Math, percentages

Hello. I am trying to find the overall percentage of winning a game. Let's say that white has a 55% chance of winning a game. In addition, white is 70 elo higher than black, giving him 60% chance of winning the game without including the advantage he has in playing the white pieces. What is whites total percent chance of winning the game before it begins? 

If you could show/explain your work, that would be fantastic, thanks!

Good question. I *believe* you just multiply the probabilities, but I'll have to think about it some more and maybe try to find similar problems (online). In general you multiply probabilities for sequential events, each of which is independent of the other and has its own probability. The position success rate A% is certainly independent of a given person's playing success rate B%, but they aren't exactly sequential events in your scenario. If they can be considered sequential events then the probability is (A*B)%.

I believe an analogous situation is a sequential set of (unpolarizing) light filters, though: if each successive filter panel causes incoming light to drop a given percentage then if all the filters are combined into a single pane the result should be the same for one step as it is for multiple steps. Therefore that logic would justify muliplying the A% filter for the position strength by the B% filter for player strength. In your example that would give .55 * .60 = .33 probability of winning.

http://math.stackexchange.com/questions/122522/probability-of-sequential-events

Bayes' Theorem:

P(win|playing as white) = P(win)*P(playing white|wins)/P(playing white)

you state that P(win)=60% given the ELO difference and let's assume P(playing white)=50% and using your figure of P(playing white|wins)=55% we have:

P(win when playing white against this opponent) = 0.6*0.55/0.5 = 66%

RiffArt,

That's a good train of thought, using Bayes' Theorem, but her question already states that the person in question is playing White, therefore P(you're playing white) = 100% = 1, which reduces your equation to...

P(you win | you're playing as White) = P(win) * P(playing as White | you win)

...and I don't see that introducing a player of different strength into that expression can be justified.

      Have you tried THIS place; http://www.chessmetrics.com/cm/  

There is no clear advantage playing with the white pieces in world class chess, let alone our chess lol .

I just took a look. That's about ratings, peak ratings, etc., so it's not applicable to this question.

I cain cownta 19 wit my shoos awf       mowur took my baby tow 

Sqod, this player should be winning more that 60% of the time, not 33% of the time.

RiffArt, your answer is closer, however, what if a 2000 player plays a 1490 player as white? .95*.55/.5=1.045=104.5% chance of winning?!

I believe the answer to my original question should lie in the 60-65% range.

 The OP's question isn't really about Chess. It's just basic probability theory: Given the information that a player with the white pieces has a 55% chance of winning (no other information given about the player's skills &c.), and that a specific player A has a 60% chance of winning over another player B, what can we say about A's chance of winning if we are told that A will have the white pieces?

As the question indicates the two are independent, I think the answer is 65%. One condition gives a 5% advantage, the other 10%. Both apply to the same event ergo 65%. Overly simple by me probably. 55-60% chance of winning though? No draws?

I agree with Fiveofswords first paragraph completely.

His second paragraph was more of a personal opinion lol.

Fiveofswords first paragraph is correct!

You can not measure in such terms.

You are trying to factor human percentages.

However, You can not use those factors because the position is what needs to be measured in order for 1 side to win, lose, or draw.

For example in the below diagram:

Lets say Magnus Carlsen is playing white in the above position.

Magnus Carlsen is ranked 2800+

The opponent who is playing black is my little cousin.

He is ranked 1200+

Magnus is using the white pieces and is the highest ranked chess player in the world.

What percentages do you give him to win with the white pieces here?

The simple fact of the matter is it is impossible for him to win with the white pieces here.

All he can do is draw or lose!

The position is beyond the point of recovery!

Which is why you can not measure chess in such human/player terms!

You have to base the measurement on the line/position!

In which case that math percentage has already been done!

They use Math statistics to find the winning percentages of a position.

Your answer lies in the knowledge of chess databases.

http://www.chess.com/explorer

http://chesstempo.com/game-database.html

http://www.365chess.com/opening.php

Not *quite* true. Here's a very handy online ELO probability calculator that should answer a number of questions people ask in this forum, including your example:

http://chess-db.com/public/winprob.jsp

If you enter the values 2800 and 1200 it does say 0.0% probability of a win for the 1200 player, but that's only because the probability is so small that when you round it off to the nearest 10th of a percent, it's closer to 0.0% than to 0.1%. Continuous probability curves never exactly hit 0 or 1, they just approach one of those values as a limit.

