Balancing Act

[Steve99]

RESOLVED

I have designed a business game that at its core has cards that show changes in value in 8 Market Sector Prices that I will call A,B,C,D,E,F,G and H.

During the game, 4 of the 14 cards are discarded at the outset and 10 cards are played over 5 rounds. I am expecting that the average total of values for each sector from any 10 cards will be about +6. Please let me know if this is not the case.

The objective is to have it possible for any two sectors to be high, low or equal together. Because of balancing issues, I usually find that the combination of values between some sectors means that if one is high the other is always low and vice versa.

The requirement therefore is this:

1. to have 14 cards
2. each of the 14 cards must contain a value for each of the 8 sectors
3. the 14 values that must be assigned to each sector are: +3, +3, -3, -3, +2, +2, +2, +2, -2, -2, +1, +1, +1, +1
4. the values for the sectors on each card must vary as much as possible. Ideally each card would contain each of the 5 values: +3, -3, +2, -2 and +1
5. no two sectors must be identical in value on every card.
6. to have each pair of sectors containing at least 1 of the 14 cards with:
   1. a +2 or +3 for both sectors
   2. a -2 or -3 for both sectors
   3. a (+2 or +3) on the first sector and a (-2 or -3) on the second sector
   4. a (+2 or +3) on the second sector and a (-2 or -3) on the first sector
7. a balanced deck that avoids the problem that if the total of one sector is high the other sector is always low and vice versa.

If there is no solution to this requirement could the same be achieved with more cards where the additional cards each had a value of +1?

As you can probably gather, I am not a mathematician so please dont hesitate to get clarification on any ambiguities in my description of the problem.

So far I have had only moderate success with creating a balanced 14 card deck. I am wondering if there is a formula for creating a 'balanced' 14 card deck for 8 sectors or wil it reqiure more cards?

Any help would be appreciated.

I will be happy to provide a copy of the game, if/when published, should I use any persons solution in entirity.

Steve Hickey.

 

[Steve99]

Step 1

There are essentially 3 groups: (i) 3's and 2's (ii) -3's and -2's and (i) 1's
So I start with A and list all 14 combinations of cards in each of the 3 groups.
I want all 5 different values to appear on each of the 14 cards, so I add these to card 1

	1	2	3	4	5	6	7	8	9	10	11	12	13	14
A	3	3	2	2	2	2	-3	-3	-2	-2	1	1	1	1
B	2
C	1
D	-2
E	-3
F
G
H

 

[Steve99]

Step 2

The second step is that I look at each pair of sectors in turn.
Consider pairing AB
I need to ensure that there is at least the following combinations between them:

• Similar group value going up
• Similar group value going down
• Opposite group value going up
• Opposite group value going down

This is realised in the following example:

	1	2	3	4	5	6	7	8	9	10	11	12	13	14
A	3	3	2	2	2	2	-3	-3	-2	-2	1	1	1	1
B	2			-3			-2		3
C	1
D	-2
E	-3

 

[Steve99]

Step 3

The third step is that I reconcile each of the other pairs that I have.
I next consider pairings AD and AE
I need to ensure that there is at least the following combinations between them:

• Similar group value going up
• Similar group value going down
• Opposite group value going up
• Opposite group value going down

This is realised in the following example:

	1	2	3	4	5	6	7	8	9	10	11	12	13	14
A	3	3	2	2	2	2	-3	-3	-2	-2	1	1	1	1
B	2			-3			-2		3
C	1
D	-2		3					2		-3
E	-3			3			-3		3

 

[Steve99]

The next step is that I reconcile each of the other pairs that I have.
I next consider pairings DE
I need to ensure that there is at least the following combinations between them:

• Similar group value going up
• Similar group value going down
• Opposite group value going up
• Opposite group value going down

This is realised in the following example:

But things start getting exponentially more difficult from here, having to consider all possible pairings from A to H and all 14 cards. Is there a simpler solution than struggling through incrementally changing and improving the overall solution?

	1	2	3	4	5	6	7	8	9	10	11	12	13	14
A	3	3	2	2	2	2	-3	-3	-2	-2	1	1	1	1
B	2			-3			-2		3
C	1
D	-2		3					2		-3	2	2	-2
E	-3			3			-3		3		2	-2	2

 

[Steve99]

Issue Resolved

I have now found a solution that I am happy with. Essentially it involved creating a 16 card deck, as the number of 2's and 3's were in multiples of 8. This resulted in each card having the following:

1 x -3
1 x +3
1 x -2
2 x +2
3 x 1

Through manual trial and error I was then able to get a good balance between all pairs of 8 elements.

So while logic helped in part I don't think this was truly a mathematical issue.