 * n] * (factorial * factorial) / d_2n; } int main() { std::cout << p(60, 5) << std::endl; }

This yields, for example, equals approximately 1/10254. In other words, buy two 60-candy packs, and observe whether they are identical. Then buy another pair of 60-candy packs, etc., until you find an identical pair. You should expect to buy about 10,254 pairs of packs on average.

(*Aside*: This is effectively the same problem as discussed here a couple of years ago.)

However, we must again account for the variability in total number of candies per pack. That is, is the probability that two packs are identical, *conditioned* on them both containing exactly candies. In question (1), we only needed the maximum *range* of possible values of , but here we need the actual probability density , so that we can integrate over all possible .

Fortunately, the variance of the approximately normally distributed pack size (we can be reasonably confident in the mean of 60) doesn’t change the answer much: the probability that two randomly purchased packs– of possibly different sizes– will be identical is somewhere between about 1/100,000 and 1/200,000.

**Expected number of packs until first duplicate**

Finally, question (3) is essentially a messy birthday problem: instead of insisting that a *particular* pair of Skittles packs be identical, how many packs must we buy on average for *some* pair to be identical? The numbers from question (2) in the hundred thousands may seem large, but a back-of-the-envelope square-root estimate of “only” about 400-500 packs to find a duplicate was, to me, surprisingly small. This is confirmed by simulation: let’s assume that every pack of Skittles independently contains a (100,0.6)-binomially distributed number of candies– for a mean of 60 candies per pack– with each individual candy’s color independently uniformly distributed. We repeatedly buy packs until we first encounter one that is identical to one we have previously purchased. The figure below shows the distribution of the number of packs we need to buy, based on one million Monte Carlo simulations, where the mean of 524 packs is shown in red.

A box of 36 packs of Skittles costs about 21 dollars… so an experiment to search for two identical packs should only cost about $300… on average.

]]>“This is supposed to be a random system. It doesn’t feel very random.” – Eric Reid, as quoted on Twitter

Last week, Eric Reid was selected to take his fifth random drug test in eight weeks with the Carolina Panthers. That might seem like a lot. It might even seem like Reid is perhaps the target of extra scrutiny by the NFL, particularly given his social activism, viewed as controversial by some, and his involvement in a current collusion lawsuit against the league.

So is the drug testing really random, or is Reid justified in his complaint? From the NFL Players Association Policy on Performance-Enhancing Substances:

“Each week during the preseason and regular season, ten (10) Players on every Club will be tested. By means of a computer program, the Independent Administrator will randomly select the Players to be tested from the Club’s active roster, practice squad list, and reserve list who are not otherwise subject to ongoing reasonable cause testing for performance-enhancing substances.”

As will be shown shortly, this is a pretty straightforward example of our very human habit of perceiving patterns where only randomness exists. But there is an interesting mathematical problem buried here as well, challenging enough that I can only provide an approximate solution.

Let’s make the setup more precise: suppose that for each of weeks we select, with replacement, a random subset of of players on a team to take a drug test. (Reid only signed with the Panthers eight weeks ago, and I am assuming that the 72 players comprise 53 active, 10 practice, and 9 reserve.) There are three reasonable questions to ask:

- What is the probability that a
*particular*player (e.g., Eric Reid) will be selected for testing or more times over this time period? - What is the probability that
*at least one*player on the team (i.e., not necessarily Reid) will be selected for testing or more times? - What is the probability that at least one player
*in the 32-team league*will be selected for testing or more times?

**Question 1: P(Eric Reid is selected 5 or more times)**

The first question is easy to answer, and is unfortunately the only question asked in most of the popular press. The probability of being selected in any single week is , and so the probability of being selected at least times in weeks is

which equals approximately 0.002, or only slightly more than one chance in 500.

But if there is a moral to this story, it’s this: *You* will almost certainly not win the lottery… but almost certainly *someone* will. That is, the second (or really the third) question is the right one to ask: what is the probability that *some* player will be selected so many times?

