Monday, 6 September 2010

Dodgy cricket odds

The cricket world has recently been rocked by allegations of match-fixing against the Pakistan team.

The News of the World set up a sting to catch sporting-agent Mazhar Majeed correctly predicting when three no-balls would be bowled during the recent Lords Test Match between England and Pakistan. Whilst this in itself is not match-fixing (it's called spot-fixing - fixing certain events in a days play to win exotic bets), it's a smoking gun pointing towards further corruption.

So what are the odds of correctly picking three no-balls in a day's Test play? Could Majeed have just been lucky?

Let's assume there are 90 overs in a day's play, and on average 10 no-balls and 3 wides. This means that 553 balls will be bowled in the day. There are two ways to work out the probability of correctly choosing 3 balls as no-balls.

1) You can choose 3 random balls from 553 in 28032676 different ways. You can choose 3 no-balls out of 10 possible no-balls in 120 different ways. So mathematically, this looks like:

2) The other way to do this to imagine a big bag of marbles, where the no-balls are black and all other balls white. The first time you pull out a marble, you have 10 chances in 553 of pulling out a black marble. On the second draw, you have 9 chances in 552 and on the third you have 8 chances in 551. This looks like:

This means there are 4 chances in a million of the sports-agent fluking his result. Essentially, he's dodgy! (It is left as an exercise for the reader to prove algebraically that the above two methods are exactly the same...)

Further reading:
Forrest, D. (2003). Sport and Gambling Oxford Review of Economic Policy DOI: 10.1093/oxrep/19.4.598
Frey, James H. (1992). Gambling on sport: Policy issues Journal of Gambling Studies DOI: 10.1007/BF01024122

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