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I have a pool of 100 different numbers, I draw 5 different numbers without putting them back into the pool. I want to calculate the chance of any single number being drawn.

So for example, what's the chance of the number 13 being drawn during those 5 draws?

I figured, the way to calculate this would be

$$\frac{1}{100}+(\frac{1}{99}*\frac{99}{100})+(\frac{1}{98}*\frac{98}{99})+(\frac{1}{97}*\frac{96}{97})+(\frac{1}{96}*\frac{95}{96})=0.0506$$

Which would mean the chance for that would be ~5.1%

Is this calculation correct? What is this calculation called and whats the formula for it?

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  • $\begingroup$ before the process is started, the chances that the 5th ball is going to be #13 is $\frac{99}{100}\times\frac{98}{99}\times\frac{97}{98}\times\frac{96}{97}\times\frac{1}{96} = \frac{1}{100}$ $\endgroup$ – Cato Jun 1 '17 at 11:46
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Correct formula: $$\Pr(E)=\frac{\text{number of draws}}{\text{number of numbers}}=\frac5{100}$$ where $E$ denotes the event that number $13$ will be drawn.

For $i=1,2,3,4,5$ let $E_n$ denote the event that number $13$ is drawn at the $i$-th draw.

Then $\Pr(E_1)=\frac1{100}$ i.e. the first term in your summation.

You can calculate $\Pr(E_2)=\Pr(E_2\mid E_1^c)\Pr(E_1^c)=\frac{1}{99}\times\frac{99}{100}=\frac1{100}$, i.e. the second term in your summation.

So these terms are okay, and if you had proceeded without making any mistakes you would have found $\Pr(E_i)=\frac1{100}$ for every $i$.

Now start wondering: can you find any reason that the probability of drawing number $13$ at e.g. the $2$-nd draw should differ from the probability of drawing number $13$ at e.g. the $4$-th draw???

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  • $\begingroup$ I think maybe reasoning that 100 balls could be placed randomly into slots 1-100, then looking to see if #13 is in slot 1-5, makes it easier to see that 5/100 is the correct answer (by symmetry #13 is equally likely to be in any given group of 5 spaces, and is effectively the same overall experiment) $\endgroup$ – Cato Jun 1 '17 at 11:49
  • $\begingroup$ @Cato That's indeed a good startoff for the stimulated "wondering" of the OP. $\endgroup$ – drhab Jun 1 '17 at 11:53
  • $\begingroup$ @Cato Wow, that made it really easy to understand! $\endgroup$ – Luke Jun 1 '17 at 11:54
  • $\begingroup$ @drhab This could still technically be seen as hypergeometric distribution right? Basically for my case it would be a population size = 100, number of successes in population = 1, sample size = 5, and number of successes in sample = 1 ? $\endgroup$ – Luke Jun 1 '17 at 11:58
  • $\begingroup$ @Luke Yes. With outcome $\frac{\binom51\binom{95}0}{\binom{100}1}=\frac5{100}$. $\endgroup$ – drhab Jun 1 '17 at 12:00
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This is binomial distribution without replacement, and is known as Hypergeometric distribution. Please find more here: https://en.wikipedia.org/wiki/Hypergeometric_distribution

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your reasoning is on the right lines, but you aren't quite right, on draw 5 (for example), it has to have failed to be drawn 4 times

may I be lazy and assum just 3 draws, I get the below example

$P = \frac{1}{100} + \frac{99}{100}\times\frac{1}{99} + \frac{99}{100}\times\frac{98}{99}\times\frac{1}{98} = \frac{1}{100} + \frac{1}{100} +\frac{1}{100} = \frac{3}{100} $

the overall chance for 3 draws is 3/100, and the chance it will have come up on any of the particular draws is 1/100 - for your 5 draws - it will be 5 / 100 -

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