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?
As Charlie justly pointed out - the summations regarding the number of remaining matches N , should be taken over N=1 to 20, instead of N=2 to 20. This would cover the case of 1 match remaining in a box, but the professor choosing the other box in which also one match remained.
so the corrected solution is the following :
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 1 and 20). The number of drawn matches, of the 40 existing ones, is therefore: 40-(N+1)=(39-N).
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 :
(1) Pan = AN / (summation over N=1 to N=20 of AN)
AN will equal the number of combinations of choosing 19 entities out of a group of (39-N), which is :
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></o:p>
E=summation over N=1 to N=20 of(Pan*N)
A simple Matlab program computing this, gives the result :
E = 1.9048
Charlie - as you can see this corrected result is even further away from your solution, the reason having been explained in my first comment to you