Math, percentages

Can I point out: Sqod's answer is the same as mine. The method is also (I believe from a quick check) equivalent.

Notably it also falls foul of bittersweet's objection that a 2000 ELO rating playing 1490 with win % of 95% would have 104.5% win chance as white, since 95% = x * 50% => x=1.9, 1.9*1.1*0.5 = 1.045

The answer to this bittersweet is that the 55% chance of winning as white that you gave is only applicable between similarly rated players (you can argue whether the true value is 53% or 54% - the method's the same). Let's take it to the extreme - a perfect chess player vs. a complete novice, then white wins 50% of the time, so as the rating difference increases, white's winning percentage decreases to the limit at 50%.

No, my method is a general *formula*. The numbers you plug in are up to you. The formula (method) should work for all numbers in a sensible range, which is why I said it was nicely generalizable. If not, *then* you can correctly claim that the method itself is faulty.

On the other hand, I'll have to think about the other objections above about using the max or min of the two supplied values. It's true that if one of your numbers already took into account the other attribute, then you would use only one of those two values. I don't think that specific objection holds here, though, since your skill level is averaged over all games with both White and Black. If two-part ratings existed, where one were your rating with the White pieces, and one were your rating with the Black pieces, then it would be a different story: you'd want to use the rating for the color in question in order to get a (slightly) more accurate answer.

P.S.---

Admittedly my formula was implied rather than given directly. In general the formula would be... (I hope I can write this without typos or logic errors!)

E(new) = E(old) * PRODUCT of all r[i]

where:

E(new) = the expected value with new conditions

E(old) = the expected value without any special conditions

p[i] = <the altering percentage from condition #i that changes the outcome>

r[i] = p[i]/E(old)

This is an incorrect application of Bayes' Theorem, because the data you used does not conform to the theorem. Your formula should read

P(win vs X|playing white) = P(win vs X)*P(playing white|win vs X)/P(playing white)

Here, X is your specific opponent. We simply do not have the information P(playing white|win vs X) to apply Bayes' theorem.

This is in fact proof that you have applied Bayes' Theorem incorrectly.

That's exactly the right thing to do to try to "break" a formula: try extreme examples, provided the data itself is within the limits that make sense. In your example here I'm assuming that you mean p[1] = 100% = 1.00, therefore:

E(old) = 0.50

r[1] = p[1]/E(old) = 1.00/0.50 = 2.00

E(new) = E(old) * r[1] = 0.50 * 2.00 = 1.00 = 100% probability of winning, which is exactly correct: with your new conditions the perfect player still plays perfectly and wins the game.

You could play with these values, like say a player has a 200% chance of winning, but right off the bat that data doesn't make sense, so neither will the result.

I answered this. Your objection vanishes once you realise that the 55% win rate for white isn't accurate as the difference in rating increases.

You're point is correct, but my method works once you accept the (unstated) assumption that P(playing white|win) is the same as P(playing white|win vs X). i.e. the players in question don't perform significantly better than expected as white or black. (in this case, assuming both players win approximately 55% of the time as white vs. a similarly rated player, and 45% as black in the same situation)

I think this is reasonable in most cases, though I'm sure there are some players for whom this doesn't hold true.

Well, I choose not to accept this assumption 

It is untrue as the counter-example shows.

Actually, I think there is not enough information in the problem to solve it rigorously. Let me first summarize without using mathematics. Your advantage as white depends on who your opponent is. If you are both highly rated and roughly equal in skill, then playing white is an advantage. But if you are highly rated and your opponent is poorly rated, the magnitude of their blunders means that the small advantage of playing white is negligible. Knowing your overall advantage of white doesn't say enough about your advantage with white in a specific situation.

In other words, the overall win percentage as white is an average over all possible opponents, and does not tell you enough about your win percentage as white against an opponent of a certain rating. This is called marginalization,

P(win|playing white) = Σ P(win|vs X, playing white) P(vs X|playing white)

where the sum is over all possible opponents X. You want to know P(win|vs X, playing white), but unless you know your win rate against all other possible opponents while playing white, you cannot extract this from P(win|playing white).

All this bickering.

You can't do this because ratings only tell you the probability of winning with colours being random. Two different players with the same rating might have much better results with white vs. black or equal results or even better results with black. You have to look at an individual player's history to figure this out.

The "counter-example" shows no such thing since it's not actually a counter-example once you amend the 55% win rate (that was given by OP) to account for the fact that the disparity in ratings is larger in this example.

