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> 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
by 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 agoThanks 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> it appears you didn't use the binomial distribution formulas for this You would use binomial distribution formulas if the problem was more complex. Say, if you cared about how many times you get covid infected, or how many times your winning ticket is drawn. The problem here is a special, simpler case. > Is there a name I could google for the formulas you used ? Any intro to probabilities book will start with such problems. If you really want to google something, the basic assumption here is that the events are "independents". You can find examples here: https://en.wikipedia.org/wiki/Independence_(probability_theory)#Examples https://en.wikipedia.org/wiki/Independence_(probability_theo...
- mech422 5y agoThanks! I appreciate the links/google fodder - it'll let me RTFM more on this stuff...
- stncls 5y ago> 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 ? Yes it is not linear. Imagine you are repeatedly throwing a dice. The probability that you get a "6" at some point is absolutely not linear with the number of throws: P(not a single "6" in N throws) = (4/5)^N so P(at least one "6" in N throws) = 1 - (4/5)^N Edit: Also, raw probabilities are constrained to the interval [0, 1]. So they are rarely linear in a parameter, because, then, they would easily escape the [0, 1] interval. Imagine that we had P(covid) = some_constant * some_parameter. There would be values of the parameter such that P(covid) > 1, which does not make sense.
- mech422 5y agoThanks! This will probably make more sense once I can RTFM the probability stuff..