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Insoluble marbles-in-urn problem?

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  Insoluble marbles-in-urn problem? John Feth <johnfeth@gmail.com> - 2011-07-03 08:14 +0000
    Re: Insoluble marbles-in-urn problem? Ray Koopman <koopman@sfu.ca> - 2011-07-04 10:44 +0000

#3466 — Insoluble marbles-in-urn problem?

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.

The answer is brute forced below is there a formal way to obtain the
answer?  I don't believe the marbles-in-urn standby, the
hypergeometric distribution, is any help at all.

Copy and paste the algorithm below into Mathematica (V6 or newer) to
find the surprising answer, estimated from a million tests in about 16
seconds.

Timing[rag=Table[x,{x,1,1000000}];For[i=1,i<Length[rag]+1,i+
+,rag[[i]]=Table[RandomInteger[{0,9}],{n,1,10}]];bug=Table[x,{x,
1,Length[rag]}];For[i=1,i<Length[rag]+1,i+
+,bug[[i]]=RandomInteger[{1,10}]];selection=Table[rag[[i,bug[[i]]]],{i,
1,Length[rag]}];freq:=Table[Count[rag[[i]],selection[[i]]],{i,
1,Length[rag]}];bull=Tally[Characters[freq]];bullsort=Sort[bull];N[(Length[rag]-
bullsort[[1,2]])/Length[rag],10]]

Below are some definitions that might make the algorithm above a
little less opaque.

rag is a table of 10 digit random strings below

(*rag=Table[x,{x,1,3}];For[i=1,i<Length[rag]+1,i+
+,rag[[i]]=Table[RandomInteger[{0,9}],{n,1,10}]];rag
{{9,6,5,3,4,9,1,7,4,3},{8,5,7,7,0,0,5,6,3,5},{1,1,8,0,9,0,4,3,4,3}}*)

bug is a table of which digit to pick from each rag[ [ ] ] above, i.e.
the 4th from the left in rag[[1]], the 2nd from the left in rag[[2]],
etc.

(*bug=Table[x,{x,1,Length[rag]}];For[i=1,i<Length[rag]+1,i+
+,bug[[i]]=RandomInteger[{1,10}]];bug
{4,2,5}*)

selection is a table of the values of the digit picked above, i.e.,
the 4th digit in rag[[1]] is a 3, the 2nd digit in rag[[2]] is a 5,
etc.

(*selection=Table[rag[[i,bug[[i]]]],{i,1,Length[rag]}]
{3,5,9}*)

freq is a table of the number selected digits in rag[[n]], i.e., there
are two 3s in rag[[1]], three 5s in rag[[2]], one 9 in rag[[3]], etc.

(*freq=Table[Count[rag[[i]],selection[[i]]],{i,1,Length[rag]}]
{2,3,1}*)

bull tallies how many times the chosen digit occurs

(*bull=Tally[Characters[freq]]
{{Characters[2],1},{Characters[3],1},{Characters[1],1}}*)

bullsort tallies the number of times the chosen digit occurs; the
chosen digit occurred once one time (9's above), twice one time (4's
above), and once three times (5's above)

(*bullsort=Sort[bull]
{{Characters[1],1},{Characters[2],1},{Characters[3],1}}*)

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#3471

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

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