HACKER Q&A
📣 mech422

How to Calculate?


a friend and I were talking about holiday plans and realized we have no idea how to go about solving this:

All other things being equal, which would be 'better'/'safer' statically(?) - 6 hours on a bus with 50 people or 2 hours on a plane with 200 ?

(forgive the covid reference - its just how it came up. Only interested in how one goes about calculating things like this..)

P.S. Is there a name for this sort of problem/math? Something I could google for more info?

Edit: Hmm - another way of saying this might be - 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?

Thanks!

  👤 stncls Accepted Answer ✓
> 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.

👤 fulafel
Are you worried about climate change, crash risk or covid, or all of them?
👤 koots
2/50 > 6/200 So 2 from a pool of 50 has more chance.
👤 mech422
Ahh - a friend pointed me to this(1) - a 'binomial distribution'. I think thats what I want ?

1: https://www.statisticshowto.com/probability-and-statistics/b...

👤 mech422
Hmm - looking at the lottery version of this...

Is it really as simple as 2 out of 50 chance vs 6 out of 200 chance ?

👤 _Microft
Try this calculator. I don’t know if they have updated it with options to account for increased infectiveness of the delta variant or for the protective effects of masks and vaccines since I found and submitted it here though. https://www.mpg.de/16015780/corona-covid-19-aerosol-infectio...