       Re: Insoluble marbles-in-urn problem?

• To: mathgroup at smc.vnet.net
• Subject: [mg120002] Re: Insoluble marbles-in-urn problem?
• From: Dana DeLouis <dana01 at me.com>
• Date: Tue, 5 Jul 2011 05:09:34 -0400 (EDT)

```.... I don't believe the marbles-in-urn
hypergeometric distribution, is any help at all.

Hi.  As the size of the urn gets large, the numbers 0-9 represent 1/10th the size.
So, as m goes to infinity...

1-Limit[PDF[HypergeometricDistribution[10,m/10,m],1],m->\[Infinity]]

612579511/1000000000

Check...

1-(9/10)^9 == %
True

As a side note, I couldn't remember the equation, so I copied the equation from Excel.
I couldn't get Mathematica to pick up on the hypergeometric function.

We note that given a size of 10, the total count is 10 times larger.

(Binomial[M,x] Binomial[-M+N,n-x])/Binomial[N,n] /.N->M*10

(Binomial[M,x] Binomial[9 M,n-x])/Binomial[10 M,n]

Limit as size of Urn gets large:

1-Limit[%,M->\[Infinity]]

1-(9^(n-x) 10^-n Gamma[1+n])/(Gamma[1+n-x] Gamma[1+x])

Given 1 out of 10...

%/.{x->1,n->10}
612579511/1000000000

%//N
0.61258

Which matches what others have given.

= = = = = = = = = =
HTH  : >)
Dana DeLouis
\$Version
8.0 for Mac OS X x86 (64-bit) (November 6, 2010)

On Jul 3, 4:14 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.
>
> 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[(Le=
ngth[ 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[], the 2nd from the left in rag[],
> 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[] is a 3, the 2nd digit in rag[] 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[], three 5s in rag[], one 9 in rag[], 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,1},{Characters,1},{Characters,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},{Characters,1},{Characters,1}}*)

```

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