I originally applied this concept to correct guessing of moves made by strong chess players but this can be generally applied elsewhere.
Whenever there is a comparison of ratios where it can be reasoned that the same ratio for the instance of where more attempts were made suggests a better performance than when fewer attempts are made, a mathematical convention could be applied.
There can be contention on who would rank higher, one who has a performance ratio of only 1 out of 2 attempts or one who has a performance ratio of 5 out of 12 attempts. Of course some weight should be given to the one who made the greater number of attempts but should the slightly less than 50% performance rate be considered better than the one who has a 50% performance rate on a lower number of attempts?
Well, finally this can be answered.
One must first come up with a reasonable convention and the one I chose is based on the following:
1/2 ~ 3/8 ~ 11/32 ~43/128
Which basically states, a performance ratio is equivalent to having 4 times the number of attempts with a positive outcome at 4 times the positive outcomes, minus 1. In this convention, per these ratios, the limit as the number of attempts goes to infinite equals (3*1-1)/(3*2)=2/6=1/3. The value of the limit, however, is not the same for differing ratios. For example, with 4/5, in conforming to the same convention, the limit would have an equivalent value of (3*4-1)/(3*5)=11/15 (I'm purposely not showing the involved math).
A fair amount of work dealing with sequences and logarithms can be reduced to the following formula to arrive at a relative score for comparison:
1/3 [ (3*a-1) *(m/b) +1],
where m = the highest number of attempts among each of the participants, a = the number of positive outcomes made by an individual, and b = the number of attempts made by an individual.
For the following example of participants, the relative rates are then computed:
6/13 ----> 6 (formula not required)
5/13 ----> 5 (formula not required)
5/12 ----> 5.39
4/10 ---->5.10
4/9 ----> 5.63
3/6 ----> 6.11
2/5 ----> 4.67
2/4 ----> 5.75
1/3 ----> 3.22
1/2 ----> 4.67
For the instances in which zero positive outcomes were made, there could be debate on whether guessing three times in error versus two times in error is worse. However, I doubt this would be a concern to go into any deep prose on the matter.
Note above, a performance ratio of 1/2 is equivalent to 2/5. In a scenario such as this, I would recommend placing the higher number of attempts at a higher rank. In applying the formula you should also be able to see that 3/12 would be equivalent to 1/3.
Note also how 3/6 is better than 2/4 and 2/4 is better than 1/2.
***********
The above represents a fast attacking of the problem of comparing performance ratios, particularly in discounting the results when the number of attempts is small. I am leaving the above to show this is how math is done. At first one may think a problem was attacked and "solved" to a reasonable extent but then in seeing a shortcoming, further work may provide a more robust solution. In the above problem, the performance ratios of 1/1 and 2/2, though not mentioned above specifically, would yield scores, respectively, of 9 and 11.17. These scores seem out of line with higher attempts, albeit at lower ratios.
In considering a performance ratio and taking into account the number of attempts made, one cannot only look at the fraction alone. This means that there should be a worked up form converting each performance ratio into a new ratio for comparison purposes.
In the desire to suppress the perfection of extremely low attempts, I feel one amazingly simple formula can be used universally now.
In the working up of such a conversion, one may, as I've done, consider the limit of the ratios of two relative or harmonious, if you will, sequences. With a performance ratio, a/b, with letting a=a(0) and b=b(0), sequences could be constructed as follows:
a(n+1) = (a(0)/b(0)+2*a(n)), and
b(n+1) = (7+2*b(n)).
The ratio of the matched sequences as n->inf reduces to the form, a*(b+1)/(b*(b+7)).
And so now, in reciting the earlier example of performance ratios and with the inclusion of several additional performance ratios, one may view the relative scoring by virtue of the formula above:
6/13 ----> 0.323
5/13 ----> 0.269
5/12 ----> 0.285
5/11 ----> 0.303
4/10 ---->0.259
6/9 ----> 0.417
5/9 ----> 0.347
4/9 ----> 0.278
3/6 ----> 0.269
2/5 ----> 0.200
2/4 ----> 0.227
3/3 ----> 0.400
1/3 ----> 0.133
2/2 ----> 0.333
1/2 ----> 0.167
1/1 ----> 0.250
Of course, the above would then be ordered from the highest scoring per applied formula but the numerical results of the formula need not be visible. Note in this formula, the performance ratio for 5/5 would result in 0.500. Also note two performance ratios above, namely 3/6 and 5/13. Though you cannot see a full decimal representation, these are identical, with the derived value of 7/26~0.269.
Heuristically, one may able to see how this functions. The formula, a*(b+1)/(b*(b+7)), could be parsed as : a/b * (b+1)/(b+7). When b is very large, the performance ratio tends to a/b as (b+1)/(b+7) tends to 1. Now for small values of b, the fraction, (b+1)/(b+7), tends to go away from 1, specifically 1/4 in the instance when b=1. This acts as the suppression for a small number of attempts.
