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Why Masterpoints Deserve Care

Principles of Masterpoint Administration

Teams Versus Pairs

Changing the massive discrepancy between team and pair events

Building a Better Formula

Criteria for a good formula, problems with the current formulas, and a better formula.

What You Can Do

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What should be the general formula for turning rank (r) and the number of contestants (n) into an award? (A field of 20 teams would count as 20 contestants.) You might think that choosing a formula is a matter of opinion, or perhaps even an arbitrary choice. But it is not. There are criteria. To give one criterion so obvious you wouldn't even think to mention it, the first place finisher should receive more masterpoints than the second place finisher.

Another obvious criterion is that coming in first out of 100 contestants is better than coming in first out of 50 contestants, and more generally, the more contestants you have, the better it is to come in first. This principle applies to every rank (for example, coming in 3rd out of 100 contestants is better than 3rd out of 50 contestants).

So what are the the less obvious criteria? And what formula satisfies those criteria?

The masterpoints awarded *per person* should not depend on the number of contestants

I rank this as *the* criterion for a formula to satisfy. Obviously, players are not more meritous just because more or less of them show up. Actually, for technical reasons associated with reliability, and for social engineering, a small advantage for more tables would be acceptable.

Imagine that their were two fields of 40 pairs each, one pair came in first in one of those fields, and now the fields are being combined into one field of 80 pairs. How many pairs in the other field of 40 are going to be better than the best pair in the first 40? There is approximately a 50% chance that there are no better pairs, a 25% chance that there is one better pair, a 12.5% chance that there are two better pairs, and so on.

The amount that this pair is expected to win in the field of 80 should equal the amount that they win in the field of 40. Again, they are not more or less meritorious just for being in a larger field. Put mathematically,

The award for first out of n/2 contestants should equal, approximately, 1/2 of the award for first out of n contestants, plus 1/4 of award for 2nd out of n contestants, plus 1/8 of the award for 3rd out of n contestants, etc.

An award should be (approximately) based on the ratio of rank to total number of contestants.

This principle is not perfect. As noted, 2nd out of 80 cannot receive as many points as 1st out of 40. And though no overalls will be awarded, we can see that 2nd out of 3 is not as good as 4th out of 6 -- 2/3 is average, while 4/6 is slightly below average. Mathematically, this suggests that the award should be based on the ratio of rank to the total number of contestants, *with* some correction factor that is relevant only when the numbers are low. Put more precisely:

(more precise) The masterpoint award should be based on the ratio of r + a to n + b, where r is rank, n is the total number of contestants, and a and b are small constants that make a difference only when r or n are small.

There are many possibilities for a and b. There are logical reasons for some selections, and one can twiddle with a and b to find a formula that looks the best. In the club formula I currently favor, the masterpoint award is based on (r - .5) / n. For the sake of understanding, this is more easily reversed to (n - r + .5)/ n. The two are mathematically equivalant, but the second gives the first place finisher the largest score. For example, finishing 1st out of 100 contestants receives a ratio score of .995. The logical justification is this. If you are first out of 100, you essentially occupy the top percentile, from 99% to 100%. Averaging these, your expected percentile rating is 99.5%, or .995. Another way to think of it is this. If there are 100 pairs, and if you came in first, you beat 99 pairs and tied yourself, giving you a score of 99.5/100.

As noted, 2nd out of 80 is not as good as 1st out of 40, and it does not receive as high of ratio -- 78.5/80 < 39.5/40. Also, 2nd out of 3 contestants receives a score of exactly one half (1.5/3), as does any other exactly middle score.

For this principle to be true in the entire field,

The percentage of awards should be approximately constant.

Suppose that the 10th place finisher out of 100 contestants receives an overall. The 20th place finisher in the field of 200 is just as meritorious, so that finisher should also receive an overall.

The difference between first and second should be bigger than the difference between second and third, and so on.

We have, as a consequence of the Ratio Rule, that an award should be based on the ratio of rank to number of competitors. However, the ratios are linear -- the difference between each is the same. Therefore, this ratio cannot be the final award.

Instead, the ratio must be put into some function. Mathematically, the function must be negatively accelerating (or positively accelerating, I forget which is which). In common sense terms, the function must make the difference between 1st and 2nd bigger than the difference between 2nd and 3rd. (This will also make the difference between 2nd and 3rd slightly bigger than the difference between 3rd and 4th, and so on.)

It would be nice if f(1) equalled infinity, because the ratio 1 corresponds to coming in first out of infinity pairs. It is not exactly clear whether f(0) should be zero or minus infinity. Zero corresponds to being last out of infinite pairs. If f(0) is minus infinity, then f(1/2) should be zero.

One function that meets the first four criteria is x/(x*(1-x)). In this formula, x stands for the ratio of rank to number of competitors, for which I have proposed (t - r + .5) / t. However, this formula did not fit my intuition of how overalls should be awarded. The final test for a formula is common sense intution. Common sense intuition can be wrong, but if a formula strongly violates common sense, it should be suspected. Also, common sense is the only basis for choosing between formulas that meet all of the above criteria.

I prefer another formula, -log(1-x). Using the ratio of (n - r + .5)/ n, this simplifies to log(n/(r-.5)). f(1) is infinity, f(0) is 0, and this fits my common sense judgment for club games. For tournament overalls, it seemed to work slightly better to use log(n/(2 * r - 1)). This simply divides the ratio by 2, so that f(1/2) = 0. I assumed that for tournaments, 10% of the competitors received overalls.

9.80 7.35 5.51 4.13 3.18 2.33

Choosing a value of k to equate the total number of masterpoints given in this regional, my formula yields

9.88 7.10 5.81 4.96 4.33

For 100 pairs, my formula yields

11.63 8.85 7.56 6.71 6.08 5.57 5.15 4.79 4.47 4.19

These numbers look reasonable to me and resemble the awards that are already being given. For the most part, the discrepancies are mathematically appropriate. The criteria for a formula are also met. First, the 50 contestants receive .642 masterpoints per person; the 100 contestants receive .650 masterpoints per person. So there is some advantage to larger fields, which is desirable, but the difference (1%) is insignificant.

Now consider the first place award. By Criterion #3, 9.88 should equal approximately 1/2 * 11.67 + 1/4 * 8.88, etc., which equals 9.75. Again the value is very close.

For club games, I assumed that masterpoints per person should equal .255, which is the value for a 10-table game. Comparing awards:

Current formula: 1.00 .70 .50 .35

My new formula: 1.04 .66 .48 .37

The expected number of masterpoints should be independent of number of contestants across the ranges of ability.

My formula holds masterpoints points per person fairly constant and keeps the expected award for first place about the same, so it would be expected to accomplish equivalence across ranges of ability. But this can also be directly calculated. Consider players who are not very good and have little chance of winning, but who, if lucky, might win an overall. Imagine that they need gold points. They care about the chance of winning an award and the size of the smaller awards.

In my formula, 10% of the contestants receive an award, so these players have the same chance of winning an award no matter how many contestants there are. Also, the bottom award remains about the same for any number of competitors.

Second, to satisfy these criteria, the formula must be of the form k * f( (n+a) / (r+b) ), where k, a, and b are constants, r is rank, and n is the number of contestants. The function f must be nonlinear.

I suggested, based on my common sense judgment, the formula log(n/(r-.5)) for club games and log(n/(2 * r - 1)) for tournament overalls.

The ACBL already uses logs in its tournament formulas (when the number of tables is very large). Therefore, there should be no practical problem with using logs, and given that logs are already being used, they presumably would be the choice for a nonlinear function.

See Cliff for a discussion of the difference between the lowest overall and not receiving any award.