harshits2808 wrote: ↑Sat Jan 20, 2018 9:40 am
daniel.is.fischer wrote: ↑Tue May 19, 2009 2:07 pm
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!
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?