# Simple, yet tough urn problem

1. Mar 5, 2012

### JolleJ

You have k urns and M (M>k) balls.

Each ball is placed in a random urn (w/ uniform distrubution).

What is the probability that every urn has at least one ball in it after the M balls are placed?

2. Mar 5, 2012

### M Quack

What is the probability that a given urn will receive no balls?

3. Mar 5, 2012

### kai_sikorski

That's not really the right approach because the urns are not independent and the OP is asking a question not about a given urn but about ALL the urns.

4. Mar 5, 2012

### kai_sikorski

Let pi(n) be the probability to have i empty urns at time n.

pi(n+1) = pi(n)(k-i)/k + pi+1(n)(i+1)/k

The initial condition is

pi(1) = δk-1

Seems like you have to solve this recurrence.

Last edited: Mar 5, 2012
5. Mar 5, 2012

### kai_sikorski

EDIT: Err this was wrong

Last edited: Mar 5, 2012
6. Mar 5, 2012

### kai_sikorski

Maybe this will work better

p0(M) = 1 - p1(M) - p2(M) - ... - pk-1(M)
= 1 - (k-M)((k choose 1) * (k-1)M + ( k choose 2) * (k-2)M + ... +(k choose k-1)1M)

Last edited: Mar 5, 2012
7. Mar 5, 2012

### M Quack

I know, I know, that was not supposed to be the complete answer. But you can build up from there. Following my route I get

$\sum\limits_{n=1}^{k-1} (-1)^{n+1} \left(1-\frac{n}{k}\right)^M \left(\begin{array}{c}k\\n\end{array}\right)$

Kai, what do you get?

8. Mar 5, 2012

### kai_sikorski

I got

$$1 - \sum_{i=1}^{k-1} \left( \begin{array}{c} k \\ i \\ \end{array} \right) \left(\frac{k-i}{k}\right)^M$$

9. Mar 5, 2012

### kai_sikorski

Applying your formula to just the case k = 3 , we get

3(2/3)M - 3(1/3)M ≈ 3 (2 M )/(3 M )

But above doesn't go to 1. The probability that every urn has a ball must clearly go to 1 if you keep adding more and more balls.

Last edited: Mar 5, 2012
10. Mar 5, 2012

### JolleJ

I been thinking sort of down the roads of what you're doing...
@Kai: Your suggestion gives negative results for m=100, k=50...

Last edited: Mar 5, 2012
11. Mar 5, 2012

### kai_sikorski

Hmm, yeah I see my problem. I'm trying to condition on there being a specific number of empty urns, but I'm calculating the probability to have more than or equal to that many.

12. Mar 5, 2012

### Dickfore

The event whose probability is required is opposite to the event:
After M independent, equally probable fillings of k urns, there is at least one (meaning either one, or two, ... , or k) being empty.

What is the probability $p_l$ of exactly l urns remaining empty? (How many M letter words can be made with k distimct letters, and how many with k-l letters if you could repeat the letters and their order matters?)

13. Mar 5, 2012

### JolleJ

Yes, but I am not having an easier time calculating this.

Knowing this would obviously solve my problem; but this is an even harder question. Harder in the sense that my need (l = 0) is a subproblem.

14. Mar 5, 2012

### awkward

15. Mar 5, 2012

### Dickfore

Label the urns from 1 to k. Then, each placement of the balls is equivalent to an M-letter word with the "letters" 1, 2, ..., k. Each of these $k^M$ words is equally probable.

Let $x(k, M)$ be the number of M letter words in which each of the k letters of the alphabet occurs at least once.

We can divide this set into two disjoint subsets: A set in which 1 enters exactly once and in which it occurs more than once. The first subset has x(k - 1, M - 1) elements. The second x(k, M - 1). Therefore, we have the recursion:
$$x(k, M) = x(k - 1, M - 1) + x(k, M - 1)$$

Obviously, $x(k, M) = 0, \ k > M$.

From here, we deduce that:
$$x(k, k) = x(k - 1, k - 1) = \mathrm{const}.$$
This constant is obviously equal to 1, since there is only one way of filling k urns with k balls (by placing exactly one ball in each urn).

Then, we have:
$$x(k, k + 1) = x(k - 1, k) + 1 \Rightarrow x(k, k + 1) = x(1, 2) + k - 1, \ k \ge 1$$
Obviously, $x(1, M) = 1, M \ge 1$. This boundaty condition is consistent with the above recursion relation if and only if $x(0, M) = 0, M \ge 0$.

