(Update: you can see a table of Kennett curse calculations here.)
When Geelong beat Hawthorn for the eleventh straight time last weekend
When Hawthorn beat Geelong in the 2013 preliminary final, ending a run of eleven straight losses to the Cats, salivating journos had the opportunity to rehash everything they’d already written about Kennett’s curse, on Friday, Thursday and every other day during the week. But of all the things written about Kennett’s curse, the one I’m stunned not to have read is: E(X) = (1 – p^r) / (p^r.(1 – p))
This roughly translates as: if two teams play a whole bunch of football matches, how many would you expect them to play before one team suffers from X: Kennett’s curse?
The answer, with the curse at 11 games, assuming that both teams have an equal chance of winning each match, is 4094 – and Geelong and Hawthorn have only played 152.
The key assumption, though, is “equal chance”. If the Geelong edge is as little as 55-45 – good enough to win 11 games out of every 20 – the calculation for E(Kennett) drops to just 1592. (If Hawthorn has the edge, it blows out to 11864… yikes.)
Surprisingly, providing a general formula for the probability of a string of wins within a longer series of matches isn’t easy, but you can do it using a ridiculously complex formula. Using it for a string of 11 wins within 150 matches, at a fifty-fifty chance of winning for each team: Kennett’s curse will occur 3.3% of the time. (Update: you can see a table of Kennett curse calculations here.)