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I'm going to make this question a little generic, but it's for a specific situation:

You have a list of 10,000 people. Every week, you randomly select 2% (200) of those people. What are the odds of one or more of those 2% getting picked the following week?

I initially thought the solution was simply, but as I run it through my head, I'm starting to think it's not so simple. I've been trying to brush up on my probability math, but as it's been about 100 years since I took probability and statistics, I just can't seem to find the way to solve it and would appreciate, not just the answer, but the method of solving the problem.

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  • $\begingroup$ Being picked in some week does not change the odds of being picked the following week. $\endgroup$
    – drhab
    Jan 12, 2015 at 14:38
  • $\begingroup$ For any single individual, yes. My odds are 2% every week. My question is, what are the odds of any of the 200 people picked one week being picked the following? Is it as simple as 2%? That just doesn't seem right to me. $\endgroup$
    – Pete
    Jan 12, 2015 at 14:43

2 Answers 2

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200 people got picked in week #1. The probability that none of them get picked in week #2 is $$\frac{{9800\choose 200}}{{10000\choose 200}}$$ Hence the probability that at least one gets picked in week #2 is $$1-\frac{{9800\choose 200}}{{10000\choose 200}}\approx 98.3\%$$

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  • $\begingroup$ You wrote "...none of them get picked in week #2". I want to know what the probability is that one or more of them gets picked in week 2? And I'd like to understand how it's calculated (the reasoning behind it.) $\endgroup$
    – Pete
    Jan 12, 2015 at 14:48
  • $\begingroup$ Okay. I get it. Thanks! $\endgroup$
    – Pete
    Jan 12, 2015 at 14:57
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As has been pointed out, this probability is $1 - {{9800 \choose 200} \over {10000 \choose 200}}$. But you may be wondering how to get a numerical value for the ${9800 \choose 200} \over {10000 \choose 200}$ term.

One way is to expand those binomial coefficients in terms of factorials, giving $$ {{9800! \over 200! 9600!} \over {10000! \over 200! 9800!}} $$ Now canceling and rearranging gives $$ {9800! / 9600!} \over {10000! / 9800!} $$ and you can explicitly write out the products to get $$ {9800 \times 9799 \times \cdots \times 9601 \over 10000 \times 9999 \times \cdots \times 9801} $$. This is still problematic in that both the numerator and denominator are very large integers. But you can rewrite this as $$ {9800 \over 10000} \times {9799 \over 9999} \times \cdots \times {9601 \over 9801} $$ and now each factor is a rational number a bit less than 1. This can be computed, for example, in Python as

x = 1 for i in range(9601, 9801): x *= i/(i+200)
which returns $0.0168786$.

Alternatively, using Stirling's approximation $n! \approx \sqrt{2 \pi n}(n/e)^n$ we get

$$ {9800!^2 \over 9600! 10000!} \sim {(2\pi \times 9800) (9800/e)^{9800 \times 2} \over \sqrt{2\pi \times 9600} (9600/e)^{9600} \sqrt{2\pi \times 10000} (10000/e)^{10000}} $$

and simplifying a bit this is

$$ {9800 \over \sqrt{9600 \times 10000}} {9800^{19600} \over 9600^{9600} 10000^{10000}}. $$

or

$$ {9800 \over \sqrt{9600 \times 10000}} \left( {9800 \over 9600} \right)^{9600} \left({9800 \over 10000} \right)^{10000} $$

which evaluates to $0.0168786$ as well, for example in Python

import math from math import sqrt (9800/9600)**9600 * (9800/10000)**10000 * 9800/sqrt(9600*10000)

Finally, if those powers are too large for whatever you're doing the computation with (they're roughly $10^{86}$ and $10^{-88}$, respectively), you can rewrite as

$$ \exp \left( \log 9800 - {1 \over 2} \log 9600 - {1 \over 2} \log 10000 + 9600 \log (9800/9600) + 10000 \log(9800/10000) \right).$$

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  • $\begingroup$ Thank you for the extensive explanation, but a great deal of it went over my head. Hopefully it will help the next guy. I chose the simpler route. I went to Wolfram Alpha and wrote: 1 - ((9800 choose 200)/ (10000 choose 200)) and let it work out the big numbers for me. Actually, my numbers were 5600 and 112, but I was trying to use round numbers to make it more general. $\endgroup$
    – Pete
    Jan 12, 2015 at 16:42

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