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Re: Insoluble marbles-in-urn problem?

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On 7/4/11 at 6:43 AM, koopman@sfu.ca (Ray Koopman) wrote:

>On Jul 3, 1:14 am, John Feth <johnf...@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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Re: Insoluble marbles-in-urn problem? Bill Rowe <readnews@sbcglobal.net> - 2011-07-05 09:11 +0000

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