Death, math, coincidence, and Chargers football

You may have read the sad news that Junior Seau, former linebacker for the San Diego Chargers, has died of an apparent suicide. Notably, Seau was a member of the Chargers' Superbowl team in 1994. Many articles reporting on his death have commented on an apparent coincidence:

Seau, who played in the NFL for parts of 20 seasons, is the eighth member of San Diego's lone Super Bowl team who has died, all before the age of 45. Lew Bush, Shawn Lee, David Griggs, Rodney Culver, Doug Miller, Curtis Whitley and Chris Mims are the others. Causes of death ranged from heart attacks to a plane crash to a lightning strike.

Without diminishing the tragedy of Seau's death -- or the deaths of his teammates -- let's analyze how coincidental this situation really is. On the way to sussing out the question of coincidence, I'll describe in layperson's terms some basic notions of probability, counting, and survival analysis.

I proceed with caution here, as I am neither a probabilist, a statistician, an actuary, a demographer... or even, really, a sports fan. I am confident that someone with more relevant mathematical training could do a better job than I do below. I welcome comments, corrections, and improvements.

Let's get going. As is often necessary with real-world problems, we will make some simplifying assumptions for tractability. According to Business Week, the average age of an NFL player in 2011 was 27 years old. We'll assume this was true in 1994. And to simplify further, we'll assume that all of the Chargers were 27 that year. Right now, it is 2012, so those players would be 45 years old.

Now we turn to life tables, clever little tools used by actuaries and demographers to calculate survival probabilities and life expectancies. There are two kinds of life tables. A period life table give the probability of death for people of different ages in a given calendar year. So for instance, a 2012 period life table would give the probability of a 90 year old dying this year, an 89 year old dying this year, and so on. A cohort life table shows the probability of death of people born in a particular year over the course of future years. So for instance, for people born in 1974 (my birth year), the cohort table would give the probability that they would die in 1975, 1976, and so on.

Since we are interested in people who were 27 in 1994, the best tool for us to use would be a cohort table for 1967 (since 1994 - 27 = 1967) but I couldn't find one. Instead, I looked at this historical period table for 1967 from the Centers for Disease Control. A period table and a cohort table are not the same thing, but nonetheless, we'll use the period table data to get a rough solution to our problem.

Take a peek at Table 5-3 (page 5-9) and focus on the first and third columns. The first column contains ages. The third column tells the story of 100,000 males. Specifically, it gives the number of the original 100,000 surviving until the age given in the first column. Of the original 100,000, 94,600 survive until age 27. These 94,600 live 27 year olds are our base population. Now we want to know the probability of living (at least) to age 45. Looking in the third column for the entry corresponding to 45, we find 89,456. This means that for those alive at 27, the probability of living to 45 would be 89456/94600 = 0.94562. In case you are unfamiliar with calculating probabilities this way, here's another example. Say a bag contains 10 balls, 3 of which are red. We reach in without looking and grab one. The probability that we choose a red ball is 3/10.

If the probability of an individual alive at age 27 in 1994 living to age 45 in 2012 is 0.94562, the probability of them dying before age 45 would be 1 - 0.94562 = 0.05438. That's because the probabilities have to sum to one (you can either live or die).

Now we have to consider a team. An NFL team has a roster limit of 53, so we'll take that as our team size. What is the probability that 8 players alive at age 27 in 1994, would have died by age 45 in 2012? This is a binomial probability question.

Let's work up to this idea. One basic idea used is that to get the probability of multiple things happening (different players living/dying) you multiply together their probabilities, so long as whether each person lives/dies is independent of the other people. For example, suppose that each day, the weather can be either rainy or sunny. And also suppose that the weather on one day is independent of weather on other days. If the probability of rain each day is 1/3, the probability that Monday and Tuesday will be rainy but Wednesday will be sunny next week would be 1/3 * 1/3 * 2/3 = 2/27. The 2/3 value is the probability of sunny weather, which is 1 - the probability of rain, or 1 - 1/3 = 2/3.

Back to our football problem. We want to know the probability of 8 players dying and the other 53 - 8 = 45 players living. With p = 0.94562 being the probability of one of the 1994 Chargers living until 45, this is something like

[p^{45} (1-p)^8 = 0.94562^{45} cdot 0.04538^8 = 6.1745 times 10^{-12}].

That's minuscule!

But wait, there's something we haven't accounted for. This is where the second idea related to binomial probabilities comes in. It's not a specific 45 players that we require to live, it's any 45 players. And similarly, it could be any 8 players who die. To account for this, we need to multiply by the number of different ways there are to select 45 live players and 8 dead players from a team of 53. This is

[frac{53!}{45!8!} = 886322710]

where the exclamation point is the factorial function. In case you don't know this function,


1! & = & 1 \

2! & = & 2 cdot 1 = 2\

3! & = & 3 cdot 2 cdot 1 = 3


and so on. So to get the overall probability that of the team of 53 Chargers in 1994, exactly 8 of them would die by 2012, we take

[6.1745 times 10^{-12} times 886322710 = 5.4726 times 10^{-3}]

or about one-half of one percent. This is small, but not negligible.

We can go even one step further. We've thus far been focused on the Chargers Superbowl team. Instead, we could ask about the probability that there would be exactly 8 deaths on any NFL team. We can figure this out with another binomial calculation, first noting that in 1994 there were 28 NFL teams. The probability that one of them would have exactly 8 deaths, as we have said, is 5.4726 times 10^{-3}, which we'll call r. The binomial calculation for exactly one team of 28 having exactly 8 deaths is

[r^1 (1-r)^{27} frac{28!}{1!27!} = 0.13213.]

To recap, that's a 13% probability that of all the NFL teams in 1994, exactly one of them would have exactly 8 players die by 2012 (with lots of simplifying assumptions thrown in). 13% is no guarantee, but neither is it so remote that we should be really surprised by it. By the way, note that when we say that one team has exactly 8 players die, this says nothing about the other teams. They could have fewer or more than 8 players die.

Let me finish with some inspirational words. As disclaimed above, I am not a probabilist. My only training in probability came when I was 11 years old. The calculation above is a very rough (an assuredly flawed) estimate based on data gathered from Googling and a few elementary notions of life tables that I picked up from the Internet. No fancy math was involved. I encourage you -- like I encourage all my students -- not to be afraid of trying to estimate things. It's do-able, it's interesting, and it's an excellent way to apply your powers of critical thinking to what's going on in the world.

Addendum: My excellent colleague Victor Addona has chimed in on this. He is a real pro who a) has a degree in statistics, b) publishes in the field of survival analysis, and c) does interesting and innovative work related to sports statistics. Victor sagely prefers a refined phrasing of my question, namely "of the 28 teams, what is the probability that the one with the most deaths would have at least 8 deaths?" He answered this question with a simulation conducted in the statistical package R, and his answer is about 19%. My very rough estimate is of 13% more like a lower bound to the answer to Victor's question. In short, the eight deaths are even slightly less coincidental than my post suggests.

[This post has been modified. Thanks to Louisa Bradtmiller for pointing out a factual error in the original post, in which I stated that the Chargers had won the Superbowl. Thanks to Victor Addona for pointing out a typo when I originally wrote down 1 - p incorrectly.]

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