Coordinator
Jan 9, 2013 at 12:45 AM
Edited Jan 9, 2013 at 12:53 AM

For background, please refer to this article at Chessbase: http://www.chessbase.com/newsdetail.asp?newsid=8751
For the conclusion, skip to the bottom.
Ivanov Borislav, unrated, scored 6.0/9 points against average opposition of 2500 Elo. Is this impossible? No, but it is unlikely.
While the present program only calculates the percentage of times an underdog can win a tournament, it can be used to suggest how likely an underdog might be a fourth place finisher that nearly won. To do so, we must make some adjustments.
First, we note that Ivanov Borislav could have won outright with one extra win. To account for this, let us assume that he is in fact not 2227 Elo, but much stronger, like 2300 Elo. Intuitively, these extra 73 points should give him a decent
chance of winning an extra game, which would have put him in first place (intuitively, I am guessing here, but that sounds reasonable). This should allow the program to work properly since, as I stated, the program only counts the percent outright wins
of a tournament, not runner up.
So, using a spreadsheet or statistic calculator, from the crosstable of players in the above link, you can calculate for all players, assuming Borislav is 2300 Elo (not 2227), the following:
X (avg) rating for all players = 2526 Elo
Sigma (Std. Dev) for all players = 98 Elo
Number of Players = 16 (input 15 into program at prompt since it adds one to total)
Number of game played in tournament: 9 games
Number of Iterations: input 1M = 1000000
On a fast PC, about 2.25 minutes later, you get the below answers (I ran the program three times):
1) Subject Player of Elo 2300 won 401 times out of 1000000 iterations, a total of 0.0% percent [works out to 1 in 2494 odds, or 1:2494]
2) Subject Player of Elo 2300 won 204 times out of 1000000 iterations, a total of 0.0% percent [works out to 1 in 4902 odds, or 1:4902]
3) Subject Player of Elo 2300 won 277 times out of 1000000 iterations, a total of 0.0% percent [works out to 1 in 3610 odds, or 1:3610]
The program gives different outputs each time, since, for ease in programming, the players are simulated as a Gaussian distribution, rather than each player Elo being input, see the other comments in this discussion and the user manual. In fact, when
running these simulations it pays to watch how high the highest simulated player isif this player is much higher than the actual highest rated player (in this case two players rated
2638), the simulation may not be accurate (I will explain this in another post and below).
So it's possible, not likely, given these assumptions, that Ivanov Borislav did indeed win as much as he did, without cheating.
Being even more generous (i.e., giving Borislav even more benefit of the doubt), we can say that in fact his Elo is closer to 2350 rather than 2227. With this assumption, the average rating X (avg) = 2529 Elo, Sigma (Std. Dev) = 90 Elo and now running
the program as above gives:
1) Subject Player of Elo 2350 won 147 times out of 1000000 iterations, a total of 0.0% percent [works out to 1 in 6802 odds, or 1:6802]{highest rated player = 2762 Elo}
2) Subject Player of Elo 2350 won 1462 times out of 1000000 iterations, a total of 0.1% percent [works out to 1 in 684 odds, or 1:684]{highest rated player =
2614 Elo}
3) Subject Player of Elo 2350 won 2013 times out of 1000000 iterations, a total of 0.2% percent [works out to 1 in 497 odds, or 1:497]{highest rated player =
2613 Elo}
I underlined outputs #2 and #3 since the highest rated player in the simulation was smaller than the actual
2638 Elo players in the Zadar Open. What this means is that we can be more sure that the simulation is accurate, since I have found in practice if there is one highly ranked player, statistically that player will 'skew' the final wins
by 'winning most all the tournaments' in the simulation. Hence, for cases #2 and #3, we can be sure this is not the case, since in fact these simulations had highest rated players that were weaker (i.e. 2613 and 2614 Elo) than the actual 2638 Elo players
in the Zadar Open.
Thus, even more favorably for Ivanov Borislav, one can say that he would have about 1:497 or 1:684 odds of winning, if his true Elo really rated 2350 Elo rather than 2227 Elo. Or roughly, taking the average, 1:590 odds of winning the Zadar
Open.
Hence, if his Elo is in fact 2227, not 2350 Elo, the odds of coming in fourth place, rather than winning (which is what we calculated above), is probably not that much different from the above (my intuition).
In conclusion, Ivanov Borislav might have approximately a 1 in 600 chance of doing as well as he did in the Zadar Open, which was fourth place, which does not defy the laws of chance, albeit it is not likely and his style of playing does seem like
that of a computer, as pointed out in the YouTube video http://www.youtube.com/watch?feature=player_embedded&v=Jr0J8SPENjM by Valeri Lilov.


