> whats gives you the best chance of having your lottery ticket drawn - 2 drawings from a pool of 50, or 6 drawings from a pool of 200?
The standard trick for this type of probability problem is to reverse the question: What is the probability to not have your ticket drawn:
P(unlucky_first_time AND unlucky_second_time AND ...)
= P(unlucky_first_time) * P(unlucky_second_time) * ...
For two drawings from 50:
P(unlucky) = 49/50 * 48/49 = 0.96
so P(lucky) = 0.04
For 6 drawings from 200:
P(unlucky) = 199/200 * 198/199 * 197/198 * 196/197 * 195/196 * 194/195 = 0.97
so P(lucky) = 0.03
In other words, your best chance of having your lottery ticket drawn is in 2 draws from 50.
Edit: Note that I assume that the ticket chosen at the first draw is taken aside and not put back in the pool. Hence the 49/50 becoming 48/49 the second time, since the pool now contains only 49 tickets.
> 6 hours on a bus with 50 people or 2 hours on a plane with 200
I believe that the original problem is very different actually.
Let's specify the question a bit more. Assume that, for each hour, for each fellow passenger, you get a probability U of getting covid. That's a strong assumption (it actually implies a specific model of contagion) but let's go with that.
(Edit: You could think of it the following way. Imagine U=1/6, then you could roll a dice and a "6" would give you covid. Then the model I propose would be: Once an hour, you go to each one of your fellow passengers and roll the dice. If you get a "6" even once, you get covid. Obviously U=1/6 is way larger than what would be realistic, but I hope you get the picture.)
Then the probability to not get covid is, in case 1:
P(no_covid) = ((1-U)^50)^6 = (1-U)^300
and in case 2:
P(no_covid) = ((1-U)^200)^2 = (1-U)^400 < (1-U)^300
So the probability of getting covid is higher for the 2 hours on a plane with 200.
Again, it's an extremely naive assumption, and it supposes a very specific contagion model.