**Question 2: P(some Carolina player is selected 5 or more times)**

This is the hard problem that motivated this post. There is some similarity to the “Double Dixie Cup” version of the coupon collector problem with group drawings, where the players are coupons, but instead of requiring at least copies of *each* coupon in drawings, here we ask for at least copies of *at least one* coupon (or the complementary equivalent, for *at most* copies of each coupon).

If we define the generating function

and to be the expansion of with all terms removed where the sum of exponents is at least , then the desired probability may be expressed as

where the denominator is simply . But unfortunately I don’t see a computationally feasible way to evaluate the coefficient in the numerator. Fortunately, we can get a good lower bound on using inclusion-exclusion and Bonferroni’s inequality:

yielding a probability that some Carolina player would be selected 5 or more times over 8 weeks.

**Question 3: P(some NFL player is selected 5 or more times)**

Finally, while this is happening, the other 31 teams in the league are subjecting their players to the same random drug testing procedure. The probability that *some* player on *some* team will experience 5 or more random drug tests over a span of 8 weeks is

In other words, it is a near certainty that some player in the league would experience the number of random tests that Reid has. Indeed, by linearity of expectation, we should expect an average of players to find themselves in a similar situation over a similar time period. How many players actually did experience multiple tests over the last couple of months would be interesting and useful data to add to the discussion.

]]>The *New York Times* publishes “mini-crosswords,” which are crossword puzzles on a relatively small grid (usually 5×5), without any black squares, so that every row and column of the grid must spell a word. The figure below shows an example of a solution to such a puzzle.

How hard is it to create these puzzles? This post is motivated by a recent *College Mathematics Journal* article (see Reference (1) below) that considers this question, and describes an approach using the Metropolis-Hastings algorithm to randomly sample instances of puzzles.

But instead of just randomly sampling one puzzle at a time, can we actually *enumerate* all possible puzzles? In particular, my idea was to reduce the problem of finding a crossword puzzle solution to that of finding a (generalized) exact cover with appropriately crafted constraints. This would be handy, because we already have code for solving exact cover problems, using Knuth’s Dancing Links (DLX) algorithm (see here and here for similar past exercises).

**Crossword as an exact cover**

To state the problem more precisely: given a positive integer indicating the size of the grid ( in the above example), and a dictionary of words each of length over alphabet , we must construct an exact cover problem whose solution corresponds to a placement of letters in all grid positions such that each row and column spells a word in .

To do this, we construct a zero-one matrix with rows, each corresponding to placing one of the dictionary words either “across” (in one of the rows of the puzzle) or “down” (in one of the columns). A solution will consist of a subset of rows of the matrix: “across” words, one in each row of the puzzle, and “down” words, one in each column, each pair of which intersect in the appropriate common letter.

To represent the constraints, we initially need columns in our matrix (where ), each indexed by a tuple (puzzle row , puzzle column , alphabet letter , across or down). For a given matrix row– representing placement of a word with letters in a particular location and orientation in the puzzle grid– we set to one those columns corresponding to the different grid locations where the word will be placed in the puzzle… where for each alphabet letter , we set the “across” column to one if and only if either

- the word is “across” and matches the letter of the word in this location (i.e., ), or
- the word is “down” and does
*not*match the letter of the word in this location (i.e., ).

If neither of these conditions is satisfied, we instead set the “down” column to one. The following Python code produces the number and list of (row, column) pairs of the resulting sparse matrix.

letters = 'abcdefghijklmnopqrstuvwxyz' m = len(words) n = len(words[0]) print(m * n * 2 * (n * 26), file=file) b = [True, False] for row, ((w, word), k, across) in enumerate( product(enumerate(words), range(n), b)): for col, (i, j, letter, horiz) in enumerate( product(range(n), range(n), letters, b)): if ((i if across else j) == k and (word[j if across else i] == letter) == (across == horiz)): print(row, col, file=file)

However, we’re not quite finished. Although the desired end result is a crossword with *distinct* words in each row and column, such as the 6×6 solution shown in (a) below, as Howard observes in the *CMJ* article, there are many valid solutions that use the same word more than once, including the extreme cases of *symmetric* “word squares” such as the one shown in (b) below.