Coordinator
Jan 9, 2013 at 2:42 AM
Edited Jan 9, 2013 at 2:54 AM

Another way of viewing this problem is to try and calculate the probability of Bulgarian Borislav Ivanov scored 6.0/9 points.
I could be wrong on this calculation, especially the part about draws, but a table can be constructed, assuming Ivanov is really rated at 2350 Elo, not 2227 Elo (giving Ivanov the benefit of the doubt that he really is stronger than his Fide rating says),
of Ivanov's wins over his opponents.
The table would look like this:
Opponent Opponent Elo Elo Difference
Probability of Ivanov winning points from this opponent
Schachinger Mario 
2426 
76 
0.4 
Jovanic Ognjen 
2538 
188 
0.26 (zero)

Kurajica Bojan 
2565 
215 
0.23 
Kuljaševic Davorin 
2561 
211 
0.235 (draw)

Zelcic Robert 
2560 
210 
0.235 
Kožul Zdenko 
2638 
288 
0.16 
Sumets Andrey 
2638 
288 
0.16 (draw)

Predojevic Borki 
2600 
250 
0.195 
Šaric Ivan 
2626 
276 
0.17 (zero)

Since Ivanov lost to Ognjen and Ivan, he got zero points and zero probability points, while Ivanov drew Davorin and Andrey, so he gets double probability points for that (I'm not 100% sure on this however, but pretty sure; these probabilities are underlined
below). Hence Ivanov probability of winning, since these events are not dependent, is the sum of the nonzero probabilities below.
0.4 
0 
0.23 
0.47 
0.235 
0.16 
0.32 
0 
0.17

Thus, taking the sum gives:
Or, taking the inverse gives: 11306.48394
Or, Ivanov's chances of winning 6 out of 9 points is about 1 in 11306.
This is a more direct and if I've not made any mistakes more accurate estimate of Ivanov's chances of winning fourth place in the Zadar Open.


Coordinator
Jan 9, 2013 at 3:07 AM
Edited Jan 9, 2013 at 3:30 AM

We can also compute the probability that Ivanov is really 2227 Elo and not 2350 Elo. From the previous post, the analysis is the same, but the probabilities are more stacked against Ivanov winning, so you get this table:

ELO opponent 
Elo Ivanov 
Difference 
Prob. Ivanov Wins 
Prob 






Schachinger Mario 
2426 
2227 
199 
0.25 
0.25 
Jovanic Ognjen 
2538 
2227 
311 
0 
0 
Kurajica Bojan 
2565 
2227 
338 
0.125 
0.125 
Kuljaševic Davorin 
2561 
2227 
334 
0.125 
0.25 
Zelcic Robert 
2560 
2227 
333 
0.125 
0.125 
Kožul Zdenko 
2638 
2227 
411 
0.08 
0.08 
Sumets Andrey 
2638 
2227 
411 
0.08 
0.16 
Predojevic Borki 
2600 
2227 
373 
0 
0 
Šaric Ivan 
2626 
2227 
399 
0.085 
0.085 
Note, again, that the draws against Davorin and Andrey count as twice the probability of winning.
Taking the sum of all nonzero rows at the last column gives the probability of Ivanov winning 6 out of 9 points against such opponents, or:
or, taking the inverse of this gives:
1 in 941176.4706, or 1 in 941000.
That is, about 1 in a million. (Technically you have to add to this amount the very small chances of getting 7, 8 and 9 points out of 9, which will increase the odds slightly, since the relevant statistics are the odds of 'at least' getting 6 out of
9 points, but, given the odds are so small, I don't think the probability will change very much at all).
Possible that Ivanov did this by himself, but not all that likely. The only way to prove it of course is to play more games so we can see his truth strength.

