# The Ride Home

geepamoogle
Posers and Puzzles 27 Apr '08 23:26
1. 27 Apr '08 23:26
Suppose I decided to take three of my friends home, charging only what the trip costs me, and that I wish to charge each friend fairly.

As it happens, they all live in the same direction, one 3 miles away, one 5 miles away, and one 7 miles away.

The trip costs me \$7. What is the fairest way to charge my three friends and why?
2. 27 Apr '08 23:44
One way is to charge each friend based on how far they were driven and how many people were in the car at the time.

The trip cost \$7 and you transported the 3rd friend for 7 miles, so it cost you \$1 per mile.

For the 1st mile there were 3 friends in the car, so they each pay \$1/3 per mile.

For the 2nd mile there were 2 friends in the car so they each pay \$1/2 per mile.

For the 3rd mile there was just 1 friend in the car so he/she pays \$1 per mile.

This gives:
1st to be dropped off pays 1/3 x 3 = \$1

2nd to be dropped off pays (1/3 x 3)+(1/2 x 2) = \$2

3rd to be dropped off pays (1/3 x 3)X(1/2 x 2)+(1x2) = \$4
3. 28 Apr '08 01:20
Originally posted by MattP
One way is to charge each friend based on how far they were driven and how many people were in the car at the time.

The trip cost \$7 and you transported the 3rd friend for 7 miles, so it cost you \$1 per mile.

For the 1st mile there were 3 friends in the car, so they each pay \$1/3 per mile.

For the 2nd mile there were 2 friends in the car so they each ...[text shortened]... \$2

3rd to be dropped off pays (1/3 x 3)X(1/2 x 2)+(1x2) = \$4
That actually was my exact reasoning.
4. joe shmo
Strange Egg
28 Apr '08 01:41
Originally posted by geepamoogle
Suppose I decided to take three of my friends home, charging only what the trip costs me, and that I wish to charge each friend fairly.

As it happens, they all live in the same direction, one 3 miles away, one 5 miles away, and one 7 miles away.

The trip costs me \$7. What is the fairest way to charge my three friends and why?
their your friends, you shouldn't charge them...
5. 28 Apr '08 03:18
Originally posted by geepamoogle
Suppose I decided to take three of my friends home, charging only what the trip costs me, and that I wish to charge each friend fairly.

As it happens, they all live in the same direction, one 3 miles away, one 5 miles away, and one 7 miles away.

The trip costs me \$7. What is the fairest way to charge my three friends and why?
Need more info. Did you come home after, thereby driving 14 miles total? I'm assuming yes.

It cost you \$7 for 14 miles. That's \$0.50/mile.

The first friend has to pitch in for 10/14 miles (his trip plus return)
The second: 12/14
The third: 14/14

So, 1st-\$5, 2nd-\$6, 3rd-\$7. That comes to \$18, but you only want \$7.

Prorate it and you charge:

1st-\$1.95
2nd-\$2.33
3rd-\$2.72

This way, all three pitch in to pay for you return from the furthest point traveled. I know I would be pretty upset if I was the 3rd friend, paying 4 times more than the 1st. When I only live a little more than 2 times further away.
6. 28 Apr '08 04:29
I considered the trip home as well, but the trip home is the reverse of the trip out, so no special route back is needed.
7. forkedknight
Defend the Universe
28 Apr '08 09:271 edit
I'm going to go with brobluto on this one, besides the fact that he calculated for people living 5, 6, and 7 miles away instead of the 3, 5, and 7 miles put forward in the problem statement.

The first person should pitch in for 6/14 miles, the second 10/14, and 14/14 for the third.

Using the same calculation as brobluto, that comes to \$1.40, \$2.33, and \$3.27 for each of them, respectively.

*Edit*
If it was on your way, and you didn't make a return trip, you'd be a jerk if you charged your friends.
8. forkedknight
Defend the Universe
28 Apr '08 09:48

Assume the cost of transporations is \$1/mile.

You're taking two of your friends home, and both of their houses are out of the way.

The first person's house is 2 miles out of they way, if you were just to take him home.

The second person's house is 3 miles out of the way.

Taking both of them together, however, you only have to drive a total of 4 miles out of your way.

What do you charge each friend for their ride?
9. wolfgang59
28 Apr '08 11:37
Originally posted by MattP
One way is to charge each friend based on how far they were driven and how many people were in the car at the time.

The trip cost \$7 and you transported the 3rd friend for 7 miles, so it cost you \$1 per mile.

For the 1st mile there were 3 friends in the car, so they each pay \$1/3 per mile.

For the 2nd mile there were 2 friends in the car so they each ...[text shortened]... \$2

3rd to be dropped off pays (1/3 x 3)X(1/2 x 2)+(1x2) = \$4
Cannot fault this except what about thye driver? If its on his way he should contribute too!

Same underlying arithmetic gives;

1st to be dropped off pays 1/4 x 3 = \$0.75

2nd to be dropped off pays (1/4 x 3)+(1/3 x 2) = \$1.42

3rd to be dropped off pays (1/4 x 3)X(1/3 x 2)+(1/2x2) = \$2.42

Driver pays the rest!
10. 28 Apr '08 23:52
Apologies if I didn't make this clear enough, but you aren't taking yourself anywhere, you are returning them to their proper homes and returning home afterwards, and would not have gone anywhere if they hadn't wanted or needed a ride home.

