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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
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