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Simple, yet tough urn problem 
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#1
Mar512, 05:22 AM

P: 35

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
Mar512, 05:50 AM

P: 673

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



#3
Mar512, 06:07 AM

PF Gold
P: 162




#4
Mar512, 06:19 AM

PF Gold
P: 162

Simple, yet tough urn problem
Let p_{i}^{(n)} be the probability to have i empty urns at time n.
p_{i}^{(n+1)} = p_{i}^{(n)}(ki)/k + p_{i+1}^{(n)}(i+1)/k The initial condition is p_{i}^{(1)} = δ_{k1} Seems like you have to solve this recurrence. 


#6
Mar512, 07:05 AM

PF Gold
P: 162

Maybe this will work better
p_{0}^{(M)} = 1  p_{1}^{(M)}  p_{2}^{(M)}  ...  p_{k1}^{(M)} = 1  (k^{M})((k choose 1) * (k1)^{M} + ( k choose 2) * (k2)^{M} + ... +(k choose k1)1^{M}) 


#7
Mar512, 07:38 AM

P: 673

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
[itex] \sum\limits_{n=1}^{k1} (1)^{n+1} \left(1\frac{n}{k}\right)^M \left(\begin{array}{c}k\\n\end{array}\right)[/itex] Kai, what do you get? 


#8
Mar512, 07:49 AM

PF Gold
P: 162

I got
[tex] 1  \sum_{i=1}^{k1} \left( \begin{array}{c} k \\ i \\ \end{array} \right) \left(\frac{ki}{k}\right)^M[/tex] 


#9
Mar512, 08:05 AM

PF Gold
P: 162

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. 


#10
Mar512, 11:05 AM

P: 35

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


#11
Mar512, 11:14 AM

PF Gold
P: 162

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
Mar512, 11:38 AM

P: 3,014

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 [itex]p_l[/itex] of exactly l urns remaining empty? (How many M letter words can be made with k distimct letters, and how many with kl letters if you could repeat the letters and their order matters?) 


#13
Mar512, 02:16 PM

P: 35




#14
Mar512, 05:01 PM

P: 327

This is the Coupon Collector's Problem.
http://en.wikipedia.org/wiki/Coupon_...or%27s_problem Oddly enough, the Wikipedia article doesn't give the formula for the probability you asked for just expected values etc. But you can find it in many probability texts. 


#15
Mar512, 08:33 PM

P: 3,014

Label the urns from 1 to k. Then, each placement of the balls is equivalent to an Mletter word with the "letters" 1, 2, ..., k. Each of these [itex]k^M[/itex] words is equally probable.
Let [itex]x(k, M)[/itex] 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: [tex] x(k, M) = x(k  1, M  1) + x(k, M  1) [/tex] Obviously, [itex]x(k, M) = 0, \ k > M[/itex]. From here, we deduce that: [tex] x(k, k) = x(k  1, k  1) = \mathrm{const}. [/tex] 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: [tex] x(k, k + 1) = x(k  1, k) + 1 \Rightarrow x(k, k + 1) = x(1, 2) + k  1, \ k \ge 1 [/tex] Obviously, [itex]x(1, M) = 1, M \ge 1[/itex]. This boundaty condition is consistent with the above recursion relation if and only if [itex]x(0, M) = 0, M \ge 0[/itex]. To reduce the recursion w.r.t. k, we perform a discrete Laplace transform w.r.t. k: [tex] X_M(s) \equiv = \sum_{k = 0}^{\infty}{x(k, M) e^{k \, s}} [/tex] Multiply the recursion by [itex]e^{(k  1)s}[/itex] and sum w.r.t. k from 1 to ∞, to get: [tex] 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 [/tex] [tex] X_{M + 1}(s) = (1 + e^{s}) X_{M}(s), \ M \ge 1 [/tex] This recursion w.r.t. M is a geometric sequence with an easy solution: [tex] X_M(s) = X_1(s) \, (1 + e^{s})^{M  1}, \ M \ge 1 [/tex] [tex] X_1(s) = \sum_{k = 0}^{\infty}{x(k, 1) \, e^{k s}} = x(0, 1) + x(1, 1) e^{s} = e^{s} [/tex] Then, take [itex]z \equiv e^{s}[/itex], and we have: [tex] \tilde{X}_M(z) = X_M(s) = \frac{1}{z} \, \left(1 + \frac{1}{z} \right)^{M  1}, \ M \ge 1 [/tex] Then, if you expand this binomial into a Laurent series near [itex]z = 0[/itex], you can identify the coefficients as [itex]x(k, M)[/itex]. We have: [tex] \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 [/tex] So, I think the answer is: [tex] P(k, M) = \left( \begin{array}{c} M  1 \\ k  1 \end{array}\right) \, \frac{1}{k^M} [/tex] 


#16
Mar512, 09:36 PM

P: 2,504

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 1P(x=0). So for 20 balls and 10 urns P(x=0) = 0.1216 and 10.1216= 0.8784 


#17
Mar612, 12:59 AM

P: 73

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 lowerlabeled urns to the left of higherlabeled 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 M1 choose k1 ways to place the urns, since k1 of the M1 distinct spots between balls must be chosen for urns 1 through k1, 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 j1). We still need urn k to go last since otherwise we would not have placed every ball into an urn. There are M+k1 choose k1 ways to place the urns now, since urn k must be placed last, and before it are M+k1 spots on which to place objects, exactly k1 of which must be occupied by urns. Thus the probability each urn contains a ball is (M1 choose k1)/(M+k1 choose k1). 


#18
Mar612, 03:15 AM

P: 673

Thanks JCVD, that explanation is very clear.



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