[Guest]

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 come 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?

[Guest]

It could be any number between 2-20. Chances are, he wouldn't choose the same box 19 times in a row, but it is possible.

[Guest]

It could be any number between 2-20. Chances are, he wouldn't choose the same box 19 times in a row, but it is possible.

No, it could be any number between ONE and 20. The previous time could have been a time to take one match from a box that held two.

Lark

[Izzy]

While it COULD be anywhere between 1-20, that's not what's EXPECTED. I'm thinking 1-2? Some sorta 50-50 going on here. Like flipping a coin 40 times.

[Guest]

The puzzle could be solved by considering a reverse scenario. That is, we consider that both match boxes have one match each and now we randomly put matches into one of the boxes until both boxes have more than 1 match.

Meaning, we need to see what is the expected number of matches we can put in box A before any match is put in box B.

Now at the beginning each matchbox has 1 match, so the first matchb can be put in either of the boxes. Now, as per the above statement, this matchbox with 2 matches is box A and the other is matchbox B.

Now, the probability that the second match is put in box A is 1/4

The prob that third match is also put in box A is 1/2*1/2 = 1/8

The prob of fourth match also in box A is 1/16

and so on

So, the overall expected number of matches is:

2/2 + 3/4 + 4/8 + 5/16 + 6/32 + 7/64 +…

This sum is equal to 3.

So, when one matchbox has only 1 match left, the expected number of matches in the other matchbox is 3

Edited January 6, 2010 by DeeGee

[plainglazed]

...there would be (almost exactly/a smidge less than) five left in the other box. Add up the odds that 19 matches will be picked out of box A in 19 lights + 19 matches in 20 lights...+ 19 matches in 38 lights you get as expected a probability of 1. But the same thing is going on with box B. Adding the probabilities of both box A and B as above, you get a probability of one (.99911967) after 34 lights. So counting backwards, if you picked 19 matches in a row from the same box there would be 20 left in the other box, 19 out of 20 there would be 19 left in the other box, 19 out of 21 there would be 18 left...19 out of 34 there would be 5 left in the other box.

[superprismatic]

If my calculations are correct, the expected number of

matches in the other box is

Sum{i from 0 to 19}[((i+19) Choose i)*(20-i)/2^(i+19)]

which is approximately 5.014827

[Guest]

If my calculations are correct, the expected number of

matches in the other box is

Sum{i from 0 to 19}[((i+19) Choose i)*(20-i)/2^(i+19)]

which is approximately 5.014827

That is the brute force method that I also thought of. I tried it on a pocket calculator but ran out of patience .

[Guest]

to answer this question, one simply needs to find the maximum of the following equation.

let n= the number of matches to begin with. in this case, 20.

this equation gives the probability of x matches remaining.

_2n−xC_n / 2^2n−x

[plainglazed]

to answer this question, one simply needs to find the maximum of the following equation.

let n= the number of matches to begin with. in this case, 20.

this equation gives the probability of x matches remaining.

_2n−xC_n / 2^2n−x

If I understand the OPs question and your response, the answer would be x as defined above. So simply posed, what is your x?

[Guest]

The puzzle could be solved by considering a reverse scenario. That is, we consider that both match boxes have one match each and now we randomly put matches into one of the boxes until both boxes have more than 1 match.

Meaning, we need to see what is the expected number of matches we can put in box A before any match is put in box B.

Now at the beginning each matchbox has 1 match, so the first matchb can be put in either of the boxes. Now, as per the above statement, this matchbox with 2 matches is box A and the other is matchbox B.

Now, the probability that the second match is put in box A is 1/4

The prob that third match is also put in box A is 1/2*1/2 = 1/8

The prob of fourth match also in box A is 1/16

and so on

So, the overall expected number of matches is:

2/2 + 3/4 + 4/8 + 5/16 + 6/32 + 7/64 +…

This sum is equal to 3.

So, when one matchbox has only 1 match left, the expected number of matches in the other matchbox is 3

Technically, you need to start with zero matches in both boxes and go until each box has at least one match. Put one match back in a random box, and call that box A. This is because when I select a box, I take a match out of it, so at the end, when I select the box with 1 match in it, that box ends up empty, so reversing that step would involve taking an empty box and putting one match in back into it. Not a big deal, but I think it will throw off your answer by 1. Also, did you put a limit on the number of matches? Remember that he only started with 20 in each box, so at some point, reversing the process will force us to put a match back into box B, ending the process. Finally, P(first match goes into A) = 1, since we assumed it to be true, so P(second match goes into A) is actually 1/2, since we already know that the first one went into A.

So anyway, using your process and my logic, we have...

P(1 match in A when B gets its first one) = 1/2

P(2 matches " " " " " " " ") = 1/4

P(3...) = 1/8

P(N...) = 1/2^N

So our expected number of matches in A when B gets its first is 1*P(1...) + 2*P(2...) + 3*P(3...) + ... + 20*P(20...) = 1/2 + 2/4 + 3/8 + 4/16 + ... + 20/1048576.

This is equal to...lemme bust out the calculator...1.99997901017.... So not quite 2. In fact, Wolfram Alpha says that if we keep increasing the number of matches originally in each box, then our answer will approach 2.