We can eliminate this duplication by adding “optional” columns to the zero-one matrix, one for each word in the dictionary, and solve the resulting *generalized* exact cover problem, so that each word may be used at most once in a solution.

**Results**

All of the source code is available here, as well as on GitHub. My initial test used the 4×4 case discussed in Howard’s paper, with his dictionary of 1826 words. He describes a process for estimating the total number of possible puzzles by repeated sampling using the Markov chain Monte Carlo approach: “We estimate that there are approximately

73,000–74,000 distinct puzzles each with no repeated words.” This is pretty accurate; it turns out that there are exactly 74,339 (each contributing two symmetric pairs of solutions to the generalized exact cover problem, for a total of 148,678 solutions).

**References:**

- Howard, C. Douglas, It’s Puzzling,
*College Mathematics Journal*,**49**(4) September 2018, p. 242-249 [DOI] - Knuth, D., Dancing Links,
*Millenial Perspectives in Computer Science*, 2000, p. 187-214 (arXiv)

]]>

The following MATLAB code implements this formula, and measures the time required to evaluate it:

tic; scale = 1; y = 0; for i = 1:10000000 x = scale; for j = 1:50 x = x * 0.5; end y = y + x; end y = y / scale; toc

Now suppose that we execute this same code again, but this time changing the “scale” factor to a much smaller value: `scale = realmin`

, corresponding to , the smallest positive normalized floating-point number. Inspection of the formula above suggests that the value of should not depend on the changed value of (as long as it is non-zero); we may spend most of our time working with much smaller numbers, but the end result should be the same.

And indeed, execution of the modified code confirms that we get exactly the same result… but it takes nearly 20 times longer to do so on my laptop than the original version with . And there are more complex– and less contrived– calculations where the difference in performance is even greater.

The problem is that these smaller numbers are *subnormal*, small enough in magnitude to require a slightly different encoding than “normal” numbers which make up most of the range of representable floating-point numbers. Arithmetic operations can be significantly more expensive when required to recognize and manipulate this “special case” encoding.

Depending on your application, there may be several approaches to handling this problem:

- Rewrite your code to prevent encountering subnormals in the first place. In the above contrived example, this is easy to do: just shift the “scale,” or magnitude, of all values involved in the computation away from the subnormal range (and possibly shifting back only at the end if necessary). This can not only result in faster code, but more accurate results, since subnormal numbers have fewer “significant” mantissa bits in their representation.
- Disable subnormal numbers altogether, so that for any floating-point operation, input arguments or output results that would otherwise be treated as subnormal are instead “flushed” to zero.

We have seen above how to manage (1). For (2), the following MEX function does nothing but set the appropriate processor flags to disable subnormals. I have only tested this on Windows 7 with an Intel laptop, compiling in MATLAB R2017b with both Microsoft Visual Studio 2015 as well as the MinGW-w64 MATLAB Add-On (*edit*: a reader has also tried this on Linux Mint 19 with MATLAB R2018b and GCC 7.3.0):

#include <xmmintrin.h> #include <pmmintrin.h> #include "mex.h" void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[]) { //_MM_SET_FLUSH_ZERO_MODE(_MM_FLUSH_ZERO_ON); _mm_setcsr((_mm_getcsr() & ~0x8000) | (0x8000)); //_MM_SET_DENORMALS_ZERO_MODE(_MM_DENORMALS_ZERO_ON); _mm_setcsr((_mm_getcsr() & ~0x0040) | (0x0040)); }

After running this MEX function in a MATLAB session, re-running the modified calculation above gets all of the speed back… but now at a different cost: instead of the correct value , every term in the sum has been flushed to zero, resulting in a final *incorrect* value of .

