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Insoluble marbles-in-urn problem?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg119982] Insoluble marbles-in-urn problem?
*From*: John Feth <johnfeth at gmail.com>
*Date*: Sun, 3 Jul 2011 04:11:34 -0400 (EDT)
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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