Q.5 A matchbox contains some matches. The number of matches is doubled, and
8 are taken away. The number of matches is doubled again, and again 8 are taken
away. This is done for one more time; the box gets empty. How many matches
were there in the matchbox originally?
Answers
Answer:
mathematician carries two matchboxes at all times: one in his left pocket and one in his right. Each time he needs a match, he is equally likely to take it from either pocket. Suppose he reaches into his pocket and discovers for the first time that the box picked is empty. If it is assumed that each of the matchboxes originally contained n matches, what is the probability that there are exactly k matches in the other box?
I'm wondering whether the following reasoning is right, because it doesn't match up with the correct probability. But here goes my reasoning:
Assuming there are k matches left in the other box, we have had to take 2n−k+1 matches to notice there are none left. The total number of ways in which we could have picked those is 2(2n−kn), for either the matchbox in the left or right pocket has k matches left inside, and in the (2n−k+1)th pick we would have found an empty box.
The total number of possibilities, taken over all possible sizes k, would then be ∑nm=02(2n−mn). So I'd assume the overall probability would be 2(2n−kn)∑nm=02(2n−mn)=(2n−kn)∑nm=0(2n−mn).
However, the mentioned solution is (2n−kn)(12)2n−k. Where is my reasoning wrong? Thanks
Answer:
qsdfghjklzxcvbnmqwertyuiop