Problem 151

[pelensing]

For four jobs, starting with an A3 paper, the expected number of times to find exactly one sheet in the envelope (excluding first and last) is 0.5, but that doesn't help you much, does it?

Can someone explain why? Somehow I think the chance is only .25. There are only two ways to put the unused sheets back in the envelope?

batch 1 a3 (take a3, cut in a4, a4 then a4 a5 a5, take a5, put a4 and a5 back
batch 2 a4 a5 (take a4, cut in a5 a5, put a5 back)
batch 3 a5 a5
batch 4 a5

or

batch 1 a3
batch 2 a4 a5 (take a5)
batch 3 a4 (cut in 2, place a5 back)
batch 4 a5

So 0.25 of the batches that count the envelope will contain 1 sheet?

[mdean]

For four jobs, starting with an A3 paper, the expected number of times to find exactly one sheet in the envelope (excluding first and last) is 0.5, but that doesn't help you much, does it?

Can someone explain why? Somehow I think the chance is only .25. There are only two ways to put the unused sheets back in the envelope?

batch 1 a3 (take a3, cut in a4, a4 then a4 a5 a5, take a5, put a4 and a5 back
batch 2 a4 a5 (take a4, cut in a5 a5, put a5 back)
batch 3 a5 a5
batch 4 a5

or

batch 1 a3
batch 2 a4 a5 (take a5)
batch 3 a4 (cut in 2, place a5 back)
batch 4 a5

So 0.25 of the batches that count the envelope will contain 1 sheet?

You listed 2 cases. In the first case, you didn't find a single sheet in the envelope except for the first and last jobs. In the second case, you found a single sheet once.

[demongolem]

I am having trouble with the problem definition for some reason. I have not started the problem, but i want to understand what is to be included first. So the first batch does involve 1 piece of paper (A1) but that is discarded. The next batch is the second batch which starts with One A2, one A3, one A4, and one A5 which can be sort of seen from the diagram. This is included. Do I have my start conditions correct? Then the last batch (and I hope I don't say too much in this regards) would always be 1 A5 and that is the 16th and final batch which is also discarded.

If this is in fact correct as stated, I really don't think I should have any problem getting the actual answer.

[TripleM]

I'm not sure what you mean by 'included'.

At the start of day two the envelope contains four pieces of paper, like you say - A2, A3, A4, and A5.

That doesn't count towards the final answer however; you want to count days where the envelope starts with one piece of paper (excluding the first where there is always one A1, and the last where there is always one A5).

[harshits2808]

For four jobs, starting with an A3 paper, the expected number of times to find exactly one sheet in the envelope (excluding first and last) is 0.5, but that doesn't help you much, does it?

That actually helped me solve it!
For some reason i thought the answer would be 1/3, then i looked carefully and banged my head on the wall for a few minutes

After that it was just a bit of change in my code and correct answer!

Thank you!

[woega]

For four jobs, starting with an A3 paper, the expected number of times to find exactly one sheet in the envelope (excluding first and last) is 0.5, but that doesn't help you much, does it?

That actually helped me solve it!
For some reason i thought the answer would be 1/3, then i looked carefully and banged my head on the wall for a few minutes

After that it was just a bit of change in my code and correct answer!

Thank you!

ah that sounds promising, my code for A3 also returns 0.33333 . banged my head against the wall a few times but didn't work what am I doing wrong?

when starting with A3 you basically start with 1 piece A4 and 1 piece A5, leaving 2 possible options:
- pick A4 first -> cut in half, now you have 2 A5 pieces (outside the envelope, the other one is still inside), pick 1 and put the rest back, so the envelope contains only two A5 pieces. Next step you take 1 of 2 -> 2 choices. So this 'branch' delivers a total of 2 paths without occurrences of single paper in the envelope.
- pick A5 first, remove it, next step only A4 (single!), cut it in half, pick an A5 and put the other one back. So here is 1 branch with a single occurrence.

total 3 branches possible, 1 of which has an occurrence of 1 paper. So expected value would be 1/3 * 1 + 2/3 * 0 = 1/3
where am I going wrong?

[mdean]

when starting with A3 you basically start with 1 piece A4 and 1 piece A5, leaving 2 possible options:
- pick A4 first -> cut in half, now you have 2 A5 pieces (outside the envelope, the other one is still inside), pick 1 and put the rest back, so the envelope contains only two A5 pieces. Next step you take 1 of 2 -> 2 choices. So this 'branch' delivers a total of 2 paths without occurrences of single paper in the envelope.
- pick A5 first, remove it, next step only A4 (single!), cut it in half, pick an A5 and put the other one back. So here is 1 branch with a single occurrence.

total 3 branches possible, 1 of which has an occurrence of 1 paper. So expected value would be 1/3 * 1 + 2/3 * 0 = 1/3
where am I going wrong?

Are each of those branches equally likely?

[woega]

Are each of those branches equally likely?

yeah thanks, got me thinking and I was finally able to solve it. gracias!

[Circling]

So here is how I understand how the "cut in half" procedure works:
Step one: Pull out a paper from the envelope. In this case, it is an A2 paper.
Step two: Cut the A2 to get two A3's
Step three: Cut one of the A3's to get two A4's
Step four: Cut one of the A4's to get two A5's
Step five: Take one of the A5's, put everything else in the envelope.

So in this case, if we had 1 A2 and 1 A5, we end up with 1 A3, 1 A4, and 2 A5's.

If this interpretation is correct, then my code should work. And yet, it returns 23 / 9 when starting with 1 A3 paper (5 / 9 if the 2 batches are excluded)

[mdean]

Your interpretation of the procedure is correct. The conclusion is not correct.

[hamsterofdeath]

i don't understand why the expected value of selected a5 in the case 00100 is 0.5
00100 does not count, we start with 00011
we pick a5 first -> 1 is found before the game ends
we pick a4 first (which gives us 2*a5) -> 1 a5 is found before the game ends
therefore f(00100) = 1

what's my mistake?

[mdean]

The problem says it excludes the first and last batch of the week. The last batch being the one with a single a5 left in the envelope.

[hamsterofdeath]

maybe i can't read, but - what exactly am i supposed to calculate for this problem? i need to ask some stupid questions now:

do i need:
* the number of cases a supervisor picks a5 out of the bag? according to my math, the solution is around 7
* the number of cases he encounters exactly 1 a5 sheet along the way? then the probability is zero because you are guaranteed to have more than one a5 sheet left over

so i seem to not get this problem at all
what is a batch?
how does it relate to a week? are weeks relevant?
how does a batch relate to the a1 sheet? is a batch 16 a5 picks? is a batch a single a5 pick?

"find the expected number of times (during each week) that the supervisor finds a single sheet of paper in the envelope."

ah, i'm supposed to count how often the envelope contains a single sheet of any size?

[mdean]

ah, i'm supposed to count how often the envelope contains a single sheet of any size?

Except for the a1 (which is the original state of the envelope at the beginning of the week) and a5. If all that is left is a single a5, it means you're at the last batch of the week as taking it empties the envelope.

Any time the envelope contains a single sheet of a2, a3, or a4, it should be counted.