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How To Estimate The Probability Of Covid 19 On Carnival’s August Cruise From Germany

Jul. 15, 2020 3:44 PM ETCarnival Corporation & plc (CCL)
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  • Even small probabilities of infection for individuals compound to large probabilities for a large number of people.
  • Carnival needs low risk passengers and screening to be near 100% effective to reduce risks.
  • This is a necessary experiment but could backfire if they allow too many on board.

This is not precise and depends greatly on our assumptions, but let’s give it a shot. First off I will note that if their screening process is 100% effective, none of this matters and no Covid will be on the ship. My purpose is only to demonstrate how probability works in surprising ways.

According to Wikepedia the AIDAperla can hold 3,286 passengers and 900 crew. Let’s assume they will sail at 40% capacity and there are 1,314 couples as passengers and 360 crew. We need this distinction because I’m going to assume each couple either has Covid together or neither does. Therefore we have 1314/2 + 360 = 1,017 possible sources of Covid to screen. I assume these are independent random samples from the population.

They are sailing from Germany which currently has approximately 240 diagnosed cases of Covid per 100,000 population. See https://www.statista.com/statistics/1110187/coronavirus-incidence-europe-by-country/

As we all know by now, many people are infected and don’t know it. According to John Hussman’s estimate in his article on Seeking Alpha July 13th in the U.S. the ratio of undiagnosed to diagnosed is roughly 2 to 1. Therefore if we assume a similar ratio in Germany then the prevalence in the population is 3 * 240/100,000 = .72%.

Now here’s the big assumption: Let’s assume the precautions they employ will screen out 80% of Covid carriers so that someone who is in fact infected only has a 20% chance of making it through the screening process which includes a questionnaire and temperature check.

This implies that overall each person on board has a probability of being infected of .72% * 20% = .144%.

What are the odds that all 1,017 couples and crew on board are Covid free? The chance of each being clean is 100% - .144% = 99.856%. The probability that all are Covid free is: .99856 multiplied by itself 1,017 times- meaning .99856 to the 1,017 power. That is 23%, which means that under the assumptions here, we estimated 77% chance that Covid will be on board.

I think the assumptions here are on the conservative side. If my assumption of screening effectiveness were dropped to say 50% then the probability of Covid on board rises to 97%.

Carnival needs to learn how to safely cruise and thus they need to try this, but they should limit the passengers to a lower risk population of young people for this experiment. A lower passenger count will improve their odds of not having an infection on board.

Analyst's Disclosure: I am/we are short CCL.

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