Comments on MSS: The curse of the duck

I can think of a batsman that bucks the trend - al...

2009-01-29T23:31:00.000+11:00

I can think of a batsman that bucks the trend - although obviously I understand that he's an outlier. England's Alastair Cook (66 innings, 2 not outs)'s mode score is 60, which he's been out on five times, compared with just one duck. I guess a few other players (like AB DeVilliers, who's never had a duck) will also fall into a category like this. But yeah, they're definitely exceptions. Really good article :).

So Mike Hussey's recent run of low scores is not p...

2009-01-21T14:04:00.000+11:00

So Mike Hussey's recent run of low scores is not poor form, it's re-affirming his greatness..

Hi Stanley - thanks for the comment, and the Digg!...

2008-12-31T16:00:00.000+11:00

Hi Stanley - thanks for the comment, and the Digg!

You are right, the batsman's scoring does follow an exponential distribution, but the statement re turning the tv is on is correct.

Imagine a coin tossed over and over again. The first time a head turns up follows an exponential distribution - like our batting dist - when the head turns up, you're out. There is a 50% chance it will turn up on the first toss. Given that 50% of the time a head has already turned up by the time we toss the coin a 2nd time, the probability that we actually toss the coin a 2nd time is 50% and the probability of that coin toss yielding a head is 50%, so overall, the probability of a head turning up on the 2nd toss is 50% times 50% - which is 25%.

The trick here is to remember that with a coin toss, at each toss, there is a 50% chance of a head turning up NO MATTER what has come before. Even if you have tossed the coin 100 times, the probability of a head on the 101st toss is 50%. It's the same the with batting, although instead of the probability involved being 50%, its around 2%.

Re moving from the exponential to the geometric dist, the exponential distribution is the continuous equivalent of geometric distribution, extending it to work for all numbers, not just integers. As you can only get out at integer values in cricket, the geometric distribution is the apt dist to use. For a decent data set like we have here, we can interchange the dists.

Thanks for reading!
marc

Very interesting analysis (I've Dugg your submissi...

2008-12-31T15:02:00.000+11:00

Very interesting analysis (I've Dugg your submission). But I'm not sure if I follow your move from the exponential distribution to the geometric...

If a batsman's scoring follows an exponential distribution, and his historical average per innings is 1/λ then:

P(he will score exactly x runs) = λexp(-λx)

If I understand correctly, your statement that "if you were to turn the television on now and find the cricket coverage, the chance that the batsman you are watching gets out on the current score is 2.2%" doesn't look right. The probability of getting out next ball isn't constant but depends on how many runs they've already made.