Game score to eliminate ties. Unique Sums - with a twist

Question

Players score points based on their rank in different colors, as shown.

The intention is to have no ties. Having ran all possible iterations, I found a few instances of ties. Rather than going with the "powers of 2" approach which results in very high values, I think I (hopefully) devised a way to create a tie breaker.

If there is a tie, players will double their highest individual score.

For instance:

Player A [2,2,3,5] and B [1,3,4,4] both have total scores of 12. But upon doubling their highest score, Player A scores 17, and B scores 16.

Is there a way to know that this is sufficient without hard analyzing each iteration?

If the provided scores for each rank within each color still allows ties after doubling the highest value, is there a better (small values preferred) set of numbers to use?

I COULD say to continue doubling your next highest score until there is a clear winner, but that isn't preferred.

Answer 1

I created a script which simulates your game. I came up with some examples providing that your technique doesnt work.

For instance:

1. A=[3 4 6 5] B=[4 3 3 7] C=[2 5 5 6] D=[1 2 4 4] , A and C have the highest score of 18 and both's best performance is 6

2. A=[3 4 4 6] B=[1 5 3 7] C=[2 3 5 4] D=[4 2 6 5], A and D have the highest score of 18 and both's best performance is 6

As you explained the points are rewarded in each color are increased by one in each color. I ran the same script adding pi points in every next color, that is in red, the players are rewarded [1 2 3 4] in blue [1+pi,2+pi,3+pi,4+pi] etc. But still this couldnt break the tie. But dublicating the maximum score in case of tie indeed work (tested for one million iterations).

After all the method of providing different points in the different colors is like a weighted average. Hence it would be more efficient if you could reward the same points span for instance: [1 2 3 4] in every color. In case of tie you could select the one that did better in a specific color (for instance the green one).

The recommendation above is based on the fact that you dont mind that the colors arent being equally importand since by doubling the maximum you are valuing the perfomance in the green color more than that in the blue, since Max(Green)>Max(Blue).

Answer 2

Your tie-breaker can still result in a tie. It should be fairly clear that this should happen whenever the tied players have the same highest score. For example, [2,3,4,5] vs [1,4,5,4].

There are different ways to avoid ties by choosing point values carefully. I'll give two, but I suspect there are many more.

One way is to make every score a multiple of 4 except for one color. The exceptional color should have one score with each residue modulo 4. That way, the residue of the total score of a player modulo 4 will only depend on the exceptional color, so it can't be duplicated. For example [4,8,12,16],[4,8,12,16],[4,8,12,16],[4,9,14,17]. (This equivalent to tie-breaking according to ranking in one color.)

Another way is to make the differences between adjacent scores in three colors multiples of 4, and in the last color make the differences exactly 3. The difference between two player's scores in the first three colors will then be a multiple of 4, but the difference in the last color must either be 3, 6, or 9, none of which is a multiple of 4. For example [1,5,9,13],[2,6,10,14],[3,7,11,15],[7,10,13,16].

If you like your tie-breaker, you could also make each score a distinct number. Then your tie-breaker of doubling the highest would work perfectly, as the doubled number would be different for each player.