# DouglasReay/MathsProblem

ec2-44-192-79-149.compute-1.amazonaws.com | ToothyWiki | DouglasReay | RecentChanges | Login | Webcomic

As I lay in bed this morning, I dreamed about a game.  Then I realised there was a maths problem with it, that I didn't know how to solve.  So, first the game, then the problem...

# The Game

## Stage 1

There are 16 counters, split into 4 of each colour Red, Blue, Green, Yellow.  Each colour has the numbers 1-4 so the counters are:
R1 R2 R3 R4 B1 B2 B3 B4 G1 G2 G3 G4 Y1 Y2 Y3 Y4

These are divided between four players (say Alice, Bob, Charlie, Dave), such that:
• Each player gets 4 counters
• Each player get one counter of each colour
• The numbers on each players counters add up to 10
• No two players end up with the same individual counter

So { R1, B2, G3, Y4 } would be a valid set, as would { R1, B1, G4, Y4 }

Once all the players have their 4 counters, they write them down on a preferences sheet of paper, and return them to a central bag.

## Stage 2

Each player gets 16 bidding tokens, marked 1-16.

The counters from the bag are then auctioned off one by one, with the winner of an auction keeping the counter.

## Stage 3

A player's score is the sum of their counters multiplied by their preference for that colour.

Eg Alice in stage one got preferences { R1, B2, G3, Y4 }.
In stage two, she won the counters { Y4, Y3, Y2, G3, G2, G1, R3 }
She would score:
`  Y4 --> 4 * 4 = 16  Y3 --> 3 * 4 = 12  Y2 --> 2 * 4 =  8  G3 --> 3 * 3 =  9  G2 --> 2 * 3 =  6  G1 --> 1 * 3 =  3  R3 --> 3 * 1 =  3  TOTAL          57`

## The Problem

The problem is with Stage 1.  It would be preferable that none of the four players know what counters the other players have.  Obviously this could be done by using a 5th party outside the game.  But is there any way for just four players, in a closed room with a supply of physical materials like pencils, paper, and bags, to distribute the 16 tiles in this manner?  What if they had access to computers with cryptography?

## The Solution

Write just the numbers on the backs of the counters, but not the colours. Turn the counters upside down and shuffle. Everyone takes a '1', then a '2', then a '3'.. It's a start, anyway. - MoonShadow
Problem with that is that he wants 1,1,4,4 and 1,3,3,3 to be valid distributions as well as 1,2,3,4.
However, I was thinking the complementary idea: for each of  red, yellow and green, you have them turned face down and let players draw one randomly. If the required number to bring their total to ten is not in {1 2 3 4}, each player complains and the numbers are redistributed. Otherwise, players take turns with the Blue Box. The Blue Box contains four boxes labelled 1, 2, 3 and 4. Players look for the counter they need in its own numbered box. If it isn't there, they complain and you start again.
It'd get the job done, but it might need a few reshuffles, which isn't desirable. --AC

Trivial 'with a computer' solution: Enumerate all the possible solutions, select one.  That works with cards, too.  --Vitenka

Also, have you played ModernArt??  Because I think you've recreated it with added unfun that doesn't seem to add much.  --Vitenka