Team events: beating the bookmakers?!

Team events: beating the bookmakers?!

By Peter Zhdanov

Let’s say you have two teams composed of four chess players each. This is typical for the Chess Olympiad, World Team Chess Championship and other important events. How high is the probability that team A will win? How probable is a draw? Team B’s victory? Here is a relatively easy step-by-step guide. As an example we will use the match Romania-Russia from round one of the recent Women’s World Team Chess Championship.

Step 1

Calculate the rating differences on each board and find out the expected score of the player.

Example: Foisor (2401) vs Gunina (2505), a rating difference of 104 points. If we consult the FIDE Handbook, we will see that it Foisor is expected to score 0.36 points per game against Gunina.

Step 2

Find out the expected probability of a win, draw and loss in each game.

Example: Foisor has White. The FIDE website tells us that she won 50% of her White games and drew 34% of them. Alternatively, you can use a ChessBase database. It offers even more accurate statistics. Since a win is worth 1 point and a draw – 0.5, we obtain a simple equation:

0.5x+ 0.5*0.34x=0.36

x is about 0.537. The probability of a win is approximately 0.5*x=0.2685, i.e. about 26.9%. The probability of a draw is, correspondingly, (36%-26.9%)*2=18.2%. The probability of a loss is (100%-probability of win – probability of draw) = 100-26.9%-18.2%=54.9%

Note: apply this formula to the weaker of the two players. Otherwise you might end up in a situation when there is no solution. For example, if the expected result is 0.9 points per game and the player draws 30% of the games and wins 40%, then it is impossible to achieve a 0.9 result while maintaining the proportion, so we should compose the equation for the second player (expected result is 0.1).

Step 3

Repeat steps 2 and 3 for all the four boards in the team.

These are the input variables:

Step 4

Each game can have three theoretically possible results: White wins, draw, Black wins. Hence, if we have 4 boards, there are 3^4=81 possible outcomes of the match. It is easy to calculate each probability.

Example 1: board 1 wins, board 2 loses, boards three and four make draws. The probability of this outcome (using data from Step 3) is: 0.269*0.612*0.034*0.204*100%= appr. 0.338%. As you can see, this is a very small probability. Example 2: and how high is the chance that Russia will win on all the four boards? 0.549*0.636*0.594*0.623*100%= appr. 12.92%.

Step 5

To calculate the probability of the match ending in a draw/victory for a certain side, we have to add up the probabilities of all the corresponding outcomes. All the 81 possible outcomes:

1111, 1110, 111=, 1101, 1100, 110=, 11=1, 11=0, 11==, 1011, 1010, 101=, 1001, 1000, 100=, 10=1, 10=0, 10==, 1=11, 1=10, 1=1=, 1=01, 1=00, 1=0=, 1==1, 1==0, 1===, 0111, 0110, 011=, 0101, 0100, 010=, 01=1, 01=0, 01==, 0011, 0010, 001=, 0001, 0000, 000=, 00=1, 00=0, 00==, 0=11, 0=10, 0=1=, 0=01, 0=00, 0=0=, 0==1, 0==0, 0===, =111, =110, =11=, =101, =100, =10=, =1=1, =1=0, =1==, =011, =010, =01=, =001, =000, =00=, =0=1, =0=0, =0==, ==11, ==10, ==1=, ==01, ==00, ==0=, ===1, ===0, ====

Where 1 stands for a win of a player on the corresponding board, = for a draw, 0 for a loss. E.g., 1111 means that all the players in the first team won their matches. 1110 means that three first players won their matches and player #4 lost.

Here are the 31 possible outcomes that result in Team 1’s victory:

1111, 1110, 111=, 1101, 110=, 11=1, 11=0, 11==, 1011, 101=, 10=1, 1=11, 1=10, 1=1=, 1=01, 1==1, 1===, 0111, 011=, 01=1, 0=11, =111, =110, =11=, =101, =1=1, =1==, =011, ==11, ==1=, ===1

Outcomes that result in draws (19):

1100, 1010, 1001, 10==, 1=0=, 1==0, 0110, 0101, 01==, 0011, 0=1=, 0==1, =10=, =1=0, =01=, =0=1, ==10, ==01, ====

The remaining 31 outcomes represent Team 2’s victories. You don’t really need to calculate them, but here is a list anyway:

1000, 100=,10=0, 1=00, 0100, 010=,01=0, 0010, 001=, 0001, 0000, 000=, 00=1, 00=0, 00==,0=10, 0=01, 0=00, 0=0=,0==0, 0===,=100, =010, =001, =000, =00=,=0=0, =0==, ==00, ==0=,===0

Step 6

Calculate the probabilities of all the outcomes of Team 1’s victories (see Step 4) and add them up. Similarly, calculate the probabilities of outcomes that result in a draw and sum them up.

Note: I am using an Excel sheet for this so that I don’t have to perform this calculation manually. Just input the variables from Step 3 into the sheet. The software will do the rest.

Example: in the example above (Romania-Russia) the chance of Romania’s triumph is about 12.91%; the probability of a tie – 17.55%. The remaining 69.54% stand for Russia’s victory.

Step 7

Translate from Math language into bookmaker’s jargon. If Romania’s chance to win is 12.91%, it means that the bookmaker is supposed to offer odds of about 100/12.91=appr. 7.75 for Romania’s victory. In the reality the coefficient will probably be lower, because the bookmakers charge a certain commission.

Similarly, the odds for a draw should be about 5.7 (100/17.55) and for Russia’s win – 1.44 (100/69.64). I have compared these coefficients to a line offered by one of the bookmakers:

Romania – 11, Draw – 4.5, Russia – 1.31

We can see that, according to our model, it didn’t make much sense to bet on Russia and/or on a draw, because the offered odds were below the expected values. On the opposite, Romania was somewhat underestimated by the bookmakers.

I have considered a few other lines, and in most cases the coefficients obtained using our model were relatively close to the ones used by bookmakers. However, there were some notable exceptions too.

Remark 1

Unlike the attention-getting title of the article suggests, this model was not designed for beating the bookmakers. It is intended for those inquisitive minds who have always wanted to know how high the chances are for “their” team to succeed. Naturally, the method described above works for any number of boards, so it is not restricted to 4 vs. 4 competitions. Initially, I have been experimenting with 2 vs. 2 models for the sake of simplicity.

Remark 2

No model is perfect. This one has an underlying assumption that players perform according to their FIDE ratings and similarly to their previous results, e.g., a “drawmaster” makes more draws than an aggressive player. Obviously, other parameters can be considered. For example, the personal score between two people; tournament standings; current shape of the players, etc.

Or, another interesting idea, use not FIDE Elo, but estimates of FIDE Elo with White and with Black. Obviously, most people perform stronger with White than with Black, so it does make a difference.

Remark 3

Even if you discover that the model suggests the odds to be 20, while the bookmaker offers 40, don’t be in a hurry to bet. While this strategy should be beneficial in the long run, your chances to succeed in a particular case are very slim. For example, Romania succumbed to Russia 0.5-3.5:

So if you had bet on Romania (11 is a higher coefficient than 7.55, see Step 7), you would have lost your money.

Acknowledgements

This article would not have been written without the support of my best friend and avid math & chess fan Nikolai Smirnov, with whom we discussed the idea in general and the potential application of the model in real life during the Chess Olympiad 2012.

Peter Zhdanov is an IT project manager, expert and author of two books on parliamentary debate, BSc in Applied Mathematics & Computer Science and a PhD student in Sociology. In chess he is a Russian candidate master, author, manager of grandmaster Natalia Pogonina and editor of the Pogonina web site.

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