If there is any moral to this story, it’s that you’re a test pilot. First, this was a very simple test setup; it’s an interesting question whether any MATLAB built-in functions might reset these flags *back* to the slower subnormal support, and whether it is feasible in your application to reset them back again, possibly repeatedly as needed. And second, even after any algorithm refactoring to minimize the introduction of subnormals, can your application afford the loss of accuracy resulting from flushing them to zero? MathWorks’ Cleve Moler seems to think the answer is always yes. I think the right answer is, it depends.

]]>

Constraining the paths to minimum length greatly simplifies the problem. So let’s generalize, and instead ask for the number of paths from to of length — so that the original problem asks for the particular case , but what if we allow longer paths where we sometimes move in the “wrong” direction away from the destination?

I think this is a nice problem, with an elegant solution only slightly more complex than the original posed in the Riddler column. As a hint, the animation below visualizes the result, where the path length increases with each frame, showing the probability distribution of the endpoint of a 2D random walk.

Perhaps as another hint, note the checkerboard pattern to the distribution; only “half” of the vertices are reachable for a particular path length , and *which* half is reachable alternates as increases.

This is a follow-up to a post from several years ago describing a C++ implementation of arbitrary-precision unsigned integer arithmetic. This weekend I extended this to also support arbitrary-precision signed integers and rational numbers. Although this started as an educational tool, it now feels a bit more complete, and actually usable for the combinatorics and probability applications of the sort that are frequently discussed here.

I tried to stick to the original objectives of relatively simple and hopefully readable code, with stand-alone, header-only implementation, as freely available in the public domain as legally possible.

The code is available here, as well as on GitHub, in three header files:

`#include "math_Unsigned.h"`

defines a`math::Unsigned`

type representing the natural numbers with all of the sensible arithmetic, bitwise, and relational operators, essentially everything except bitwise one’s complement… although more on this shortly.`#include "math_Integer.h"`

defines an`Integer`

type with a sign-and-magnitude implementation in terms of`Unsigned`

, with all corresponding operators, including bitwise operators having two’s complement semantics assuming “infinite sign extension.”`#include "math_Rational.h"`

defines a`Rational`

type implemented in terms of`Integer`

numerator and denominator.

This was a fun exercise; there were interesting challenges in developing each of the three classes. As discussed previously, the unsigned type handles the actual arbitrary-precision representation (implemented as a `vector<uint32_t>`

of digits in base ), where division is by far the most complex operation to implement efficiently.

The implementation of the signed integer type is relatively straightforward… except for the bitwise operators. Assuming a sign-and-magnitude representation (using an `Unsigned`

under the hood), it is an interesting exercise to work out how to implement bitwise *and*, *or*, *xor*, and *not*, so that they have two’s complement semantics even for negative operands. In the process, I had to add an “AND NOT” operator to the original underlying unsigned type (there is actually a built-in operator `&^`

for this in Go).

With this machinery in place, the rational type is the simplest to implement. The only wrinkle here is that a few additional constructors are needed, since user-defined conversions from the more primitive integral types (e.g., `Rational`

from `Integer`

, `Integer`

from `int32_t`

, etc.) are only implicitly applied “one level deep.”

**Example application: Are seven riffle shuffles enough?**

To test and demonstrate use of these classes, consider riffle shuffling a standard poker deck of 52 playing cards. How many shuffles are sufficient to “fully randomize” the deck? A popular rule of thumb, attributed to Bayer and Diaconis, is that seven riffle shuffles are recommended. (See a longer list of references here, along with some simpler counting arguments that *at least six* shuffles are certainly *necessary*.)

