Considering the question of the 26-losses-and-a-draw sequence suffered by Stephen Crockett, whose curious playing record we have been looking at this past week, it occurred to me that we have a precedent in a story many chess players learn soon after they learn the moves. It's described on Wikipedia as the "wheat and chessboard" problem. I'm sure most readers know how it goes.
Sometimes it ends well, sometimes it doesn't.
Now in our current example, nobody is calling for anybody's execution, while controller of the Grand Prix isn't quite "a high-ranking advisor", but let us perform our own version of the problem, just to get ourselves a starting-point figure to work with.
It is, as I say, a starting-point figure, no more than that, so to obtain it, I've taken it that we wish to calculate the probability of a random player1 obtaining such a score in a twenty-seven game sequence2 against evenly-matched opponents.3 I've neglected the question of colours and I've chosen to estimate the probability of each result as follows:
Win 40% Draw 20% Loss 40%.
Obviously the reader is welcome to employ different values and thereby obtain a different result.
A different result, that is, from 1.37 billion to one.
What a figure. How do we get there? Well, our random player, who has to score no more than 0.5 out of 27 (but 0.0 will do) has 27 chances to get his or her draw, which draw has a probability of 0.6. (Or 3/5 if you prefer.)
That's 27 x 0.6, which is 16.2.
In the other 26 games, they lose, each instance of which which has a probability of 0.4, or 2/5. Over 26 games that's 0.4 to the power4 of 26, which is where we start remembering the grains of wheat on the chessboard, since the number that emerges is so high.
Because it's easier on your calculator you may like to consider this as constituting 1 over 2.5^26, and our overall calculation as 16.2 ÷ 1/2.5^26, which is 16.2 ÷ 22204460492.5. No really.
Now just as to get the fraction of (say) 5/20 down to betting odds, we divide 20 by 5 and then take off 1 to get us 3-1, so with our calculation we divide 22204460492.5 by 16.2, which is 1370645709.4, take off 1, which is 1370645708.4, lose the 0.4 because frankly who cares, and that gets us
which is quite a number.5 A number higher than 1.37 billion. That number, to one.
Now let me reiterate, these aren't the odds against Mr Crockett performing the feat, they're the odds, given the values suggested,6 against a random player performing the feat (or worse) against evenly-matched opponents.
With any given player, Mr Crockett not excepted, it doesn't work precisely that way, for all sorts of reasons, not least because chess games aren't entirely independent events. The result of one can be affected by the result of a previous one, and - as has been suggested by Mr Crockett's defenders - perhaps some people are more likely to be affected than others by a bad start to a tournament and can go on to lose a rack of other games.
No doubt. No doubt. But, when we say "more likely" - how much "more likely" do we mean?
Likely enough to bring down a figure of 1.37 billion to one to something we can actually believe?
How likely is that?
Two points here: two inescapable points.
The first is that there is no "normal" way this sequence can have taken place. Either there is something exceptional about the player, or there is something exceptional about the way he has conducted the games.
The second is that the odds against the sequence are simply too great - far too great, massively too great - for us to take the first of those two explanations for granted.
If you want your competitions to have any integrity, and if you want your grading system to have any integrity, and you have a set of results that - things being equal - is so unlikely that we can measure the odds at a billion to one against, then you simply cannot assume that things were equal. It's absurd.
So if you're looking for explanations, then they have to be good ones. They can't just be handwaving. They have to have enough evidence behind them to make up the gap between more than a billion to one, on one hand, and something we think is credible, on the other.
Because that's a hell of a big gap, even by the standards of credibility gaps.
So do we have an argument for the credibility of this sequence that rests on more than assertion? Anything that rests, reliably, on the record of other players, in similar circumstances? Because if not, our evidence for Mr Crockett's own explanation for Mr Crockett's record is simply Mr Crockett's word.
Which is OK, if it's OK with you. But on the other side of the argument, there's a straightforward explanation, one that fits all the facts known to us, and one that would fit a pattern of results whereby a player who loses the first game in a tournament goes on to lose most or all of the rest.
And on that side of the argument there's some really big numbers.
Big numbers, like more than a billion to one.
- - -
1 Of course another way to work would be to calculate the odds of each result as predicted by the ECF grades then obtaining. I've chosen so far not to do this, though anybody else is welcome to. This is partly because I particularly wanted to look at the odds in general, for illustrative purposes, partly because I don't entirely trust the integrity of the grading list. For what it is worth, if I have read the sequence right then Mr Crockett was graded on average of slightly more than 1.5 points higher than his opponents with listed grades: however, at least one of the two players without a listed grade may well have been a stronger player, perhaps substantially so.
2 For the actual sequence scroll to collapsed.
3 Of course where the opponents vary in strength, the odds against such a disastrous sequence would actually increase, provided we assume (as I think we can) as much variance towards weaker opponents as stronger. (To illustrate this, examine the equation 1/2 x 1/2 x 1/2 = 1/8, which is 7-1 against, whereas, say, 1/4 x 3/4 x 1/2 gives 3/32, or more than 10-1.)
4 ^ apparently means "to the power of". No, we didn't use it when I was at school, we used them superscript things, as we did not call them then.
5 If you didn't follow that, I don't blame you. Maybe try doing the calculation yourself but as if it was only, say, four games rather than twenty-seven, so you see how it works.
6 I tried the calculation again, inputting the skewed values of win 25%, draw 25%, loss 50%. It brought the odds down loads. To, ah, a mere 3.3 million to one.
[Thanks to Bat for his help with the maths.]
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