Bittrsweet,

Yes, that rapid drop in percentage is why I was so hesitant to say definitively that the percentages should be multiplied. That seems way too low. However, I'm still leaning toward that solution, even if it goes against my intuition. I just don't see that Bayes' Theorem applies: that formula is for situations where you can provide another value for when the conditional probability is swapped from P(A|B) to P(B|A). It doesn't matter if the numbers used from that method came out closer to what one would expect: if it's the wrong logic and wrong equation, the resulting number is meaningless. Also Bayes' Theorem is used only for conditional probabilities, and your opponent's rating has nothing to do with the opening (or equivalently, color played) you choose, so those two probabilities are definitely independent, not conditional. 

http://www.mathgoodies.com/lessons/vol6/independent_events.html

OK, I'll keep looking online for analogous situations where a mathematical solution was provided. The question very much intrigues me because it seems so simple yet the numbers and formulas don't seem to apply.

To answer this I think we need more assumptions and maybe even empirical data  (and in that sense, I guess the question is really about Chess, though I think the question is more interesting in the general case). 

There are other similar questions related to general rating systems and such that are quite fascinating, some time ago I sat down one evening and fooled around a bit with this, maybe I'll post some notes tomorrow (it's getting quite late here). Anyway, if anyone has some good references to this kind of stuff please post them!

I'm pretty sure I just figured it out. I used the assumption that the expected value of the outcome of a perfectly even game was 50% = 0.50 points of the games played, which is what would happen when playing an opponent of equal strength in a game where there were no opening advantage.

Since each of the percentages given in the original question (55%, 60%) boost the expected value to slightly above 50%, I figured out the (multiplicative) factor that would produce 55% from 50%, and also the (multiplicative) factor that would produce 60% from 50%, respectively.

If...

55% = x * 50%

...then...

0.55 = x * 0.50

0.55/0.50 = x

1.1 = x

Therefore 55% is 50% boosted by the factor 1.1.

(2)

If...

60% = x * 50%

...then...

0.60 = x * 0.50

0.60/0.50 = x

1.2 = x

Therefore 60% is 50% boosted by the factor 1.2.

Therefore the expected value of 0.50 from playing an equal opponent with no color advantage should be boosted successively by factors of 1.1 and 1.2 (in any order) to take into account color advantage with playing strength advantage. Therefore the expected value with both of those boosts would be...

 0.50 * 1.1 * 1.2 = 0.66 = 66%

Oddly, I've never seen or heard of this method of figuring out such problems, but it generalizes nicely to any such problem. The key is to multiply your *expected value* (here 0.50) instead of 1.00, since that makes a huge difference in the answer. My light filter analogy was partly flawed since that model allowed only for light to be *reduced*, never boosted. If I'd started out with a 50% filter, then the analogy would have been quite accurate.

He has a 60 % winning chance.It is the maximum winning chance that was spoted with considering one aspect and ignoring other aspect in the data collecting.So,this winning percentage always true in any system regardless other aspect exist in data collecting.If it isn't,then the 60% winning chance is fails in the beginning of data collecting. 

100% CHANCE of of losing...or winning because ya jus don't know.

Sqod's method gives the correct answer. The higher rated player will win 66% of the time with white and 54% of the time with black, averaging to the original 60% advantage overall.

Sqod method is incorrect because he has based his method on numbers which the OP Bittrsweet gave him.

First lesson:  Know the source of your information to see how reliable it is!

The OP Bittrsweet made those percentages up with no base!

Which in laymens terms means the percentages and answers you came up with are completely useless.

If all you wanted to do was pick random percentages than I could of did that very simply.

Stronger player chance of winning: 51%

Weaker player chance of winning: 49%

The simple bases for the above percentages is:

You would expect the stronger player to win more than 50% of his games against the weaker player.

You would expect the weaker player to lose or come up short less than 50% of his games against the stronger player

However, even those percentages are completely useless because even strong players knowingly enter into inferior positions.

At which point the position itself has a percentage.