Balls falling in buckets: "If X balls are dropped randomly into Y buckets..."

jma

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Oct 6, 2017
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Balls falling in buckets: "If X balls are dropped randomly into Y buckets..."

Hello,

I have a question:
0 0

If X balls are dropped randomly into Y buckets (and buckets can hold an unlimited number of balls), what is the probability that n out of Y buckets are empty?


Backstory: I schedule events. My boss has about 100 events a year and wants me to make sure that at least one is scheduled for every week. But it's very hard to pick particular dates for events (I have to coordinate across a lot of people's schedules). Last week my boss got mad because there were no events scheduled that week. I'd like to be able to show my boss that the probability of having at least one event every week across the whole year is low, so she realizes I'm doing my job as well as I can.



Thank you!
 

tkhunny

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Are you saying that you are unable to see an open slot in the future and make an effort to fill it? Seems like this is the best solution, rather than excusing a break in the action.

1 Bucket. Drop 2 balls. What is the probability that they all go into the same bucket?
2 Buckets. Drop 2 balls. What is the probability that they all go into one bucket. Is this different from the probability that they leave one bucket empty?
3 Buckets. If the buckets are equally probable then the probability that EACH ONE leaves one bucket empty is?

You can solve the problem in this way. Most importantly, don't think about how many buckets there are. Missing the same on EVERY TIME is more important.
 

Jonathan

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Aug 8, 2015
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I'm not an expert in statistics but feel the Poisson distribution https://en.wikipedia.org/wiki/Poisson_distribution is needed here.
For an average of 2 events a week
Prob of k events in 1 week = (e^(-2)*(2^k))/k!
So no of weeks (in 1 year) with k events = (e^(-2)*(2^k)*52)/k!
0 events in 7.04 weeks
1 event in 14.07 weeks
2 events in 14.07 weeks
3 events in 9.38 weeks
4 events in 4.69 weeks
5 events in 1.88 weeks
6 events in 0.63 weeks
more than 6 events in 0.24 weeks
 

tkhunny

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M Buckets

First Ball
p(we don't get the bucket that should be empty) = (M-1)/M

Second Ball
p(we don't get that same bucket) = (M-1)/M

...

Xth Ball
p(we don't get that same bucket) = (M-1)/M

Summary:
p(We miss that bucket EVERY time) = [(M-1)/M]^X

360 Buckets and 150 balls: [(360-1)/360]^150 = (359/360)^150 = 0.659 -- Pretty likely that we'll have an empty bucket, but this exposes a little problem with the setup. We should specify that X > M, otherwise, there WILL be an empty bucket. With 150 < 360, there MUST be empty buckets.

365 Buckets and 400 Balls: [(365-1)/365]^400 = (364/365)^400 = 0.338
 
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