The intuitive way to think about this is the heuristic “small numbers produce more extreme outcomes”. Both choices have the same expected number of deaths. But the 50% option is higher variance than the 5% option. Our goal is to maximize the likelihood of getting the “0 deaths” outcome, which is an extreme outcome relative to the mean. So we can conclude the 50% option is better without doing any math.
You got the wrong answer, but I do like the idea of comparing variances, and at least for this distribution, whichever has greater variance will have more weight on 0. But in this case, the variance of the 50% option is 0.5 and the variance of the 5% option is 0.95. And indeed the 5% option is preferable. (Binomial(n,p) has variance np(1−p), if the means np are the same then whichever has lower p will have higher variance.)
That’ll teach me to post without thinking! Yes, you’re right that np(1−p) is the better way to deal with variance here. (Or honestly, the (1−1n)n method from the above comment is the slickest way.)
I had been thinking of a similar kind of situation, where you have a fixed p and varying sample sizes n. Then, the smaller n gives more extreme outcomes than larger n. Of course, this isn’t applicable here.
The intuitive way to think about this is the heuristic “small numbers produce more extreme outcomes”. Both choices have the same expected number of deaths. But the 50% option is higher variance than the 5% option. Our goal is to maximize the likelihood of getting the “0 deaths” outcome, which is an extreme outcome relative to the mean. So we can conclude the 50% option is better without doing any math.
You got the wrong answer, but I do like the idea of comparing variances, and at least for this distribution, whichever has greater variance will have more weight on 0. But in this case, the variance of the 50% option is 0.5 and the variance of the 5% option is 0.95. And indeed the 5% option is preferable. (Binomial(n,p) has variance np(1−p), if the means np are the same then whichever has lower p will have higher variance.)
That’ll teach me to post without thinking! Yes, you’re right that np(1−p) is the better way to deal with variance here. (Or honestly, the (1−1n)n method from the above comment is the slickest way.)
I had been thinking of a similar kind of situation, where you have a fixed p and varying sample sizes n. Then, the smaller n gives more extreme outcomes than larger n. Of course, this isn’t applicable here.