1

I am trying to understand the hypergeometric distribution

I looked at this example, I can see that it makes sense when we force each object to be distinct, though I am wondering why this is correct when each element is not distinct.

For example if I am choosing from red und black balls the formula $\Pr(X = k) = \frac{\binom{K}{k} \binom{N - K}{n-k}}{\binom{N}{n}}$ states that I would have to choose k out of the red balls and n - k out of the black balls. But in reality there is just one possibility to choose $\binom{K}{k}$ since every chosen red ball is the same und not distinct. It would just be a set $(r, ... , r)$ k times without being able to say what $r_i$ with $0 \leq i \leq k$ actually is.

We could even choose $\Omega := {(,,,...,),(,,...,),...,(,,...,)} $ as sample space. Why is it not done in this way?

So why do we have to enforce distincness of the objects when they are actually not distinct to get the disired result and why is this still correct?

OuttaSpaceTime
  • 147
  • 3
    "But in reality there is just one possibility to choose $\binom{K}{k}$ since every chosen red ball is the same and not distinct" By that logic, the black balls are also the same and not distinct and there is only one way, not $\binom{K}{k}$ ways... The point is that not every way is equally likely to occur. In order to ensure that we are working in an equiprobable sample space... we pretend that the balls are all uniquely labeled and in fact distinct in order to have the convenience of working in an equiprobable sample space to make the math easier. – JMoravitz Mar 04 '21 at 13:10
  • We don't care that the balls in reality are indistinguishable to our eye, they are in fact distinct objects by the simple fact that they occupy different physical spaces in our universe. If they were to have labels on them which make them easier for us to tell apart at sight, the presence of the labels doesn't change the probabilities at play, it just helps us to better organize our thoughts. – JMoravitz Mar 04 '21 at 13:11
  • 2
    For further intuition... compare to the simple case of $n=1$... that is the case of us selecting a single ball out of an urn with $K$ red balls and $N-K$ black balls. Clearly, the probability of picking a red ball isn't $\frac{1}{2}$ regardless what $N,K$ are... if you have one red ball and a billion black balls... you will be more likely to pick the black. You shouldn't have reason to doubt in this case for $n=1$ so why doubt in the event that $n>1$? It follows from the same intuition. – JMoravitz Mar 04 '21 at 13:15
  • this makes sense so far, especially looking at the sample space. Though I am still wondering might there not be an other way to define this distribution? I mean the sample space $\Omega$ could just be ${(r,r,r,...r,r),(r,r,...rb),...,(b,b,...,b)}$ as well. – OuttaSpaceTime Mar 04 '21 at 13:43
  • Your last example in comments of $n+1$ possible outcomes can work, but these events are not equally probable and finding their probabilities takes you back to the hypergeometric distribution – Henry Mar 04 '21 at 13:46
  • 1
    I suggest reading my answer to Why is flipping a head then a tail different than flipping a tail then a head?. The punchline is that the sample space we use is a choice and we choose the one that helps facilitate organization and/or calculations. – JMoravitz Mar 04 '21 at 13:58
  • thank you very much! Your answer added great details to my question and made it quite clear. – OuttaSpaceTime Mar 04 '21 at 14:10

0 Answers0