This recommendation is based on analysis of the Gilbert-Shannon-Reeds model of a single riffle shuffle, and of the total variation distance between probability distributions and , where is the distribution of arrangements of the deck after GSR riffle shuffles, and is the desired uniform distribution where every arrangement is equally likely. We can compute this total variation distance exactly as a function of the number of shuffles, as demonstrated in the following example code:

#include "math_Rational.h" #include <iostream> using namespace math; Integer factorial(int n) { Integer f = 1; for (int k = 1; k <= n; ++k) { f *= k; } return f; } Integer binomial(int n, int k) { if (0 <= k && k <= n) { return factorial(n) / factorial(k) / factorial(n - k); } else { return 0; } } Integer power(int base, int exp) { Integer n = 1; for (int k = 0; k < exp; ++k) { n *= base; } return n; } Integer eulerian(int n, int k) { Integer r = 0; for (int j = 0; j < k + 2; ++j) { r += (power(-1, j) * binomial(n + 1, j) * power(k + 1 - j, n)); } return r; } Rational total_variation_distance(int cards, int shuffles) { Rational q = 0; for (int r = 1; r <= cards; ++r) { Rational a = Rational( binomial((1 << shuffles) + cards - r, cards), Integer(1) << (cards * shuffles)) - Rational(1, factorial(cards)); q += (eulerian(cards, r - 1) * (a < 0 ? -a : a)); } return q / 2; } int main() { int cards = 52; for (int shuffles = 0; shuffles <= 15; ++shuffles) { std::cout << shuffles << " " << total_variation_distance(cards, shuffles).to_double() << std::endl; } }

The following figure shows the results. Total variation distance ranges from a maximum of one (between discrete distributions with disjoint support) to a minimum of zero, in this case corresponding to an exactly uniform distribution of arrangements of the deck.

We can see the sharp threshold behavior, where total variation distance transitions from near one to near zero over just a few shuffles, first dropping below 1/2 at seven shuffles.

]]>However, after a delay of several minutes, a flight attendant came on the PA and asked for two– specifically two– volunteers to give up their seat, in exchange for a flight later that afternoon. Two people immediately jumped up, left the airplane, and *then* we were ready to go… now with two empty seats.

The problem was *weight*: due to a changing forecast of bad weather, both in Baltimore and en route, we had taken on additional fuel at the last minute (e.g., to allow for diverting to a possibly now-more-distant alternate airport), resulting in the airplane exceeding its maximum takeoff weight. Something had to go, and apparently two passengers and their carry-on bags were a sufficient reduction in weight to allow us to take off.

What I found interesting about this episode was the relative *precision* of the change– 175 (or even 174) passengers bad, 173 passengers good– compared with the *uncertainty* in the total weight of the passengers, personal items, and carry-on bags remaining on board. That is, how does the airline know how much we weigh? Since Southwest does not ask individual passengers for their weight, let alone ask them to step on an actual scale prior to boarding, some method of estimation is required.

The FAA provides guidance on how to do this (see reference below): for large-cabin aircraft, the assumed average weight of an adult passenger, his or her clothing, personal items, and a carry-on bag is 190 pounds, with a standard deviation of 47 pounds. The figure below shows the resulting probability distribution of the *total* weight of all 175 passengers on the initially completely full flight:

It’s worth noting that the referenced Advisory Circular does provide a more detailed breakdown of assumed average passenger weight, to account for season of travel (5 more pounds of clothing in the winter), gender, children vs. adults, and “nonstandard weight groups” such as sports teams, etc. But for this summer flight, with a relatively even split of male and female passengers, the only simplifying assumption in the above figure is no kids.

The point is that this seems like a significant amount of uncertainty in the *actual* total weight of the airplane, for less than 400 pounds to be the difference between “Nope, we’re overweight” and “Okay, we’re safe to take off.”