And I would agree with the \$4/\$2/\$1 division, personally with the following reasoning. For the first three miles, you are driving all 3 friends, and thus each should share in paying for those three miles. The next two miles are shared by the two friends who remain in the car, while the last 2 miles are the last friend's alone.

As for the return trip, it follows the same route back in reverse, and therefore each mile would belong to those people who made it necessary to travel that far in the first place.

That is my opinion in the least. Of course, if you were a really good friend, you'd do this gratis and save yourself the trouble, right? (Heheh)
11. 28 Apr '08 23:57
Originally posted by forkedknight

Assume the cost of transporations is \$1/mile.

You're taking two of your friends home, and both of their houses are out of the way.

The first person's house is 2 miles out of they way, if you were just to take him home.

The second person's house is 3 miles out of the way.

Taking both of them together, however, ...[text shortened]... o drive a total of 4 miles out of your way.

What do you charge each friend for their ride?
I see two potential solutions to this one, and I don't know which would be "fairer".

Both start with charging two people for the extra mileage to take them home, \$1 per mile.

However, you do regain a mile by driving both back and to be fair, you ought to refund them this amount (or don't charge it to being with).

The question then is, who gets what part of the savings?

ONe answer is to split it 50/50, because the extra mile is due to some pathway between their residences.

You would therefore charge one \$2.50 anf the other \$1.50.

However, you could split it proportional to what you're charging them, amking it a \$2.40/\$1.60 split.

Personally, I lean towards the latter.
12. forkedknight
Defend the Universe
29 Apr '08 03:49
Originally posted by geepamoogle
I see two potential solutions to this one, and I don't know which would be "fairer".

Both start with charging two people for the extra mileage to take them home, \$1 per mile.

However, you do regain a mile by driving both back and to be fair, you ought to refund them this amount (or don't charge it to being with).

The question then is, who gets w ...[text shortened]... 're charging them, amking it a \$2.40/\$1.60 split.

Personally, I lean towards the latter.
so extending this to 3 friends:
-one who is 6 miles out of the way
-one who is 10 miles out of the way
- and one who is 14 miles out of the way

and the total distance out of your way is 14 miles, how do you split it up?

I would take the same argument, leaving you with my previous answer to the original problem...
13. 29 Apr '08 07:09
Originally posted by forkedknight
so extending this to 3 friends:
-one who is 6 miles out of the way
-one who is 10 miles out of the way
- and one who is 14 miles out of the way

and the total distance out of your way is 14 miles, how do you split it up?

I would take the same argument, leaving you with my previous answer to the original problem...
1: 1,40
2: 2,39
3: 3,27

30/14=ans
(30=total distance friends, 14=total distance)
6/ans x0,5
10/ansx0,5
14/ansx0,5
14. 29 Apr '08 12:39
Originally posted by forkedknight
I'm going to go with brobluto on this one, besides the fact that he calculated for people living 5, 6, and 7 miles away instead of the 3, 5, and 7 miles put forward in the problem statement.

The first person should pitch in for 6/14 miles, the second 10/14, and 14/14 for the third.

Using the same calculation as brobluto, that comes to \$1.40, \$2.33 ...[text shortened]... on your way, and you didn't make a return trip, you'd be a jerk if you charged your friends.
Yes. my bad. That's what I get for doing it late at night!
15. 29 Apr '08 17:28
Originally posted by forkedknight
so extending this to 3 friends:
-one who is 6 miles out of the way
-one who is 10 miles out of the way
- and one who is 14 miles out of the way

and the total distance out of your way is 14 miles, how do you split it up?

I would take the same argument, leaving you with my previous answer to the original problem...
Excellent point, which means I would either have to revisit one of my two answers.

So let me attempt, then, to relook at your problem in the exact same manner, if indeed I can.

I will therefore suppose only two friends, one who lives 1 mile out (2 miles round trip), another 1.5 miles out (3 miles round trip), and 1.5 miles apart (so that it's 4 miles to take both home).

So now, let me examine how it would be divided depending on who gets taken home first.

Suppose the nearer goes home first..

That's 1 mile split, 1.5 miles for the further, and 1.5 miles to account for for my trip back home. When everyone was in a straight line of sorts, this was easier, as each was the reverse of a mile going, not the case this time around. I will answer this with the notion that I ought to divide it proportion to the miles for the trip there, assuming each was the only one.

So 0.5 + 0.0 + 0.6 miles for the nearer, and 0.5 + 1.5 + 0.9 for the further, or a \$2.90/\$1.10 split, further and nearer.

Now what if I take the further home first?

That would be 1.5 split, 1.5 to the nearer, and 1 divided as before.

Nearer is .75 + 1.5 + .4 = \$2.65, and the further \$1.35.

This analysis would seem patently unfair, as shown by its inconsistency.

I will have to revisit this when I have more time.