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?
Charlie's solution is mistaken on 2 accounts :
1. The opening situation, at which point we are asked about the remaining matches, is that in one box is still one match left ( and not zero matches), therefore the group of drawn matches we observe should be (39-N) and not (40-N).
2.The probability of having N matches left in the other box, is a conditional probability- given that 19 matches have been drawn from one box. Therefore, the ensamble out of which we must calculate the different possible combinations of drawing 19 matches out of a group of (39-N), does not include all possible outcomes of sequences having the length of (39-N), as Charlie assumes. The ensamble to be observed is the ensamble of the group of outcomes that conforms with the opening situation, i.e.- (39-N) drawn matches, 19 of which are from one box !! Choosing the ensamble of 2^(39-N), includes for example, for N=8, all possible sequences having the length of 31 drawn matches, regardless of the fact that there must be 19 of one box.
I am posting a new print out of my solution , which will further clarify the situation.
re Charlie's solution
