AbigailII:
Finally, there's a tactical element. Suppose you're in a 6 player section. Your current score is 4 out of 4, your opponent has 4 out of 4 as well. There's one other section, and it's already know there's a single winner there. Suppose you, your opponent, and the winner of the other section are all equal in stength, so you estimate a 50% chance of winning any game between any two of you. If you play for a win, you have a 25% chance of winning the tournament - 50% chance of beating the current opponent, and 50% chance of beating the other winner. But if you go for a draw, your chance of winning the tournament increases to 50%! (25% of being a solo winner, 25% of joined winner).


First and since you accept equal strength of the opponent, we must assume that the opponent should also have equal cleverness in his decisions. So he would accept the draw as then he would have more chances to win the tournament (50% instead of 25% as you said-i don't agree with the 50%-number but only with the conclusion (that he would have more chances to win)).......

So after accepting the draw both 2 players would advance to the final, so the final would have 3 players of equal strength that would take the same decisions in draw cases.

But there is one important thing also: What a tie does? Forgetting about Sonneborn-Berger criterions of ranking we have the following cases:

We assume that in the final ranking if there is a tie of 2 or 3 players, then all players win!
That means and since all players have the same smartness, they will agree all games to a draw and all be the winners with a probability of 100%! If they risk one or more games by not offering a draw or rejecting one then they would have less than 100% chance of winning. So they would all offer a draw from the beginning and their equal-generally opponents will accept and win......

But since you assume that the 50% number, is the probability of winning at the final between 2 players, by having the statement: "and 50% chance of beating the other winner.", you assume that a tie repeats the game. Having that in mind.......:

We assume that in the final ranking if there is a tie between 2 or 3 players, then all series of games are repeated.
This means that the probability a player has to win the final is 1/3 or 33.3...%
If he didn't offered a draw at the first round he would then had 50%·50% = 25% chances of winning at the final.
So the player who offered the draw to achieve better chances to win the final, was correct in his decision. The same conclusion with yours but the probability to win is 33.3...% and not 50%.


And we can find a general statement with this observation: If we have for example 2 groups of N equal players each, that would compete for the final.
Then if a specific player tries to win to advance in the final he would have (1/N)·1/(1+X) chances to win the tournament, assuming a tie repeats the games (X (0<X<=N) number of players from the other group that promoted to the final).
But if the specific player offered draw in every game and his opponents accepted and did the same to all of their games(same IQ), they would then have 1/(N+X) chances to win the tournament, assuming a tie repeats the games.
And since (1/N)·1/(1+X) < 1/(N+X) the smart players of the one group that drawn all their games will have better chances to win the tournament than they would have if they fight for having wins.....
But all these are impractical cases as there are no equal players.....

As for the r being smaller than s, in your below example, i think that this is not a good thing to happen in such a game-site as it should be s=r. And the fact is that at Braiking and specifically at Backgammon it is r=s=8 for players rated above 2100. In fact for players with more than 2100, the system is easy: "+8 -8 =0" with equal(in a +-400 range BKR points) BKR points and "+1 -16 =-14" for players with not equal(in a range more than +-400 BKR points) BKR points.......

I don't like this system at all but...........I would prefer the range factor to be smaller and the rating change to be more wild.....
For example:

A game against opponents that their BKR difference is less than or exactly 100:
A win for the "stronger" player = "+10 -10 =0"
A win for the "weaker" player = "+10 -10 =0"

A game against opponents that their BKR difference is less than or exactly 200 and more than 100:
A win for the "stronger" player = "+8 -12 =0"
A win for the "weaker" player = "+12 -8 =2"

A game against opponents that their BKR difference is less than or exactly 300 and more than 100:
A win for the "stronger" player = "+6 -12 =0"
A win for the "weaker" player = "+12 -6 =2"

and so on........

That would tend to rise the BKR average but that's not bad at all and it happens in many Chess lists........

Lacking fantasy.........??????
RYTFBE!