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?
(In reply to
re: Corrected Solution after Charlie's comments by Charlie)
The probability P20 you computed answers the following question :
"The Professor has drawn randomly exactly 19 matches. What is the probability that all of them were drawn from the same box."
The probability P20 should answer a completely different scenario, namely :
"The professor has drawn randomly an unknown number of matches, stopping when either one of the boxes is left with one match. What is the probability that when that occurred, 20 matches were left in the other box."
The probability of having 20 matches left in one box (say "A") should therefore not be calculated, as you did, only over the cases of 19 draws, but instead, over the ensemble of all possible outcomes which result in the given situation of one left match in box "B" and an unknown number between 1 to <st1:metricconverter w:st="on" ProductID="20 in">20 in</st1:metricconverter> box "A.
<o:p> As for your analysis of the case of P1 - you assumed that there are 2^38 different options, whereas in reality there are only 2^37. Using this you would get my results (within computational errors).</o:p>
<o:p>Explanation : After the 37th draw, the situation inside the boxes must be either A=2, B=1 or A=1, B=2. In order for the 38th draw to double the options from 2^37 to 2^38, there must be 2 possible ways to continue each series - either by drawing from A or by drawing from B, but in reality, each of the resulting options after the 37th draw does offer only one possible continuation: all the combinations that resulted in A=2, B=1 force a continuation of drawing from A only, and all the rest force a draw from box B. The other half of the choices would cause the stopping of the process and the addition of this case to the number of results which left 2 matches in the "other" box.</o:p>
<o:p> </o:p>
re(2): Corrected Solution after Charlie's comments