Essentially what you need to do is let P(win|white)=f(x) where x is the rating disparity and have f(x)=55% with no disparity and tending to 50% as |x| increases. The precise function isn't what the OP is concerned with.

I have said much the same as you do in this paragraph already to reject the "counter-example".

As to your second point (made in the last two paragraphs. This seems to be a different point to the preceding one, not just re-worded), you may not be able to know your win % as white vs. a particular opponent but you can (given the function f(x) above) calculate your mean win % as white vs. opponents of a certain grade. Note the OP is categorising the players based solely on their grade, suggesting that this is what we're being asked for, rather than trying to incorporate elements individual to a player such as being more capable as black/white/in certain openings.

Determining the function f(x) for an individual player would be tricky but we can get a good estimate of what it should look like by looking at games played by all players of the same rating and the results, colours and rating differential.

Anyway, I'm just a lurker here and thought I'd provide some nice maths to solve the OP's problem. I'm going back to lurking. If anyone's actually interested in the maths and has questions I'll be happy to answer if you send me a msg but please read the whole thread first. Many of the "objections" raised here have already been answered and so I've ignored them :)

Why lurk? I welcome everyone with thoughtful things to say to contribute more. You make good points, but I don't think every point was 100% correct 

For example, I do believe the marginalization equation I gave is a mathematically accurate way of saying there is insufficient information in the OP, and such mathematical detail is exactly what was requested. As with all equations, the "vs X" part of the equation has a very general meaning. It can mean either a particular opponent, or against a class of opponents with a certain rating. The latter is exactly the f(x) you describe.

But I should point out, knowledge of f(x) is cheating, since you are essentially telling OP that to answer their question, you need to know the answer beforehand. You need to already know the win rate playing as white if "white is 70 elo higher than black".

Can this be answered as just a math question? I don't think anyone has gotten it right yet. For example, something that ordinarily has a a 50% chance of occuring recieving two % boosts, 55% and 60%. There has yet to be a good equation for this.

I already posted the equation in a P.S. at the end of one of my posts above:

----------

E(new) = E(old) * PRODUCT of all r[i]

where:

E(new) = the expected value with new conditions

E(old) = the expected value without any special conditions

p[i] = <the altering percentage from condition #i that changes the outcome>

r[i] = p[i]/E(old)

----------

In your specific example the equation becomes...

E(old) = 0.50

p[1] = 55%

p[2] = 60%

r[1] = p[1]/E(old) = 0.55/0.50 = 1.1

r[2] = p[2]/E(old) = 0.60/0.50 = 1.2

E(new) = E(old) * PRODUCT of all r[i] =

0.50 * r[1] * r[2] =  0.50 * 1.1 * 1.2 = 0.66 = 66%

Seems simple and good, I like it!

We have all been conditioned through many years of school and many textbooks that all maths questions can be solved, if we simply squiggle enough symbols on a page. The seemingly simple problem you posed seems solvable.

But this is not the case. There is insufficient information to solve it rigorously. This I demonstrated with the marginalization formula. Assumptions are required to arrive at an answer. Some assumptions are more reasonable than others, and the answer is only valid in the domain where the assumptions hold.

I have already solved the answer in the simplest of terms.

Stronger player 51% chance of winning

Weaker player 49% chance of winning

You can not factor an equation because there to many missing unknown's.

Furthermore, The assumptions which are based on the player in question are of second hand importance compared to the position in question.

People always way, way, way, way overthink this.

Look at how much rating you will gain from winning and lose from losing.

That is all the information you will need.

The chance of winning is w/(w + l) where w is the amount of rating you gain from winning and l is the rating you lose from losing.

It fundamentally has to work that way or else you wouldn't have the rating you have.

The excecption is when you do very lopsided battles, like 1600 vs. 2800. The ELO system breaks down at that point because there aren't enough decimal places, and people that different aren't really supposed to play rating games. In general, if carlsen played lots and lots of rating games against a very low rated player, either he would eventually lose all of his rating because it would disallow him from gaining any, or his rating would rise without limit because he would always gain 1 rating point.

If you see another opponent across the board and you want to know your "chances" of winning, just look on the right where it says "rating vs. rating (win a| loss b) and it's very very very simple to calculate.

No, one of the weaknesses of Elo, Glicko, etc. is that they do not account for the advantage of playing first. This in fact is the motivation of the OP, who asks if it possible to combine information from Elo with the knowledge that white has x advantage to predict more accurately the chances of winning.