Sunday, January 05, 2014

Cup Double

With Charlton having already played Oxford at home then Huddersfield away in the opening rounds (for us) of the Capital One Cup this season, what are the ex ante chances (see Note 1) of the exact same thing happening in the FA Cup? 


Well if my maths is right, it can be estimated as follows:


- probability of Oxford reaching the Third Round: 0.25 (see Note 2)


- probability of Charlton drawing Oxford at home in the Third Round: 0.008 (see Note 3)


- probability of Huddersfield reaching the Fourth Round: 0.4 (see Note 4)


- probability of Charlton reaching the Fourth Round (having been drawn at home to Oxford): 0.73 (see Note 5)


- probability of Charlton drawing Huddersfield away in the Fourth Round: 0.016 (see Note 6)


Therefore a reasonable approximation of the probability of playing Oxford at home then Huddersfield away in the first two rounds of the FA Cup after it has already occurred in the Capital One Cup is:


0.25 x 0.008 x 0.4 x 0.73 x 0.016 = 0.00000934 or 107,065/1


To put this in perspective this is less probable than tossing a coin 16 times and getting consecutive heads, or less probable than selecting seven random cards from a shuffled pack and each card being from the same suit.


Indeed whilst the probabilities will obviously be different for any given three club combination, I wonder if it has ever happened before?



1.       ‘Ex ante’ in this context means the estimated probability before the First Round of the FA Cup (featuring Oxford) had been drawn.

2.       Given Oxford United are riding high in League Two, before the First Round draw is made (see Note 1), I estimate the probability of them winning their First and Second Round ties to be 50% respectively, and thus 50% x 50% = 25% or 0.25.

3.       There are 63 other teams in the FA Cup Third Round draw, and an equal chance of being drawn home or away to any of them ie. 1/(63 x 2) = 0.008.

4.       As a midtable Championship side, this is a simple estimate of Huddersfield’s ex ante probability of reaching the Fourth Round in any given season – given they were given a relatively straightforward tie (Grimsby away), their actual probability increased significantly but this is not relevant for the calculation which considers the position ex ante (see Note 1).

5.       Current bookmakers odds of approx 4/6 imply a 60% probability (ie. 6/10) that Charlton win the tie at the first time of asking, whilst draw odds of 14/5 imply a 26% probability (ie. 5/19) that the tie needs a replay.  If a replay is required, I estimate the odds that Charlton still win the tie from that position (possibly after penalties of course) to be 50%.  Thus the estimated probability that Charlton reach the Fourth Round is 60% + (50% x 26%) = 73% or 0.73.

6.       There are 31 other teams in the FA Cup Fourth Round draw, and an equal chance of being drawn home or away to any of them ie. 1/(31 x 2) = 0.016.


At 11:16 PM, Anonymous Anonymous said...

probably lol

At 4:53 PM, Blogger Blackheath Addicted said...

NYA, great you're back, hope the interregnum was good. Unfortunately all the calculations overlook the overriding principle, namely that Charlton's chances of progression in any cup competition are close to bugger all.

At 6:09 PM, Blogger Peter Parker said...

I still cannot believe we drew Oxford and Huddersfield AGAIN in the same season, who would've thought it possible??

At 6:14 PM, Blogger New York Addick said...

...not only draw them but in the same (equivalent) rounds and also home then away.

It's even more amazing given (as noted above) that Oxford had to win two matches even to have the chance to play us.

At 6:46 PM, Blogger kappacino said...

NYA ... you need to get out more :-)

At 9:14 PM, Anonymous Anonymous said...

Or it could mean the draw is fixed - what were the odds of Arsenal being "randomly" drawn to play Spurs?

Great posts, by the way.


At 10:13 PM, Blogger New York Addick said...

Choose any two clubs and the odds are 63/1 obviously.

However broaden the concept out to say the 'big six' clubs (Arsenal, Spurs, Liverpool, Man U, Man C, Chelsea) and you have 15 possible pairings, and a commensurately higher chance that two are drawn together.

It is an example of the 'birthday paradox' which showed that if you have a room full of 23 people, it is more likely than not that at least two share the same birth date. It feels instictively wrong but it's a function of the sheer number of pairs.

At 10:24 PM, Blogger New York Addick said...

...indeed if my maths is correct, there's a 21% chance that ( at least) two of those six clubs draw each other in the Third Round:

1 - ((62/63)^15)

At 10:34 PM, Blogger New York Addick said...

....and if you are still with me, it is thus also more likely than not that over the course of three seasons that two of those six teams meet in the Third Round at least once. See below:

Probability two of the six teams meet: 21% (see above)
Probability two of the six teams don't meet: (1 - 21%) = 79%
Probability two of the six teams don't meet for three seasons in a row: 79% x 79% x 79% = 49.3%
Probability two of the six teams DO meet in the course of three seasons: (1 - 49.3%) = 50.7%


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