Tournament Math: Antianxiety Medication?
For those of you going to the World Open this week, let me try to ease your anxiety so you can relax and have some fun.
Objectively speaking, there is a range of possible tournament results from zero to nine (there are nine total games at the World Open), and a distribution of probabilities that you will get one of those results.
People tend to get anxious when they start to focus on the tail ends of the distribution ('What if I crash and burn and get 0/9', or 'Oh boy what if I get 9/9 and bring home the million dollar prize?'). Both of these extremes are unlikely (unless you are playing in the wrong section), and tend to evoke extreme emotions that may distract you from the game itself.
Getting practical with the mathtical
To inject reality into the situation, let's look at the math. The following graph shows the probability distributions of overall tournament scores, assuming for each game you have a fixed probability of winning (given in each panel), and that the games are independent. Basically this is the scenario of 'How many heads will you get when you throw nine biased coins?' Obviously these assumptions are simplistic, but the general point holds even if we were to make things more complicated.
The bottom distribution shows the probability of getting different overall tournament scores if the probability of a win in an individual game is 0.8 (this number means if you were to play 100 games, you'd win about 80 of them). If you are lucky enough to have this for each game, then you are either a sandbagger and should be playing in a higher section, or just at the top of your game. Either way, you can expect to score between six and nine points at the tournament, and fewer than six would be disappointing.
(Note while most chess players probably subjectively feel that the probability of winning is much higher than 0.5 in any given game, clearly if you are in the right section and the rating system works, this cannot be the case for most players).
The middle panel shows the more realistic case of a fair coin, when the probability of winning an individual game is 0.5. Note the distribution of tournament scores peaks where it should, around 4 or 5 games won. Technically, the mean number of wins is 4.5, so you should do something like win four and draw one. For those playing in the right section, this is probably what you should realistically expect. Don't worry too much about going 0/9, and don't put too much pressure on yourself to win nine games. Yes, it would be nice, but relax and have fun. If you were just destroying everyone you would be in the wrong section.
The top panel shows the distribution of overall results when the probability of winning an individual game is 0.2. In this case, you indeed have a very low chance of winning many games, and can expect to get between zero and four points at the tournament. This might be a scenario when you are playing up a section and really trying to learn, so buck up you have a lot to learn from the people you are playing!
From slices to pies
The above probability histograms are really just slices in a three-dimension space with probability of winning on the x-axis, tournament score on the y-axis, and overall probability on the z-axis. This more general plot is shown below.
Despite the fact that it is awesomely cool, the above plot is hard to interpret. Hence, let's represent the same information as a contour plot below. In the contour plot, the x- and y-axes are as above, but the z-axis is represented by "isoprobability" lines. That is, the parts of the probability mountain with the same height (probability) are traced together in one line of a single color--the bar on the right gives the mapping from line color to probability.
We can see that as the probability of winning an individual game goes up, the center of the distribution of overall tournament results shifts up as well. Interestingly, this relationship looks quite linear, and we can quantify the expected points as a function of individual win probability:
Expected Net Points=9xP(one win)I drew this linear function as a dotted line in the diagram.
That equation makes sense: if P(one win)=0.5, then Net Points expected is exactly 4.5. If P(win)=0, then you will win zero games. Etc..
So, the take home message is: Chill Out! If you are playing in the right section, then there is a distribution of likely scenarios. Given an honest assessment of your skills within that section, you get a picture of what reasonable (as opposed to extremely optimistic or pessimistic) expectations are. Don't focus on the tail ends of the distribution. Just forget all that crap, go in and play chess. When you start to focus on the tails, and get anxious, focus on the whole distribution, bring yourself back in line with mathematical reality.
That said, of course go in there and take home some scalps, but most importantly have fun with the game.