Re: Insoluble marbles-in-urn problem?

*To*: mathgroup at smc.vnet.net*Subject*: [mg120004] Re: Insoluble marbles-in-urn problem?*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Tue, 5 Jul 2011 05:09:55 -0400 (EDT)

On 7/4/11 at 6:43 AM, koopman at sfu.ca (Ray Koopman) wrote: >On Jul 3, 1:14 am, John Feth <johnf... at gmail.com> wrote: >>There is a huge urn full of marbles, each marked with a single >>digit: 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. The marked marble >>quantities are uniformly distributed between all of the digits and >>the marbles are thoroughly mixed. You look away, choose 10 >>marbles, and put them in a black velvet bag. >>When you have some time, you look away, open the bag, and remove >>one marble. You close the bag, look at the digit on the marble, >>open a beer perhaps, and calculate the probability that there is at >>least one more marble in the bag with the same digit. >P(at least one more marble with the same digit) = >1 - P(no more marbles with the same digit) = >1 - (9/10)^9 = .612579511 This solution assumes an infinite urn. If the urn and the number of marbles in it is assumed finite, then the solution is: pDiff = 1 - Product[(9 n - k)/(10 n - k - 1), {k, 0, 8}]; which gives the same result for the limit case of n = infinity In[37]:= Limit[pDiff, n -> Infinity] == 1 - (9/10)^9 Out[37]= True