Re: Insoluble marbles-in-urn problem?

• To: mathgroup at smc.vnet.net
• Subject: [mg120003] Re: Insoluble marbles-in-urn problem?
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Tue, 5 Jul 2011 05:09:44 -0400 (EDT)
• References: <201107041044.GAA02477@smc.vnet.net>
• Reply-to: drmajorbob at yahoo.com

```> 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}]

Yes, and taking that to the limit (a VERY large urn), we have

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

(9 n (-8 + 9 n) (-7 + 9 n) (-6 + 9 n) (-5 + 9 n) (-4 + 9 n) (-3 +
9 n) (-2 + 9 n) (-1 + 9 n))/((-9 + 10 n) (-8 + 10 n) (-7 +
10 n) (-6 + 10 n) (-5 + 10 n) (-4 + 10 n) (-3 + 10 n) (-2 +
10 n) (-1 + 10 n))

612579511/1000000000

{N@%, % == 1 - (9/10)^9}

{0.61258, True}

Bobby

On Mon, 04 Jul 2011 05:44:24 -0500, Bill Rowe <readnews at sbcglobal.net>
wrote:

> 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
>
> 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.
>
>

--
DrMajorBob at yahoo.com

```

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