## 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:

$\frac{\binom {10}3}{\binom {553}3} = \frac{120}{28032676} = 4.28 \times 10^{-6}$

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:

$\frac{10}{553} \times \frac{9}{552} \times \frac{8}{551} = 4.28 \times 10^{-6}$

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...)

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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