Whenever there is a comparison of ratios where it can be reasoned that the same ratio for the instance of where more attempts were made suggests a better performance than when fewer attempts are made, a mathematical convention could be applied.
There can be contention on who would rank higher, one who has a performance ratio of only 1 out of 2 attempts or one who has a performance ratio of 5 out of 12 attempts. Of course some weight should be given to the one who made the greater number of attempts but should the slightly less than 50% performance rate be considered better than the one who has a 50% performance rate on a lower number of attempts?
Well, finally this can be answered.
One must first come up with a reasonable convention and the one I chose is based on the following:
1/2 ~ 3/8 ~ 11/32 ~43/128
Which basically states, a performance ratio is equivalent to having 4 times the number of attempts with a positive outcome at 4 times the positive outcomes, minus 1. In this convention, per these ratios, the limit as the number of attempts goes to infinite equals (3*1-1)/(3*2)=2/6=1/3. The value of the limit, however, is not the same for differing ratios. For example, with 4/5, in conforming to the same convention, the limit would have an equivalent value of (3*4-1)/(3*5)=11/15 (I'm purposely not showing the involved math).
A fair amount of work dealing with sequences and logarithms can be reduced to the following formula to arrive at a relative score for comparison:
1/3 [ (3*a-1) *(m/b) +1],
where m = the highest number of attempts among each of the participants, a = the number of positive outcomes made by an individual, and b = the number of attempts made by an individual.
For the following example of participants, the relative rates are then computed:
6/13 ----> 6 (formula not required)
5/13 ----> 5 (formula not required)
5/12 ----> 5.39
4/10 ---->5.10
4/9 ----> 5.63
3/6 ----> 6.11
2/5 ----> 4.67
2/4 ----> 5.75
1/3 ----> 3.22
1/2 ----> 4.67
For the instances in which zero positive outcomes were made, there could be debate on whether guessing three times in error versus two times in error is worse. However, I doubt this would be a concern to go into any deep prose on the matter.
Note above, a performance ratio of 1/2 is equivalent to 2/5. In a scenario such as this, I would recommend placing the higher number of attempts at a higher rank. In applying the formula you should also be able to see that 3/12 would be equivalent to 1/3.
Note also how 3/6 is better than 2/4 and 2/4 is better than 1/2.
***********
The above represents a fast attacking of the problem of comparing performance ratios, particularly in discounting the results when the number of attempts is small. I am leaving the above to show this is how math is done. At first one may think a problem was attacked and "solved" to a reasonable extent but then in seeing a shortcoming, further work may provide a more robust solution. In the above problem, the performance ratios of 1/1 and 2/2, though not mentioned above specifically, would yield scores, respectively, of 9 and 11.17. These scores seem out of line with higher attempts, albeit at lower ratios.
In considering a performance ratio and taking into account the number of attempts made, one cannot only look at the fraction alone. This means that there should be a worked up form converting each performance ratio into a new ratio for comparison purposes.
In the desire to suppress the perfection of extremely low attempts, I feel one amazingly simple formula can be used universally now.
In the working up of such a conversion, one may, as I've done, consider the limit of the ratios of two relative or harmonious, if you will, sequences. With a performance ratio, a/b, with letting a=a(0) and b=b(0), sequences could be constructed as follows:
a(n+1) = (a(0)/b(0)+2*a(n)), and
b(n+1) = (7+2*b(n)).
The ratio of the matched sequences as n->inf reduces to the form, a*(b+1)/(b*(b+7)).
And so now, in reciting the earlier example of performance ratios and with the inclusion of several additional performance ratios, one may view the relative scoring by virtue of the formula above:
6/13 ----> 0.323
5/13 ----> 0.269
5/12 ----> 0.285
5/11 ----> 0.303
4/10 ---->0.259
6/9 ----> 0.417
5/9 ----> 0.347
4/9 ----> 0.278
3/6 ----> 0.269
2/5 ----> 0.200
2/4 ----> 0.227
3/3 ----> 0.400
1/3 ----> 0.133
2/2 ----> 0.333
1/2 ----> 0.167
1/1 ----> 0.250
Of course, the above would then be ordered from the highest scoring per applied formula but the numerical results of the formula need not be visible. Note in this formula, the performance ratio for 5/5 would result in 0.500. Also note two performance ratios above, namely 3/6 and 5/13. Though you cannot see a full decimal representation, these are identical, with the derived value of 7/26~0.269.
Heuristically, one may able to see how this functions. The formula, a*(b+1)/(b*(b+7)), could be parsed as : a/b * (b+1)/(b+7). When b is very large, the performance ratio tends to a/b as (b+1)/(b+7) tends to 1. Now for small values of b, the fraction, (b+1)/(b+7), tends to go away from 1, specifically 1/4 in the instance when b=1. This acts as the suppression for a small number of attempts.
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