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Matching Matchbox Muse (Posted on 2010-03-30) Difficulty: 3 of 5
Professor X smokes a pipe. He carries two identical matchboxes, originally containing 20 matches each. When he lights his pipe, he chooses a matchbox at random and lights his pipe with one match and discards the used match.

There will eventually arise an occasion when he first selects a matchbox with only one match in it. At this point, what is the expected number of matches in the other box?

No Solution Yet Submitted by K Sengupta    
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Solution Solution | Comment 3 of 15 |

The event which stops the process is the occurrence of a single match in either box, while in the other one remain N matches ( N being between 2 and 20). The number of drawn matches, of the 40 existing ones, is therefore:  


<o:p> </o:p>


AN=The number of possible different sequences of drawn matches from the 2 boxes, given the number of drawn matches from box A or box B, is 19, and the total group of drawn matches is (39-N).

The probability of the occurrence of a specific N will therefore be :

<o:p>      (1)  Pan=AN/”(AN)   (” is over N=2:20) </o:p>
<o:p></o:p>     <v:shapetype id=_x0000_t75 stroked="f" filled="f" path="m@4@5l@4@11@9@11@9@5xe" o:preferrelative="t" o:spt="75" coordsize="21600,21600"><v:stroke joinstyle="miter"></v:stroke><v:formulas><v:f eqn="if lineDrawn pixelLineWidth 0"></v:f><v:f eqn="sum @0 1 0"></v:f><v:f eqn="sum 0 0 @1"></v:f><v:f eqn="prod @2 1 2"></v:f><v:f eqn="prod @3 21600 pixelWidth"></v:f><v:f eqn="prod @3 21600 pixelHeight"></v:f><v:f eqn="sum @0 0 1"></v:f><v:f eqn="prod @6 1 2"></v:f><v:f eqn="prod @7 21600 pixelWidth"></v:f><v:f eqn="sum @8 21600 0"></v:f><v:f eqn="prod @7 21600 pixelHeight"></v:f><v:f eqn="sum @10 21600 0"></v:f></v:formulas><v:path o:connecttype="rect" gradientshapeok="t" o:extrusionok="f"></v:path><o:lock aspectratio="t" v:ext="edit"></o:lock></v:shapetype><v:shape id=_x0000_i1025 style="WIDTH: 80.25pt; HEIGHT: 33pt" type="#_x0000_t75" o:ole=""><v:imagedata o:title="" src="file:///C:DOCUME~1DanLOCALS~1Tempmsohtml11clip_image001.wmz"></v:imagedata></v:shape>

AN will equal the number of combinations of choosing 19 entitiesout of a group of 39, which is :

<o:p> </o:p>

AN=2*(39-N)!/(20-N)!/19                !

The factor 2  takes care of the possibility that the box with one match may be either box A or box B.

Substituting eq.(2) into eq.(1) gives Pan as function of N, and allows the computation of the requested expectance E as :


<o:p>E=”(Pan*N)        (”is taken over N=2:20             </o:p>

<v:shape id=_x0000_i1027 style="WIDTH: 78.75pt; HEIGHT: 33.75pt" type="#_x0000_t75" o:ole=""><v:imagedata o:title="" src="file:///C:DOCUME~1DanLOCALS~1Tempmsohtml11clip_image005.wmz"></v:imagedata></v:shape>

The very large numbers of the factorials make the computation somewhat tricky, so I wrote a small Matlab program which takes care of this problem :

<o:p> </o:p>


for N=2:20<o:p></o:p>


for m=2:20<o:p></o:p>


    for q=1:18<o:p></o:p>









<o:p> </o:p>

The resulting expectance   is :

<o:p> </o:p>


  Posted by Dan Rosen on 2010-06-11 15:44:29
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