4 ms·

> The author explains that the efficacy number is derived from the the number of the number of people in the control/placebo group infected divided by the total

> The author explains that the efficacy number is derived from the the number of the number of people in the control/placebo group infected divided by the total number of infections in the study. Wouldn't that make a vaccine that does nothing 50% effective, instead of 0% effective (which is what one would expect)?

Eric_WVGG 6y ago
Well, let's see… say you had 2000 participants in the trial, 1000 with a junk vaccine, and 1000 with placebos. Furthermore, let's say you had 200 infections, split equally between the junk and the placebo. So yeah, 100 / 200 = 50% effective. A coin toss, which is to say, no better or worse than not getting the vaccine. Again, this isn't measuring "what one would expect" (how many vaccinated people, exposed to the virus, get the disease), it's "what proportion of vaccinated people get sick relative to unvaccinated people." (again IANAD)
- StavrosK 6y ago
  Kind of seems odd, given that you can never have 0% efficacy (unless your vaccine somehow caused every vaccinated person to get the disease instead of a non-vaccinated person). It seems like you're wasting half your "spread", basically. If you measured efficacy as "1 - (vaccinated infections / unvaccinated infections)", it would be much more intuitive, and you could even get a bit of negative efficacy in the case where your vaccine did nothing.
- iggldiggl 6y ago
  No, 50 % means the infection rate (or whatever you're actually measuring) in the vaccine group is half that of the control group. In your example the infection rate are exactly the same, giving an efficacy of 0 % (± whatever confidence interval the size of your trial and placebo groups gives you). See e.g. https://blogs.sciencemag.org/pipeline/archives/2020/09/21/the-vaccine-protocols https://blogs.sciencemag.org/pipeline/archives/2020/09/21/th...

4 ms·

> The author explains that the efficacy number is derived from the the number of the number of people in the control/placebo group infected divided by the total

by StavrosK 6y ago

Eric_WVGG 6y ago
Well, let's see… say you had 2000 participants in the trial, 1000 with a junk vaccine, and 1000 with placebos. Furthermore, let's say you had 200 infections, split equally between the junk and the placebo. So yeah, 100 / 200 = 50% effective. A coin toss, which is to say, no better or worse than not getting the vaccine. Again, this isn't measuring "what one would expect" (how many vaccinated people, exposed to the virus, get the disease), it's "what proportion of vaccinated people get sick relative to unvaccinated people." (again IANAD)
- StavrosK 6y ago
  Kind of seems odd, given that you can never have 0% efficacy (unless your vaccine somehow caused every vaccinated person to get the disease instead of a non-vaccinated person). It seems like you're wasting half your "spread", basically. If you measured efficacy as "1 - (vaccinated infections / unvaccinated infections)", it would be much more intuitive, and you could even get a bit of negative efficacy in the case where your vaccine did nothing.
- iggldiggl 6y ago
  No, 50 % means the infection rate (or whatever you're actually measuring) in the vaccine group is half that of the control group. In your example the infection rate are exactly the same, giving an efficacy of 0 % (± whatever confidence interval the size of your trial and placebo groups gives you). See e.g. https://blogs.sciencemag.org/pipeline/archives/2020/09/21/the-vaccine-protocols https://blogs.sciencemag.org/pipeline/archives/2020/09/21/th...