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	Rugged defense Information

How long does it take to catch up the leader ?

Since low scores grow faster, and big scores grow constantly, is it that a late player could catch up a long range player in the historical ladder ?

Well this is virtually impossible if long range player has systematically a better win pct (with scens of approximately equal size played by both). Let's look at it from closer.

Suppose we deal with large scores, so that your winnings at each match does not depend anymore on your previous score.

Suppose Player A has a win percent w_a , has played z_a equivalent standard matches in the period and has a score of S_a at the start of the period. The period is for instance one month or any period needed to end these z_a standard matches. Suppose also that the outcomes of these matches are approximately equal on average to the win percent at the start of the period, ie the player plays as usual. Then at the end of the period, Player A has won z_a w_a w_a / 25 points.
Same for Player B.

Let's say S_b = S_a - D, and D > 0, we suppose Player B has a lower score and we want to know how many standard matches he has to play to catch up Player A. For this possible, he needs of course to have a better win percent on average. So we can say also that w_b = w_a + d, and d > 0, to express that B has a better win percent.

B catches up A if :

=> S_b + z_b w_b w_b / 25 > S_a + z_a w_a w_a / 25

ie if :

=> w_a w_a ( z_b - z_a ) + ( 2 w_a d + d² ) z_b - 25 D > 0

Since we suppose that both players have fixed win percent, this is a function of z_a and z_b the number of standard matches played by both players. We can also fix z_a since we want to know what B has to do, knowing what A did.

First suppose they play about the same number of standard matches in the period, ie z_a = z_b = z .
Then we want find starting from what z the quantity + ( 2 w_a d + d² ) z - 25 D is positive.

This is the case when

z >	25 D d ( 2 wa + d )

Example, suppose the difference of score is D=2000, the difference of win percent is d=10, and suppose Player A has a win percent of 70 (then Player B has 80), then if they play about the same number of matches it would need z = 33 standard matches to catch up...

Now if A and B do not play the same number of standard matches and knowing that A has played z_a matches, B has to play a number of matches z so that :

z >=	wa wa za + 25 D wa ( wa + 2 d ) + d2

Example, suppose as above and A plays only 20 matches, then B has to play z=23 matches.