**Reference:**

- Federal Aviation Administration Advisory Circular AC-120-27E, “Aircraft Weight and Balance Control,” 10 June 2005 [PDF]

You have once again been captured by mathematically-inclined pirates and threatened with walking the plank, unless you can win the following game: some number of black balls and white balls will be placed in an urn. The captain will randomly select and discard a ball from the urn, noting its color, then repeatedly draw and discard additional balls *as long as they are the same color*. The first drawn ball of a *different* color will be returned to the urn, and the whole process will be repeated. And repeated, and repeated, until the urn is empty. If the last ball drawn from the urn is black, you must walk the plank; if it is white, you will be set free.

You can choose any positive numbers and of black and white balls, respectively, to be placed in the urn at the start. How many of each should you start with to maximize your probability of survival?

]]>“Doctors say he’s got a 50/50 chance of living… though there’s only a 10% chance of that.”

I’ve lately had occasion to contemplate my own mortality. How long should I expect to live? The most recent life table published by the Centers for Disease Control (see the reference at the end of this post) indicates an expected lifespan of 76.5 years for a male. This is based on a model of age at death as a random variable with the probability density shown in the following figure.

The expected lifespan of 76.5 years is (using the red curve for males). In other words, if we observed a large number of hypothetical (male) infants born in the reference period 2014– and they continued to experience 2014 mortality rates throughout their lifetimes– then their ages at death would follow the above distribution, with an average of 76.5 years.

However, I have more information now: I have already survived roughly four decades of life. So it makes sense to ask, what is my *conditional* expected age at death, given that I have already survived to, say, age 40? In other words, what is ?

This value is 78.8 years; I can expect to live to a greater age now than I thought I would when I was first born. The following figure shows this conditional expected age at death , as well as the corresponding expected *additional* lifespan , as a function of current age .

For another example, suppose that I survive to age 70. Instead of expecting just another 6.5 years, my expected additional lifespan has jumped to 14.5 years.

Which brings us to the interesting observation motivating this post: suppose instead that I *die* at age 70. I will have missed out on an additional 14.5 years of life on average, compared to the rest of the septuagenarians around me. Put another way, at the moment of my death, I perceive that I am dying 14.5 years earlier than expected.

But this perceived “loss” *always* occurs, no matter when we die! (In terms of the above figure, the expected value is always positive.) We can average this effect over the entire population, and find that on average males die 12.2 years earlier than expected, and females die 10.8 years earlier than expected.

**Reference**:

- Arias, E., United States Life Tables 2014,
*National Vital Statistics Reports*,**66**(4) August 2017 [PDF]

Following are the probabilities for the United States 2014 period life table used in this post, derived from the NVSR data in the above reference, extended to maximum age 120 using the methodology described in the technical notes.

