RaleighCraps
RaleighCraps
Joined: Feb 20, 2010
  • Threads: 79
  • Posts: 2501
March 1st, 2011 at 5:24:46 PM permalink
In school, I was always one of the first ones done with every test. I didn't get the highest scores, but I bet I had the highest scoring avg per minutes spent taking the tests.
So one day the teacher hands us all a surprise test and says this is a timed test, and speed plus following directions counts most. SWEET!

Teacher says go, and flip the test over. Typical instruction nonsense about reading carefully, answer accurately, and take the time to read all the questions first, before you begin answering......blah blah.
I jump right in and am on a heater. Questions are simple and I am flying. Get to page 2 and sense some people are really struggling, as they don't seem to be answering any questions. Pretty soon hardly anyone is answering questions.... Uh oh. What did I miss?

Last question says, "Once you have read this question, you are done. You do not need to answer any other questions" So, had I followed the instructions, and read through all the questions first............
Always borrow money from a pessimist; They don't expect to get paid back ! Be yourself and speak your thoughts. Those who matter won't mind, and those that mind, don't matter!
thecesspit
thecesspit
Joined: Apr 19, 2010
  • Threads: 53
  • Posts: 5936
March 1st, 2011 at 5:27:36 PM permalink
I reckon I scored 5-10% higher on almost every test I did at school by reading through the questions first and planning my order of attack...

I'd divide my time on "so long per mark" with time to check answers at end. If a question took longer than the marks were worth, I'd stop it (unless almost done) and move onwards. I'd almost always do one of the low scoring questions first, then do the big mark questions before going back to the short answers.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
RaleighCraps
RaleighCraps
Joined: Feb 20, 2010
  • Threads: 79
  • Posts: 2501
March 1st, 2011 at 5:32:06 PM permalink
Quote: thecesspit

I reckon I scored 5-10% higher on almost every test I did at school by reading through the questions first and planning my order of attack...

I'd divide my time on "so long per mark" with time to check answers at end. If a question took longer than the marks were worth, I'd stop it (unless almost done) and move onwards. I'd almost always do one of the low scoring questions first, then do the big mark questions before going back to the short answers.



Too much work for me. I just wanted the damn things over with, and 93 paid the same as 100, so the extra effort wasn't worth it..... :-)
Always borrow money from a pessimist; They don't expect to get paid back ! Be yourself and speak your thoughts. Those who matter won't mind, and those that mind, don't matter!
spinswizard
spinswizard
Joined: Feb 14, 2020
  • Threads: 2
  • Posts: 8
July 28th, 2020 at 3:19:48 AM permalink
Intuitively you can check that the possible outcomes are the following:
2 probab: 1/4
2+3=5 probab: 1/4*1/3
2+5=7 probab: 1/4*1/3
2+9=11 probab: 1/4*1/3
2+3+5=10 probab: 2*1/4*1/3*1/2 = 1/4*1/3
2+3+9=14 probab: 2*1/4*1/3*1/2 = 1/4*1/3
2+5+9=16 probab: 2*1/4*1/3*1/2 = 1/4*1/3
2+3+5+9=19 probab: 3*1/4*2*1/3*1/2 = 1/4

And then from here since 2 and 19 have the same probability to occur we can take their average 21/2=10.5 with probab 1/2
And in the same way for the others 5+6=21, 7+14=21, 11+10=21 => avg = 10.5 with probab 1/2.

I am summing them 2 by 2, just to show that the distribiution is symmetric and i think its nice to see quite visually why that the EV is just (2(collect) + 19(sum of all) )/2, like the user gog stated before me :)

and then by induction this is true for a game with n buckets and even repeated values of the prizes :)
ThatDonGuy
ThatDonGuy 
Joined: Jun 22, 2011
  • Threads: 94
  • Posts: 4187
July 28th, 2020 at 8:20:17 AM permalink
Quote: RaleighCraps

Last question says, "Once you have read this question, you are done. You do not need to answer any other questions" So, had I followed the instructions, and read through all the questions first............


I have seen a number of these. Usually, the first two instructions are, "Read everything before doing anything," and, "Write your name at the top of the paper," and the last instruction is, "Now that you have read all of the instructions, do only the first two." Usually, one of the other instructions is something like, "If you are the first one to read this instruction, say out loud, 'I am the best at following instructions.'"

I wouldn't be surprised if someone did a variation on this where one of the instructions in the middle is, "Ignore the final instruction."
charliepatrick
charliepatrick
Joined: Jun 17, 2011
  • Threads: 31
  • Posts: 2047
July 29th, 2020 at 2:13:43 AM permalink
It's the first time I've seen this puzzle.

One neat way of solving it is consider the buckets placed in a row, and then going through all the permutations.

Person one makes their picks from left to right, they will pick any prizes to the left of the "Collect" bucket and the "Collect" bucket.
Person two goes from right to left, they will pick any prizes to the right of the "Collect" bucket and the "Collect" bucket.
Together they pick all the prize buckets and 2 x "Collect" = 3+5+9+2+2 = 21.

If they work through all the permutations, every time the total of the two will be 21, and each person will have covered all the permutations. Hence the average for two people is 21, so the average for one person is 10.5.


On the "ignore the final instruction" there's a puzzle you probably know where it starts with "You are driving a bus and at the first stop 15 people get on...."
spinswizard
spinswizard
Joined: Feb 14, 2020
  • Threads: 2
  • Posts: 8
August 2nd, 2020 at 1:45:59 AM permalink
Hii I came up with a pretty neat formula for calculating the EV w.r.t. any number of buckets, including multiple "collect"-s. It goes like this:
Sum of all buckets (incl. all collects) - S
Count of collects - h
Value of Collect - c

(S + c) / (h +1) ;

I've come to it via the general formula with the permutations you suggested, which is the following
h *( SUM i from 0 to n-h ((S-c(h-1)) *Binomial [n-h-1; i-1] + c* Binomial [n-h-1; i]) * i! * (n-1-i)! /n! )

Its a bit long but the end result is so satisfactory! Maybe you will be interested to see :)

https://www.wolframalpha.com/input/?i=SUM+h*%28%28S-c*%28h-1%29%29+*+Binomial%5Bn-h-1%2C+j-1%5D+%2B+c*+Binomial%5Bn-h-1%2C+j%5D%29+*+j%21+*+%28n-1-j%29%21+%2Fn%21+0+to+n-h+
Last edited by: spinswizard on Aug 2, 2020

  • Jump to: