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 >> 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. > > -- DrMajorBob at yahoo.com

**References**:**Re: Insoluble marbles-in-urn problem?***From:*Bill Rowe <readnews@sbcglobal.net>