Age P(all) P(male) P(female) =========================================================== 0 0.005831 0.006325 0.005313 1 0.000367843 0.000391508 0.000343167 2 0.000246463 0.000276133 0.000216767 3 0.000182814 0.000206546 0.000157072 4 0.000156953 0.000183668 0.000129216 5 0.000141037 0.000160804 0.000120255 6 0.000125127 0.000142914 0.000106328 7 0.000112203 0.000128008 0.0000963806 8 0.000100276 0.000112117 0.0000884231 9 0.0000913317 0.0000992073 0.0000834481 10 0.0000883454 0.0000932456 0.0000824477 11 0.0000952854 0.000103156 0.0000874072 12 0.000119095 0.000137857 0.000101304 13 0.000164729 0.000203286 0.000124134 14 0.000227209 0.000294457 0.000155893 15 0.000293617 0.000391501 0.000190617 16 0.000362946 0.000491412 0.000227306 17 0.000442117 0.000609009 0.000265957 18 0.000529115 0.000743227 0.000302594 19 0.000616971 0.000881116 0.000338207 20 0.00070566 0.00101964 0.000373784 21 0.000786255 0.00114394 0.000408332 22 0.000847887 0.0012343 0.000438875 23 0.000886654 0.00128494 0.000465417 24 0.000909534 0.00130883 0.000490933 25 0.000929392 0.0013228 0.0005174 26 0.000952147 0.00134161 0.000545804 27 0.000977786 0.00136426 0.00057515 28 0.0010063 0.00139366 0.000605432 29 0.00103864 0.00142878 0.00063566 30 0.00107285 0.0014657 0.000668788 31 0.00110792 0.00150147 0.000705793 32 0.00114483 0.0015361 0.000745674 33 0.00118454 0.00156959 0.000792356 34 0.00122898 0.00160582 0.000845811 35 0.00128495 0.00165443 0.000908956 36 0.00135141 0.00171632 0.000981746 37 0.00142538 0.00178654 0.00106021 38 0.00150387 0.0018631 0.00114136 39 0.00158781 0.00194786 0.00122516 40 0.00168489 0.00204935 0.00131843 41 0.00179886 0.00217314 0.00142304 42 0.00192761 0.00231898 0.00153401 43 0.00207581 0.00249327 0.00165615 44 0.00224899 0.00270038 0.00179419 45 0.00243725 0.00292846 0.00194116 46 0.00265083 0.0031884 0.00210855 47 0.0029103 0.00350311 0.00231349 48 0.00321588 0.00387243 0.0025555 49 0.0035468 0.00427362 0.00281771 50 0.00387592 0.00467212 0.00307957 51 0.00420287 0.00507013 0.00333606 52 0.00454693 0.00549737 0.00359937 53 0.00492128 0.00597165 0.0038758 54 0.00532664 0.00649003 0.00417147 55 0.00575619 0.00703581 0.00448667 56 0.00619215 0.00758275 0.00481225 57 0.00662626 0.00813029 0.00513737 58 0.00705499 0.00866988 0.00545972 59 0.00748745 0.00921066 0.00578812 60 0.00794918 0.00978879 0.0061402 61 0.0084469 0.010399 0.00653116 62 0.0089597 0.010994 0.00696556 63 0.00947691 0.0115436 0.00744933 64 0.0100035 0.0120603 0.00797895 65 0.0105466 0.0125691 0.00854801 66 0.0111425 0.0131347 0.00916566 67 0.0118165 0.0137895 0.00985298 68 0.0126025 0.0145881 0.010625 69 0.0135386 0.0155721 0.0115183 70 0.014622 0.016711 0.0125556 71 0.0157853 0.0179169 0.0136863 72 0.0169733 0.0191484 0.0148415 73 0.0181664 0.0203416 0.0160437 74 0.0193544 0.0214907 0.0172763 75 0.0205581 0.0226235 0.0185559 76 0.0219039 0.023887 0.0199909 77 0.0233782 0.0252875 0.0215559 78 0.0249405 0.0266573 0.0233399 79 0.0266659 0.0281283 0.0253501 80 0.0283006 0.0295587 0.0272207 81 0.0298041 0.0307938 0.0290306 82 0.0311707 0.0318902 0.0307088 83 0.0326118 0.0329808 0.0325375 84 0.0338734 0.0336728 0.0344093 85 0.0348103 0.0342521 0.0357896 86 0.0356915 0.0345144 0.0373244 87 0.036144 0.0342741 0.0384714 88 0.0361034 0.0334914 0.0391388 89 0.0355212 0.0321521 0.0392438 90 0.0343716 0.0302738 0.0387212 91 0.0326583 0.0279093 0.0375332 92 0.0304192 0.0251463 0.0356786 93 0.0277276 0.0221028 0.0331989 94 0.02469 0.0189177 0.030181 95 0.0214386 0.0157381 0.0267542 96 0.0181203 0.0127037 0.0230805 97 0.0148823 0.00993297 0.0193396 98 0.011857 0.00751144 0.0157101 99 0.00914934 0.00548606 0.0123497 100 0.00682791 0.0038652 0.00937893 101 0.0049216 0.00262443 0.0068711 102 0.00342273 0.00171608 0.00484973 103 0.00229461 0.00108015 0.00329442 104 0.00148199 0.000654343 0.00215221 105 0.000921789 0.000381551 0.00135156 106 0.000552114 0.000214241 0.000815772 107 0.00031851 0.000115918 0.00047333 108 0.000177059 0.0000604924 0.000264139 109 0.0000949139 0.0000304833 0.000141878 110 0.0000491106 0.0000148533 0.0000734294 111 0.0000245565 0.00000700891 0.0000366663 112 0.0000118822 0.00000320819 0.0000176914 113 0.00000557214 0.00000142697 0.00000826206 114 0.00000253659 0.000000617876 0.00000374136 115 0.00000112287 0.000000260924 0.00000164594 116 0.0000004842 0.00000010766 0.00000070483 117 0.000000203758 0.0000000434809 0.000000294373 118 0.0000000838243 0.0000000172192 0.000000120143 119 0.0000000337717 0.0000000066978 0.0000000480077 120 0.0000000216956 0.00000000409582 0.0000000304297]]>

