Ebola is just one plane ride away… But how likely is it, really?

Note: Please see caveats to this analysis here.

In the midst of this year’s outbreak (the largest in documented history), there’s been a lot of talk lately about Ebolavirus hopping on a plane from West Africa and making its way to the United States. Today, I want to quantify the probability of this actually happening. [Spoilers: The likelihood is basically nonexistent. Surprised? You shouldn’t be!]

Let’s break it down in English first. Given that there’s already an ongoing outbreak in West Africa, the following three events need to be fulfilled for someone to bring Ebola to the United States. In short, the person in question needs to:

A. Successfully acquire the virus;
B. Get on an airplane while infected with the virus;
C. Either transmit the virus to someone who will fly all the way to the United States, or fly there on his or her own

At the intersection of these three events, we have our event of interest:

X. Exportation of Ebolavirus to the United States from West Africa

By multiplying the probabilities of Events A through C and then dividing the product by the probability of the ongoing outbreak itself, we can find out just how likely it is that our event of interest will happen as a result of this year’s outbreak.

In reality, this is just an application of Bayes’ Theory, which lets us calculate the probability of Event X given that Event Y has already occurred. The formula is actually pretty simple:

P(X|Y) = [P(Y|X)P(X)]/P(Y)

For our problem, Events X and Y are as follows:

Event X: Exportation of Ebolavirus to the United States from West Africa
Event Y: Outbreak happens in West Africa in 2014

P(Y|X) and P(Y) are actually pretty easy to calculate:

P(Y|X) = 1,
Because (based off of our current understanding), Event Y must occur for Event X to occur

P(Y) = 1/38 = .0263,
Because since 1976, 38 years have gone by and only one of them (2014) has amounted in an outbreak of Ebola in West Africa

P(X) is where things get interesting. As noted above, P(X) = P(A)P(B)P(C). So, let’s approach Events A through C one at a time:

Event A: Successfully acquire the virus
There are three countries in which an individual can successfully acquire the virus – at varying rates. We can calculate a rate – or probability – for each country by dividing number of cases reported by population:

P(A_Guinea) = 427/11,450,000 = .0000373
P(A_Liberia) = 249/4,190,000 = .0000594
P(A_SierraLeone) = 525/5,979,000 = .0000878

Event B: Get on an airplane while infected with the virus
During the outbreak thus far, only one person (that we know of) has gotten on a plane while infected with Ebola, so:

P(B) = 1/1201 = .000833

Event C: Either transmit the virus to someone who will fly all the way to the United States, or fly there on his or her own
This event is a little trickier to calculate and requires a little number-fudging (though I’m working on a more accurate approximation for the near future). There are two things that we need to consider when we approach Event C:

1. How likely is it that someone on a plane out of West Africa is going all the way to the United States?
2. If the infection person isn’t going to the United States, how likely is it that a person with Ebola will transmit it to someone who is (while on the plane)?

Truth is, we need really high-resolution flight-by-flight data out of each country to properly estimate Event C, which I hope to get my hands on soon. (Thanks HealthMap!) For the time being, we can try to assess the probability of Event C by breaking it apart into two mini events: (1) transmission in close quarters and (2) taking a route that either flies through an airport (LHR, BRU, etc.) with transfers to the United States or goes all the way to the United States.

This paper on the Uganda 2000-2001 outbreak gives us a rough probability estimate of Mini Event 1. In the study, 1/20 people contracted Ebola without direct contact with a sick person.

Calculating the probability of Mini Event 2 is somewhat more complex. To address this, we can figure out how many routes out of each country’s major international airport go through an airport with transfers to the United States. A quick search on Kayak gives us the following information:

1. To get from Conakry International Airport (in Guinea) to the United States, passengers need to fly through Paris, Casablanca, or Dakar. 9 different routes out of Conakry (of 22 total) fly to one of these three destinations.
2. To get from Roberts International Airport (in Liberia) to the United States, passengers need to fly through Brussels or Accra. 6 different routes out of Roberts (of 21 total) fly to one of these two destinations.
3. To get from Lungi International Airport (in Sierra Leone) to the United States, passengers need to fly through Conakry, Guinea or Monrovia, Liberia. 8 different routes out of Lungi (of 27 total) fly to one of these two destinations.

This said, it’s hard to know exactly how many of these passengers actually go all the way to the United States instead of ending their journey in any one of these “pit-stop” destinations. While we’re waiting on flight-by-flight statistics, we can use a bit of a fudge factor to estimate this. According to the US Office of Travel & Tourism Industries, only .5% of total arrivals into the United States in March 2014 were from the African Continent… Not perfect, but let’s use that figure for now.

So, the probability of Event C for each of the three countries is as follows:

P(C_Guinea) = (9/22)*(1/20) + (9/22)*(5/1000) = .0225
P(C_Liberia) = (6/21)*(1/20) + (6/21)*(5/1000) = .0157
P(C_SierraLeone) = (8/27)*(1/20) + (8/27)*(5/1000) = .0163

Now that we have estimates P(A), P(B), and P(C), we can calculate P(X):

P(X) = P(B)*[P(A_Guinea)*P(C_Guinea) + P(A_Liberia)*P(C_Liberia) + P(A_SierraLeone)*P(C_SierraLeone)] = .000833[0000373*.0225 + .0000594*.0157 + .0000878*.0163] = .00000000267

Plugging this value of P(X) into the Bayesian model we framed earlier, we arrive at our final destination:

P(Exportation of Ebolavirus to the United States from West Africa | Outbreak happens in West Africa in 2014) = P(X|Y) = [P(Y|X)P(X)]/P(Y) = [(1)(.00000000267)]/(.0263) = .0000102%

This is a teeny, tiny number. Of course, it’s still possible… But based off of the information currently available, it seems highly unlikely.

Nevertheless, it’s crucial to remember that just because Ebola might (under highly improbable circumstances) be exported to the United States, it doesn’t mean that there’ll be any further transmission. We have ready access to the resources and facilities needed to properly contain the infection in case it does end up finding its way on American soil. Though there’s no way to truly estimate the likelihood of an Ebola outbreak here at home, I think it’d be safe to call it practically zero.

Disclaimer: The model developed above is parameterized using highly imperfect and constantly changing information; final approximation subject to change.

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One thought on “Ebola is just one plane ride away… But how likely is it, really?

  1. Pingback: Follow-Up: Ebola is just one plane ride away | Mens et Manus

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