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

enter image description here

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.

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    Isnt it possible for two players to have scored the same highest score in different or even the same colors? – 2017-02-14
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    The high score in each color is unique. Only one player will score the 1st place value for each color. So a person can be first in say Red (4), Third in Blue(3), first in Green (6) and 4th in Yellow(4). Player's highest scores might be the same, but I'm trying to find an elegant way to show that if their highest individual scores are the same, they are not tied for the most total points with anyone else. – 2017-02-14
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    They may have the same highest score across two different colors, but I'm thinking their totals won't be tied then. And if they are tied, a different player has a higher score. (That's what I'm thinking at least. I'm technically looking for proof otherwise, or a different set of scores that fulfills the conditions) – 2017-02-14
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    You should provide then some details about how the points are splitted between the players to investigate if your technique works. What about this one A=[ 4 3 2 1 ] , B= [1 4 3 2] , C=[ 2 1 4 3 ] , D = [3 2 1 4] ? – 2017-02-14
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    You'll notice that the values within the different color columns slightly vary such that the example you gave isn't possible (that is to say that the values 2 and 1 don't exist for the green column and 4, 3, 2 and 1 do not exist for the yellow column). – 2017-02-14

2 Answers 2

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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).

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    Thank for sticking with me johny. I appreciate your tenacity. The thing you mention about having colors not being equally important might have to be the way to go. It bums me out a bit, but I can balance to make it noticeably harder to rank in the more valuable color I suppose. – 2017-02-14
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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.

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    I like your approach And I think it will help keep the final scores to a digestible total. I'm a bit stuck on the final facet of the question. In your example [2,3,4,5] vs [1,4,5,4] clearly those players are tied. And in fact, a third player could be tied with a score of 14 [3 2 3 6]. That third player has the MINIMUM total score available after considering the other two players' scores. Yet with the 6, the third player is the clear winner. So certainly there are instances of ties prevailing after my tie breaker, but how can I find if they are even scores to be considered for the win? – 2017-02-14
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    @johny has provided two distinct examples of the tied scores indeed being the highest sum , and having the same highest individual score. So clearly the numbering I'm using isn't the way to go. – 2017-02-14
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    Your idea to make each score a distinct number is actually simple and elegant. I believe I'd actually only have to make one color's scores unique and higher than the others. Red 4 3 2 1 | Green 4 3 2 1 | Blue 4 3 2 1 | Yellow 8 7 6 5 So certainly, there are plenty of ties to be had between sums in the R,G and B columns, but because the values in yellow are unique and higher than the others, this is essentially the "tie breaker" color. Does that look like that would work? – 2017-02-14
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    Ah, I didn't realize you meant ties for first place, but as you say, johny has already answered this. And, yes, making yellow higher than all the other colors as you suggest will also work fine with your tie-breaker. – 2017-02-14