Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2011

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Insoluble marbles-in-urn problem?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg119988] Re: Insoluble marbles-in-urn problem?
  • From: Bill Rowe <readnews at sbcglobal.net>
  • Date: Mon, 4 Jul 2011 06:44:24 -0400 (EDT)

On 7/3/11 at 4:11 AM, johnfeth at gmail.com (John Feth) 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?

Yes. I will assume in your urn the number of marbles marked with
each digit is exactly the same. Assume there are n sets of
marbles marked with the digits 0 through 9 in the urn. Draw out
1 marble. Then you problem becomes 1 minus the probability of
not drawing another marble marked the same in 9 additional draws.

In the second draw, there are 10 n -1 marbles in the urn and 9 n
marbles marked differently than the first. Hence the probability
of getting a differently marked marble is 9n/(10n-1).

On the third draw there are 10n-2 marbles in the urn and 9n-1
marbles different than the first marble drawn assuming the
second marble is different than the first. So, the probability
of the not matching the first marble in two additional draws is

9n/(10n -1) (9n-2)/(10n-2). Continuing in this manner it is
clear the probability of not drawing another marble like the
first in 9 additional draws is

Product[(9 n - k)/(10 n - 1 - k),{k,0,8}]

Hence the probability of at getting at least one more marble
like the first in 9 additional draws is

1 - Product[(9 n - k)/(10 n - 1 - k),{k,0,8}]

Note, to create a simulation of this problem you do not want to
simulate your urn of marbles with RandomInteger[{0,9},10 n}]
since this will not ensure there are exactly the same number of
marbles marked with each digit. Nor do you need your simulated
urn to have a random order. It is sufficient to make a random
choice regardless of whether the array representing the urn is
ordered or not.

So, for 5 sets

In[13]:= pDiff = Product[(9 n - k)/(10 n - k - 1), {k, 0, 8}];

In[14]:= N[pDiff /. n -> 5]

Out[14]= 0.431337

That is there is a 57% chance of getting at least one marble
like the first in the next 9. Simulate an urn with 50 marbles by:

In[15]:= urn = Flatten@Table[Range[0, 9], {5}];

Repeat the simulation 1000 times

In[16]:= Total@
  Table[Boole[
    MemberQ[Rest@#, First@#] &[RandomSample[urn, 10]]], {1000}]

Out[16]= 581

That works out to 58.1% which is reasonably close to the
expected value of 57%.

Note if I make urn

In[18]:= urn = RandomInteger[{0, 9}, 50];

In[19]:= Tally[urn]

Out[19]= {{9, 5}, {6, 6}, {1, 5}, {2, 7}, {8, 6}, {0, 5}, {3,
5}, {5,
   4},
    {4, 4}, {7, 3}}

There is not exactly 5 of each digit. So, the simulated
probability will vary according to how the first marble drawn is marked.



  • Prev by Date: Re: How to write a "proper" math document
  • Next by Date: Re: Help with old Mathematica file
  • Previous by thread: Re: Insoluble marbles-in-urn problem?
  • Next by thread: Re: Insoluble marbles-in-urn problem?