To reduce the recursion w.r.t. k, we perform a discrete Laplace transform w.r.t. k:
$$X_M(s) \equiv = \sum_{k = 0}^{\infty}{x(k, M) e^{-k \, s}}$$
Multiply the recursion by $e^{-(k - 1)s}$ and sum w.r.t. k from 1 to ∞, to get:
$$e^{s} \, \left[ X_M(s) - x(0, M) \right] = X_{M - 1}(s) + e^{s} \, \left[ X_{M - 1}(s) - x(0, M - 1) \right], \ M > 1$$

$$X_{M + 1}(s) = (1 + e^{-s}) X_{M}(s), \ M \ge 1$$
This recursion w.r.t. M is a geometric sequence with an easy solution:
$$X_M(s) = X_1(s) \, (1 + e^{-s})^{M - 1}, \ M \ge 1$$

$$X_1(s) = \sum_{k = 0}^{\infty}{x(k, 1) \, e^{-k s}} = x(0, 1) + x(1, 1) e^{-s} = e^{-s}$$

Then, take $z \equiv e^{s}$, and we have:
$$\tilde{X}_M(z) = X_M(s) = \frac{1}{z} \, \left(1 + \frac{1}{z} \right)^{M - 1}, \ M \ge 1$$

Then, if you expand this binomial into a Laurent series near $z = 0$, you can identify the coefficients as $x(k, M)$. We have:
$$\tilde{X}_M(z) = \frac{1}{z} \, \sum_{k = 0}^{M - 1}{\left( \begin{array}{c} M - 1 \\ k \end{array}\right) \, \frac{1}{z^k}} = \sum_{k = 1}^{M}{\left( \begin{array}{c} M - 1 \\ k - 1 \end{array}\right) \, \frac{1}{z^k}} \Rightarrow x(0, M) = 0, x(k, M) = \left( \begin{array}{c} M - 1 \\ k - 1 \end{array}\right), 1 \le k \le M, x(k, M) = 0, k > M$$

So, I think the answer is:
$$P(k, M) = \left( \begin{array}{c} M - 1 \\ k - 1 \end{array}\right) \, \frac{1}{k^M}$$

16. Mar 5, 2012

### SW VandeCarr

The OP defined M>k, but I prefer n=M, n>k as more standard notation for the binomial distribution.

Then we have n events with 1/k probability. The binomial distribution will have a non zero probability for zero events for any value of n given P=1/k. An event is a ball in an urn. The mean (expected) value would be n/k assuming all balls go into an urn.

EDIT: The probability that every urn has at least one ball should be 1-P(x=0). So for 20 balls and 10 urns P(x=0) = 0.1216 and 1-0.1216= 0.8784

Last edited: Mar 6, 2012
17. Mar 6, 2012

### JCVD

This is just a standard exercise from enumerative combinatorics.

Label the urns 1 through k. How many ways can we place the M balls in the k urns so that each urn gets a ball? Line up the balls and urns in a straight line, with lower-labeled urns to the left of higher-labeled urns and with the leftmost object being a ball, the rightmost object being an urn, and no two urns adjacent. Put all balls to the left of urn 1 into urn 1, then put all remaining balls to the left of the urn 2 into urn 2, and repeat until each ball is in an urn. There are M-1 choose k-1 ways to place the urns, since k-1 of the M-1 distinct spots between balls must be chosen for urns 1 through k-1, with urn k going on the very end.

Now how many total ways are there to put the balls in the urns? We repeat the process above, except we take away the requirements that the leftmost object be a ball (so that urn 1 can be empty) and that urns cannot be adjacent (so that for j>1 urn j can be made empty by putting it directly after urn j-1). We still need urn k to go last since otherwise we would not have placed every ball into an urn. There are M+k-1 choose k-1 ways to place the urns now, since urn k must be placed last, and before it are M+k-1 spots on which to place objects, exactly k-1 of which must be occupied by urns.

Thus the probability each urn contains a ball is (M-1 choose k-1)/(M+k-1 choose k-1).

18. Mar 6, 2012

### M Quack

Thanks JCVD, that explanation is very clear.

Well spotted. I was trying to calculate the opposite probability, that there is at least one empty urn. But even that I got wrong.

Last edited: Mar 6, 2012
19. Mar 6, 2012

### SW VandeCarr

I got P= 0.8784 for n(M)=20, k=10 using the binomial distribution. Does anyone want the check JCVD's solution against this for these values?

Last edited: Mar 6, 2012
20. Mar 6, 2012

### JolleJ

Thank you very much JCVD ! What a great way to do it.