🕸📝Fergus Duniho wrote on Thu, Apr 26, 2018 12:09 AM UTC:
How would you measure the diversity of games played between two players? Suppose X1 and Y1 play five games of Chess, 2 of Shogi, and 1 each of Xiang Qi, Smess, and Grand Chess. Then we have X2 and Y2, who play 3 games of Chess, 3 of Shogi, 2 of Xiang Qi, and 1 each of Smess and Grand Chess. Finally, X3 and Y3 have played two games each of the five games the other pairs have played. Each pair of players has played ten games of the same five games. For each pair, I want to calculate a trust value between a limit of 0 and a limit of 1, which I would then multiply times the maximum adjustment value to get a lower adjustment value.
Presently, the formula n/(n+10) is used, where n is the number of games played between them. In this case, n is 10, and the value of n/(n+10) is 10/20 or 1/2. One thought is to add up fractions that use the number of games played of each game.
This follows the same pattern, though the trust values are lower. To clearly see the difference, look at X2 and Y2, and compare 2/22, which is for two games of Xiang Qi, with 2/21, which is for one game each of Smess and Grand Chess. 2/22 is the smaller number, which indicates that it is giving lower trust scores for the same game played twice.
Since it is late, I'll think about this more later. In the meantime, maybe somebody else has another suggestion.
How would you measure the diversity of games played between two players? Suppose X1 and Y1 play five games of Chess, 2 of Shogi, and 1 each of Xiang Qi, Smess, and Grand Chess. Then we have X2 and Y2, who play 3 games of Chess, 3 of Shogi, 2 of Xiang Qi, and 1 each of Smess and Grand Chess. Finally, X3 and Y3 have played two games each of the five games the other pairs have played. Each pair of players has played ten games of the same five games. For each pair, I want to calculate a trust value between a limit of 0 and a limit of 1, which I would then multiply times the maximum adjustment value to get a lower adjustment value.
Presently, the formula n/(n+10) is used, where n is the number of games played between them. In this case, n is 10, and the value of n/(n+10) is 10/20 or 1/2. One thought is to add up fractions that use the number of games played of each game.
X1 and Y1
5/(5+10)+2/(2+10)+1/(1+10)+1/(1+10)+1/(1+10) = 5/15+2/12+1/11+1/11+1/11 = 17/22 = 0.772727272
X2 and Y2
3/13 + 3/13 + 2/12 + 1/12 + 1/11 = 6/13 + 2/12 + 2/11 = 695/858 = 0.81002331
X3 and Y3
2/12 * 5 = 10/12 = 0.833333333
The result of this is to put greater trust in a diverse set of games than in a less diverse set, yet this is the opposite of what I was going for.
How would this change if I changed the constant 10 to a different value? Let's try 5.
X1 and Y1
5/(5+5)+2/(2+5)+1/(1+5)+1/(1+5)+1/(1+5) = 5/10+2/7+3/6 = 1 2/7 = 1.285714286
Since this raises the value above 1, it's not acceptable. Let's try changing 10 to 20.
X1 and Y1
5/(5+20)+2/(2+20)+1/(1+20)+1/(1+20)+1/(1+20) = 5/25+2/22+1/21+1/21+1/21 = 167/385 = 0.433766233
X2 and Y2
3/23 + 3/23 + 2/22 + 1/22 + 1/21 = 6/23 + 2/22 + 2/21 = 2375/5313 = 0.447016751
X3 and Y3
2/22 * 5 = 10/22 = 0.454545454
This follows the same pattern, though the trust values are lower. To clearly see the difference, look at X2 and Y2, and compare 2/22, which is for two games of Xiang Qi, with 2/21, which is for one game each of Smess and Grand Chess. 2/22 is the smaller number, which indicates that it is giving lower trust scores for the same game played twice.
Since it is late, I'll think about this more later. In the meantime, maybe somebody else has another suggestion.