[jon.kiparsky]

This came up at the bar last night, and I thought some of you might appreciate it.

What's the best way to fairly select one person from a group, without using tools? By "fair" I mean a selection such that no one person can dominate the choice. By tools, I mean dice, lots, coins, darts, or other randomizers. There is a sheet of paper and a pen, if that's useful.

If it helps, here's a scenario:
You're at the bar with N people, and it comes time to settle up. As the bill arrives, you realize that there are two calamaties to deal with. The first is that nobody has brought cash - it's all cards. The second is that the bar won't split up the tab - it's all going on one card. Nobody particularly minds being the one selected to pay the tab, as long as they feel that they were selected fairly.
How do you select one person fairly?

Here's my solution:

Spoiler

First, count off around the table so that each person has a unique number from 0..N
Each person picks a number - any integer will do. The piece of paper and pen are passed around the table, and each person writes their number on the paper, folding it so their number is obscured.
When each person has selected a number, take the total mod N. That person has been selected.

EDIT: As Skydiver points out, the range should have been 0..N-1, not 0..1

Anyone got a better one? (sepp2k, I'm looking at you!)

This post has been edited by jon.kiparsky: 17 May 2013 - 09:00 AM

[Skydiver]

It's not completely fair. Since people count off starting with 1, rather than N (do you ever remember counting off starting with zero in grade school?), the person who gets to be the N-th person will be exempt from ever getting the bill because and number modulo N, will result in a value from 0 to N-1.

[jon.kiparsky]

You didn't count off from zero? What sort of weird grade school did you go to?

[modi123_1]

I am assuming you want some sort of math solution? Typically it falls on "who makes the worst warface gets their card abused".

[baavgai]

You have paper?

Tear it up, each person puts their name on it, and have the waitress pick from the hat.

Seriously, man, step away from the computer.

[modi123_1]

I believe that constitute a 'tool'.

It's just a thought exercise, maaaaaaaaaaan.

[jon.kiparsky]

baavgai, on 17 May 2013 - 11:17 AM, said:

You have paper?

Tear it up, each person puts their name on it, and have the waitress pick from the hat.

Seriously, man, step away from the computer. />

If you want to go that route, you can save the paper, have each person put their card in the hat. back atcha!

[andrewsw]

You could put their cards in the hat (assuming someone brought a hat?) .. but I would put in my library card

jon.kiparsky, on 17 May 2013 - 04:26 PM, said:

If you want to go that route, you can save the paper, have each person put their card in the hat. back atcha!

Jinx again!

[jon.kiparsky]

andrewsw, on 17 May 2013 - 11:27 AM, said:

You could put their cards in the hat (assuming someone brought a hat?) .. but I would put in my library card />

That would suck in three weeks when you have to bring the beer back... in the same condition you got it...

[modi123_1]

Not too hard when you are drinking Coors Light.

[andrewsw]

You could chose a random 0-9 number and the first person whose card ends in that number would pay. But credit card numbers aren't random are they?

[jon.kiparsky]

What if you have more than one person whose card ends in that number?
Also, where do you get your random number from?

[modi123_1]

From adding the digits of the current date together plus the numerical values of first letters of their full names.

[jon.kiparsky]

But is it an African swallow, or a European swallow?

[modi123_1]

Only if today falls on a value produced by the Lucas sequence for the month.