Originally posted by forkedknightThis has been done above.
I would just divide what it would cost each of them to rent their own boat by the total of what renting 4 different boats would and have each pay that percentage of the fare.
I challenge somebody to find a solution for every possible version.
ie A, B, C, D, AB, AC, AD, BC, BD, CD, ABC, ABD, and BCD; Where each version favors those named and is mathematically plausible and "fair"
Originally posted by JS357Hold on, I just noticed something. How did you split the initial ten dollars it takes to rent the boat?
Here's way that could be just a beginning of the argument.
[hidden] A causes B to go an extra mile. He pays that for B. So he pays for 4 miles or $6.67.[/hidden]
[hidden] B causes no one to go more miles so he pays the 4 miles he caused or $6.67.[/hidden]
[hidden] C causes no one to go more miles so he pays the 3 miles he caused or $5.00.[/hidden]
[hidde ...[text shortened]... to pay for fewer miles because it can also be said that B saves both C and D 4 miles![/hidden]
If you use your same method but split the ten dollars equally then you get this:
A =6.50
B =6.50
C =5.50
D =6.50
Originally posted by tomtom232That works, too, and can be paid without pennies.
Hold on, I just noticed something. How did you split the initial ten dollars it takes to rent the boat?
If you use your same method but split the ten dollars equally then you get this:
A =6.50
B =6.50
C =5.50
D =6.50
OTOH, it's 6.25 each.
Interesting to see everyone is coming up with differing methods. How would a system be judged to be "fair"? Everyone agrees? Or is it the least unpalatable? Could anyone ever come up with a system where everyone agreed?
I propose the following : I pay this time and my next 3 rides are free when I next meet you lot at the pub. 🙂
Originally posted by ThomasterThis is one of the reasons why I chose the method I did. The split should be independent of the path chosen if there are multiple paths of the same distance.
What if the boat travels D -> C -> B -> A? Would it change the calculation?
That path (D->C->B->A) is not of the same distance, but if you imagine the scenario where A,B,C,D all live the same distance from the party:
A @ (0,3)
B @ (3,0)
C @ (0,-3)
D @ (-3,0)
Then how would you split it?
2 edits
Originally posted by forkedknightI was thinking of calculating the amount that each would have to pay if alone and having them each save an equal amount such that the result is 25.
This is one of the reasons why I chose the method I did. The split should be independent of the path chosen if there are multiple paths of the same distance.
That path (D->C->B->A) is not of the same distance, but if you imagine the scenario where A,B,C,D all live the same distance from the party:
A @ (0,3)
B @ (3,0)
C @ (0,-3)
D @ (-3,0)
Then how would you split it?
A = 3.75
B = 4.75
C = 7.75
D = 8.75
D saves six cents more than the rest because I rounded 18.06 to 18. You can limit this by adding .03 to D and subtracting .01 from the rest but D still saves 2 cents more.