Can You Calculate the Odds of These Unlikely Events?

[michaelwright]

TL;DR

Out of 4M people, what's the probability of randomly choosing 3 who all play a particular sport for a specific team?

Hi folks - I need some help with a tricky probability. Here's the situation:

Let's say there are 4M internet users in Age Group A. (The total set)
Of those 4M, there are 1,000 users who play a specific sport.
Those 1,000 are spread evenly over 125 teams, so 8 players each.

1. What's the probability of selecting 3 random users (from the 4M) who all play that sport?
2. What's the probability of all 3 random users being members of the same team?

This isn't a school problem - I'm a 47 year old writer, and I need to figure out this absurdly improbable number. (The improbability is the point.) Any advice / guidance / direction here would be appreciated! (I have a moderate familiarity with combinations & permutations, and their notations, so do feel free to use whatever formulae / notations are needed. An online calculator for this type of problem would be a great help, too!)

Any questions, please feel free to let me know. Thanks! ~ MW.

 

[hutchphd]

Do you want an absolutely exact number? The answers are:

1. (1000/4M)³
2. (8/4M)³

Pretty small. This assumes you can select the same person twice. If not, the results are slightly smaller

 

[michaelwright]

Thanks, Hutch! I know the numbers are insanely small - that's the point. Ideally I'd love to have the formulae for getting these answers (assuming you cannot select the same person twice.) It'd be great, as well, if there is an online calculator into which I can plug these *and similar) numbers in for my own calculations. Do let me know your thought here, and/or any questions you may have. Thanks again & in advace!

 

[Dale]

Do you want an absolutely exact number? The answers are:

1. (1000/4M)³
2. (8/4M)³

Pretty small. This assumes you can select the same person twice. If not, the results are slightly smaller

For the second one don’t forget the factor of 125 since there are 125 teams.

 

[Dale]

I'd love to have the formulae for getting these answers (assuming you cannot select the same person twice.)

For the first case you are picking players from any team, so there are 1k out of 1M possibilities. So the probability of getting one on the first try is $\frac{1k}{1M}$. Now the probability for getting one on the second try is a little different. Now there are only 999 players left and only 999999 in the population, so the probability of getting one on the second try is $\frac{1k-1}{1M-1}$. And similarly for the third.

The final probability is the product of the three individual probabilities or
$$\frac{1k}{1M} \frac{1k-1}{1M-1} \frac{1k-2}{1M-2}$$

A similar approach can be used for the second case, except that you need to multiply by the number of teams. So you get
$$125 \frac{8 }{1M } \frac{8 -1}{1M -1} \frac{8 -2}{1M -2} $$

 

[hutchphd]

For the second one don’t forget the factor of 125 since there are 125 teams.

Thank you sir. That was an oops.

 

[michaelwright]

Thanks, all... (BTW, the total set is 4M, not 1M, but your advice works regardless - I can update the number later.) Do you happen to know of an online calculator that will give me final figures? I know these values will be absurdly small - that's the point - but I need relatively exact figures for comparison. Thanks again, and do let me know if there's anything I can use. Thanks!

 

[PeroK]

Thanks, all... (BTW, the total set is 4M, not 1M, but your advice works regardless - I can update the number later.) Do you happen to know of an online calculator that will give me final figures? I know these values will be absurdly small - that's the point - but I need relatively exact figures for comparison. Thanks again, and do let me know if there's anything I can use. Thanks!

You can do all the calculations in an Excel spreadsheet. It's only multiplication.

The first one you should be able to do without a calculator: it's 1/(64 billion).

 

[Dale]

Do you happen to know of an online calculator that will give me final figures? I know these values will be absurdly small - that's the point - but I need relatively exact figures for comparison.

I wouldn’t trust most web calculators for this computation. Unless you use an arbitrary precision math library the numbers will be dramatically wrong. It is better to calculate this by hand.