4

As stated in title, there are $z$ things to pick from, and you get $y$ picks, with replacement. What's the probability of picking such that you get at least one of each of $x$ things? Assume $x \leq y$ and $x \leq z$, and order doesn't matter. It seems like it should just be an extension of stars and stripes or balls in boxes but I'm having trouble getting it right.

Stars and stripes seems like it would be choosing types for $y-x$ things, since $x$ are set, which means distributing $z-1$ bars in those $y-x$ things for $\binom{y-x+z-1}{z-1}$ for an overall probability of that of $z^y$, which I reduced to $\frac{(y-x+z-1)!}{(z-1)! (y-x)! z^y}$. However as $y$ goes to infinity that seems to go to $\frac{y^z}{z^y}$, when intuitively it should go to 1.

For boxes I'm not sure of how to represent "at least 1 of each," and the inverse is not really that simple either.

Thanks!

Community
  • 1
colblitz
  • 41
  • Does it matter which $x$ items? That is, are there $x$ items specified in advance that you want to get, or do you just want to get any $x$ different items among your picks? – paw88789 May 11 '15 at 10:21
  • The $x$ items are specified in advanced (would getting from there to any $x$ different items just be a factor of z choose x?) – colblitz May 11 '15 at 13:25
  • See my comment following the answer of Valentin. – user2661923 Sep 12 '23 at 02:14
  • You're asking about a probability without specifying a distribution. Usually, when people don't specify a distribution, they intend to imply that something is distributed uniformly; but here it's not quite clear what's uniformly distributed. The first sentence sounds as if each pick is equally likely to pick any of the $z$ things; but then it's unclear what you mean by "order doesn't matter". The answer by Valentin seems to interpret this to mean that it's in fact the (unordered) multisets of items you picked that are equiprobable. Please clarify the question. – joriki Jan 13 '24 at 21:05

1 Answers1

0

Think about it in terms of balls and bins. You have to distribute $y$ identical balls among $z$ bins. $x$ predefined bins should contain at least one ball each (WLOG, first $x$ bins). You can firstly put one ball into each of the $x$ first bins, and then distribute remaining $y-x$ balls among $z$ bins. As you know from stars and bars, the total number of ways of doing it equals $z + y - x - 1 \choose z$. The total number of ways of distributing your balls is $z + y - 1 \choose z$. The answer is therefore ${z + y - x - 1 \choose z} / {z + y - 1 \choose z}$.

Valentin
  • 265
  • I could be mistaken. It seems that you are assuming that each Stars and Bars solution is equally likely. I suspect that that assumption is inaccurate. This is why Stars and Bars, while a good tool for Combinatorics problems, is not typically appropriate for Probability problems. Again, I could be mistaken. – user2661923 Sep 12 '23 at 02:13