5 ms·

Ask HN: How to Calculate?

• koots 5y ago
  2/50 > 6/200 So 2 from a pool of 50 has more chance.
  • mech422 5y ago
    I _think_ thats for a single drawing? From the document my friend linked, I think for multiple drawings you use the binomial thing ?
    • Jensson 5y ago
      It is a good approximation when you have very small probability like this. For larger probabilities you can do the same calculation and their relative position is still true even though the probability isn't, since both will be approximately exponential with growth factor roughly equal to their fraction, and exponentials adds linearly.
      • mech422 5y ago
        thanks very much!
• mech422 5y ago
  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 5y ago
  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-infection https://www.mpg.de/16015780/corona-covid-19-aerosol-infectio...
  • mech422 5y ago
    Thanks - but I'm actually looking for how to solve problems like this myself and google fodder to look for more info. I think this is some sort of probability, but I'm not sure how the multi-hour stuff plays in...
    • Someone 5y ago
      You start with a model. Depending on how good you want to match real-world data, that may involve answering questions such as - if somebody sneezes, how many virus particles get out? (may depend on person’s age, mask-wearing, etc) - how far (may depend on person’s height, lung volume, etc) - how long do they survive? (may depend on temperature, humidity, etc) - how likely are they to get to uninfected people? (may depend on what people are there. Kids in a kindergarten, for example, tend to get closer together than workers in an office) As you can see, that can get complex fairly fast. If you want it simpler, you can start with such things as - If you’re in a room with P persons who are spreading virus for an hour (or a minute, a second, etc), how many virus particles will you receive? - What’s the probability you’ll become ill if you receive P virus particles in a given time period? - if you become ill, how serious will it be? - if you receive P particles in a time period and don’t get ill, and repeat that immediately, is the probability of getting ill in that second period the same as in the first, higher, or lower? You can make each as simple or complex as you want. Once you have a model, calculations can be easy or very complex. That’s what the spreadsheet given in that link does for one model. If you’ve done the calculations you’re not done. You have to validate your model to know whether the numbers it spits out actually make sense.
      • mech422 5y ago
• mech422 5y ago
  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/binomial-theorem/binomial-distribution-formula/ https://www.statisticshowto.com/probability-and-statistics/b...
• stncls 5y ago
  > 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.
  • mech422 5y ago
    Thanks for the detailed response! On a quick look, for the second (bus/plane) example, it appears you didn't use the binomial distribution formulas for this? Is there a name I could google for the formulas you used ? edit: I'm puzzled by the fact that the probability term was RAISED to the number of hours, rather then multiplied by it? so the chance isn't linear with the amount of time ? Thanks!
    • stncls 5y ago
• fulafel 5y ago
  Are you worried about climate change, crash risk or covid, or all of them?
  • mech422 5y ago
    not really any of the above per se, I just wanted to know how to figure the probabilities..