wins lower then loses?

You said that people with a ranking 200 higher ought to win 75% of the time against their opponents.  What's the ratio (I mean, win percentage) for 300 pts higher, like the thread creator has been saying?

Ok, and now a math problem:

Player A has a rating n points higher than player B.  The two play a total of y games, and Player A wins all of them.  Assuming n non-negative, find an expression for the probability that PLayer A wins all y games.  For n = 300, how many games must Players A and B play against each other for that probability to be less than 0.1?

 (Assume that the games are unrated)

 Ok, now that you've solved that , let's take a look at how many games the thread creator has played against players with ~300 pts higher than he has.  Does he have any probable reason to be alarmed?

 I don't remember the exact numbers. I've read them before a number of times but couldn't find them again during the 5 minutes I just spent on Google. They are out there, though, if you're interested. The answer would be: higher than 75%. Something like 80-90% probably, but I don't really remember.

Ok, and now a math problem: Excellent. I have a couple degrees in math. I love math problems! (Statistics isn't my specialty though)

Player A has a rating n points higher than player B.  The two play a total of y games, and Player A wins all of them.  Assuming n non-negative, find an expression for the probability that PLayer A wins all y games.  For n = 300, how many games must Players A and B play against each other for that probability to be less than 0.1?

Since we are given that player A has in fact already won all of them, the probability of this event having occurred is 1 .  The real question is, what is the a priori probability?

Let me take a stab here (I know I'm probably going to screw something up - I'm too lazy to dig up my first-year stats textbook). Given that player A's rating is n points higher than player B's, let the probability of player A winning one game against player B be p. Thus, the probability of player B winning a game against player A is 1-p. The probability of player A winning k out of y games against player B can be determined by applying the binomial theorem:

(y choose k)*(p^k)*[(1-p)^(y-k)]

You are asking for the special case where y=k. In this case, the formula simplifies to p^y. 

how many games must Players A and B play against each other for that probability to be less than 0.1?

I think we need to know p and y before we can answer this. 

 Ok, now that you've solved that , let's take a look at how many games the thread creator has played against players with ~300 pts higher than he has.  Does he have any probable reason to be alarmed?

He likely has no reason to be alarmed. 

Ok, I should have said:  Assuming that chess games are independent, what is the probability that player A will win any y consecutive games. 

I also didn't know the term a priori probability. 

Isn't there also a way to do this by Poisson approximation (especially for the more general solution, where A wins k of y games)? In this case, lambda = p*y, right?

I'll post again when I figure out the probability of winning based solely on ratings. 

Another fun variation of the problem assumes that each player receives a new ranking after each game.  If player A keeps winning, then Player B should simply expect to keep losing, right? Or does the probability not diverge due to the ever-widening gap in ranking?  Let me rephrase; I realize that a probability cannot exactly diverge:  If the players keep playing and A wins every game, what is the expected number if games they must play for B to win at least one, assuming that each game is separately ranked, given that Player A  begins with n points more than Player B?

http://www.chesselo.com/probabil.html

Also, if he plays against players with more that 350 points higher than he has, he has a >50 percent chance of winning 1 in 6 games.  

 The rating system is very complicated, though, because the score is calculated based on your expected result, so if you win one in 6 games against a player  with a score 350 points higher than you have, your result (1.0) compared with your expected result (.11(by definition for a player with 350 pts more than you)*6 = .66), your rating will change by k (where k is a constant (depends on your ranking and how many games you've played, usually 10, 15, or 20)) * (result-expected value), so you'll end up gaining a couple points.  

here's the table I used:

http://www.chesselo.com/probabil.html