Will COVID Vaccination Prevent Mutation? Napkin Calculation #7

Preface: This is not an anti-vaccine argument. The conclusion implies we will need more vaccines and vaccinations, not less.

Will COVID vaccination prevent mutation? This has been touted as a reason to get the whole world vaccinated.  There are other good reasons; we challenge only the idea it will prevent mutations. The challenge is presented in the form of a napkin calculation. Since it is more involved than previous calcs, the last half of this is  labeled for techies only. The first half is for everybody. Let’s start with a simple visualization — pesticide resistance.

In areas with rat infestation, bait laced with warfarin, was used as  rat poison, in a cycle like this:

  • See rats.
  • Set the bait.
  • Almost all the rats die.
  • Problem solved, for a while.
  • Rats come back, repeat with bait.
  • Eventually, after many cycles, the bait loses  effectiveness, failing to kill rats.

If rats could actually be extirpated, as has been managed on a few small islands, this example would not be useful. But rats are mobile, repopulating areas where they were exterminated. Some new rats have random mutations.   Controlling rats with this poison confers a survival advantage to a warfarin-resistance mutation. From (Wikipedia) Warfarin,

The use of warfarin itself as a rat poison is now declining, because many rat populations have developed resistance to it,[94] and poisons of considerably greater potency are now available. Resistance is due to an autosomal dominant on chromosome 1 in Norway rats.[94]

Pretty soon the poison resistant rats take over, so another poison must be used. Predicates, with analogy to COVID:

  • If warfarin made rats go extinct, the mutation would not have occurred.
  • If warfarin were not poisonous, the mutation would still randomly occur, but would not confer a survival advantage, so the majority of rats would never acquire it.
  • In between these extremes, warfarin could be mildly poisonous, effective against rats most of the time.
  • By analogy, COVID vaccines are effective against COVID to varying degrees. A breakthrough infection is analogous to a rat that takes the bait and survives.
  • Unlike rats, with COVID, breakthrough infections to date have only partly been the result of mutations. Mutations occur in the course of infections, but it is impractical to identify the individuals.

An escape mutation allows the virus to evade the vaccine. Per each exposure, there is a chance of a single escape mutation. How is this a function of vaccine efficacy? In this napkin calc model,

  • The calc is for a single, vaccinated individual, for whom the chance of infection is a constant.
  • Everyone in the whole world is completely vaccinated.
  • The complexities of the real world are ignored.
  • If the  vaccine is 0% effective, an escape mutation may occur, but without survival advantage that would cause it to out-compete viruses without the mutation.
  • If the vaccine is 100% effective, there are no cases, hence no mutations. Most epidemiologists seem focused on the neighborhood.
  • In between these extremes, what level of vaccine efficacy is most likely to produce an escape mutation? Intuitively, it’s a number not right up against 0% or 100%, but somewhere in between.

Napkin calcs are not to be trusted. We use this one for a legitimate purpose, to undermine the assertion that practical levels of vaccination and vaccine efficacy can make a dent in variants of concern. The number is 50%.  In our simplified world, a vaccine that works half the time is most likely to result in a COVID escape mutation in a single person.

The actual chance of  occurrence of an escape is very small, like 1 in 500 million.  Get vaccinated!!!

If you are not a techie, the takeaway for you  is 1/2. A vaccine that works half the time is close to the real situation, which does not involve people fully vaccinated with a perfect matching vaccine. You should now go have a beer. What follows is for techies only. Or maybe I should say, like CNN: Warning. Contains disturbing numbers. Viewer discretion is advised.

For Techies Only

Our napkin calc lacks a model, so we use a common trick of the statistics game: We impose a statistical distribution motivated by intuition, and see how well it compares to the real world. In this case, comparison is impossible. We use a Gaussian  distribution with the tails chopped off. Definitions:

P( something )  = probability of something.

“|” means “given”, or “conditional on.”

“^”,  means raised to the power of.

break = “breakthrough”

escape = occurrence of escape mutation

Conditional probability: P( escape )= P( escape | break )*P( break )

Our trial Gaussian- inspired distribution is

P( escape) = C*exp[  – (break – alpha )^2/beta^2 ] + D

where alpha, beta, C, and D are to be determined.

Boundary conditions:

If P( break ) = 0,  P( escape ) = 0   (case of perfect vaccine)

If P( break ) = 1,  P( escape ) =0   (case of vaccine dud)

For a vaccine with small break,  P( escape ) = P( break ). In words: If a vaccine is almost perfectly protective, an infection is likely to have an escape mutation. This provides an enhanced break = 0 boundary condition for the first order expansion of

exp – [ ((break – alpha)/beta)^2 ]

At break = 1,  to 2nd order, the boundary condition for the expansion is,

1 – ( 1 – alpha )^2/beta^2 = 0

This gives

alpha = beta = 1/2

With these boundary conditions, the goal is to find the value of break that maximizes P( 0 < break < 1).

Insert alpha and beta into the full exponential. Determine D so that at  break = 0 and break = 1,

C*( exp[ – (break – alpha )^2/beta^2 ] + D ) = 0

, re-satisfying the boundary conditions. We get

 D  =  -1/e

It is not necessary to determine C by normalization; we only want the maximum.


P( escape ) = C*[exp – ( (2*break – 1)^2)  – 1/e ]

which is maximum at break = 1/2. This implies a vaccine with 50% breakthrough is the most likely to produce escape mutations. Since this is a really tough problem, the most we can say is,

Or something like it.