The title question came up recently, which I think makes for an interesting combinatorics exercise. The idea is that, given the extent of “borrowing” of past musical ideas by later artists, are we in danger of running out of new music?

We can turn this into a combinatorics problem by focusing solely on *pitch *and* rhythm*: in a single bar of music, how many possible melodies are there, consisting of a sequence of notes and rests of varying pitch and duration? The point is that this number is *finite*; it may be astronomical, but *how* astronomical?

This is not a new question, and there are plenty of answers out there which make various simplifying assumptions. For example, Vsauce has a video, “Will We Ever Run Out of New Music?,” which in turn refers to a write-up, “How many melodies are there in the universe?,” describing a calculation based on a recurrence relation that effectively requires cutting segments of a bar exactly in halves– undercounting in a way that I suspect was not intentional, implicitly prohibiting even relatively simple melodies like “I’ll Be Home for Christmas.”

But the most common simplifying assumption seems to be a lack of treatment of *rests*— that is, only counting melodies consisting of a sequence of *notes*. Rests are an interesting wrinkle that complicates the counting problem: for example, a half note is different from two consecutive quarter notes of the same pitch, but a half rest *sounds the same* as two consecutive quarter rests. The objective of this post is to add this “expressive power” to the calculation of possible melodies.

**Solution**

Consider a single bar in 4/4 time, consisting of a sequence of whole, half, quarter, eighth, and sixteenth notes and/or rests, with notes chosen from 13 possible pitches, allowing melodies within an octave of the 12-pitch chromatic scale, but also allowing an octave jump (e.g., “Take Me Out to the Ball Game,” “Over the Rainbow,” etc.).

We can encode the choice of a single note of possible pitches with the following generating function, weighted by duration:

and all possible rests with

Then the generating function for the number of possible melodies is

Intuitively, a melody consists of zero or one rest, followed by a sequence of zero or more sub-sequences, each consisting of a note followed by zero or one rest. The coefficient is the number of one-bar melodies given the above constraints… but this counts two melodies as distinct even if they only differ in relative pitch. The number of possible melodies consisting of sequences of *intervals* confined to at most an octave jump is

where the +1 accounts for the single “silent melody” of a whole rest. The result is 3,674,912,999,046,911,152, or about 3.7 billion billion possible melodies.

**Results**

The machinery described above may be easily extended to consider different sets of assumptions: different time signatures, longer or shorter lists of possible notes to choose from, dotted notes and rests, triplets (e.g., the *Star Wars* theme), etc. The figure below shows the number of possible one-bar melodies for a variety of such assumptions.

As might be expected, dotted notes and/or rests do not affect the “space” of possible melodies nearly as much as note *duration*: halve the shortest allowable note value, and you very roughly double the number of “bits” in the representation of a melody. If we extend our expressive power to allow 32nd (possibly dotted) notes and rests, then there are 6,150,996,564,625,709,162,647,180,518,925,064,281,006 possible melodies.

Of course, these calculations only address the question of how many melodies are *possible*— not how many of such melodies are actually appealing to our human ears.