What are the odds?

(Pic from Geo Australia. I own a set of polyhedra dice, and had just bought another set whilst writing this post up. They were used in a game known as Dungeons and Dragons during my schooling years. I was more interested in the dice than the game.)

We've reached the business end of the football season. Only the bravest will think Man Utd still has a chance of catching runaway leaders Chelsea. Here's an attempt to quantify just how small the chances of Man Utd overhauling Chelsea and putting one over Mr. Boh Ho Seh are.

Background information

Man Utd has eight matches left, of which three are away. One of those three away matches is against Chelsea. Chelsea has seven matches left, of which four are away. The most vital statistic is of course Man Utd is 12 points in arrears. Here are the remaining matches for the two clubs:

For Man Utd,
West Ham, Bolten (a), Arsenal, Sunderland, Tottenham (a), Middlesbrough, Chelsea (a) and Charlton

For Chelsea,
Birmingham (a), West Ham, Bolten (a), Everton, Blackburn (a), Man Utd and Newcastle (a)

Assumptions

Ignoring the match against Chelsea, I assume the chances of winning each of the remaining seven matches for Man Utd to be uniform, i.e. the same. This is to simplify calculations. I also have to make the simplifying assumption of ignoring draws. These same assumptions apply to Chelsea as well.

And I'm ignoring the currently healthier goal difference of Chelsea.

Model

Now here's the model.

The match against Chelsea, I consider special enough to warrant separate consideration. I am therefore going to calculate the overall probability hinging on its outcome. The remaining seven matches for Man Utd, I model with a Binomial Distribution, with one set of parameters. The remaining six matches for Chelsea, I model with another Binomial Distribution, with another set of parameters.

Now, Man Utd is playing at away for this match. That gives Chelsea a distinct advantage. As such, I assign a probability of 0.3 that Man Utd wins it.

If that fortuitous event takes place, it means Man Utd will be 9 points in arrears, i.e. they have to win 3 more matches in their remaining matches than Chelsea. Keep in mind Man Utd do have a game in hand over Chelsea. If however, Man Utd loses it, it means they will be 15 points behind, i.e. they have to win 5 more matches in their remaining matches than Chelsea (we shall see later that this probability is virtually zero).

Now to estimate the parameters for the Binomial Distributions of both clubs. The matches that are away from home will get lower predictions for the team's winning probabilities, as is expected.

For Man Utd, here are my estimates of their chances of winning each of their remaining matches (given in square brackets after the team's name):

West Ham [0.8], Bolten (a) [0.7], Arsenal [0.5], Sunderland [0.95], Tottenham (a) [0.65], Middlesbrough [0.9] and Charlton [0.85]

Here are my corresponding estimates for Chelsea:

Birmingham (a) [0.9], West Ham [0.8], Bolten (a) [0.7], Everton [0.8], Blackburn (a) [0.7] and Newcastle (a) [0.75]

Now to make them uniform (see Assumptions above), I take the average of these for both teams. That gives 0.764 for Man Utd and 0.775 for Chelsea. In other words, for the seven remaining matches of Man Utd, we are modelling it with the Binomial(7, 0.764) distribution. For Chelsea, it will be the Binomial(6, 0.775) distribution.

Results

So what we're after boils down to this:

Probability(Man Utd wins the league) = Probability(Man Utd beats Chelsea) x Probability(Man Utd wins three more matches than Chelsea for their respective remaining matches) + Probability(Man Utd loses to Chelsea) x Probability(Man Utd wins five more matches than Chelsea for their respective remaining matches)

For Probability(Man Utd beats Chelsea), we already know this from above. It's 0.3. Probability(Man Utd loses to Chelsea) is therefore just 1 - 0.3 = 0.7.

For the other probabilities, they're a bit more tedious:

Probability(Man Utd wins three more matches than Chelsea for their respective remaining matches) = Probability(Man Utd wins 7) x Probability(Chelsea wins 4 or fewer) + Probability(Man Utd wins 6) x Probability(Chelsea wins 3 or fewer) + ...you get the idea... + Probability(Man Utd wins 3) x Probability(Chelsea loses all 6)

Probability(Man Utd wins five more matches than Chelsea for their respective remaining matches) follows the same logic.

Here's where the two Binomial Distributions come into play. I won't go through all the calculations because they are similar. So just one example should suffice. Let's consider say, Probability(Man Utd wins 4):

Probability(Man Utd wins 4 and loses 3) = 7C4 x 0.764^4 x (1-0.764)^3 = 0.157

Keep in mind also that the probabilities for Chelsea are actually sums of individual probabilities. Let's consider Probability(Chelsea wins 4 or fewer) as an example:

Probability(Chelsea wins 4 or fewer) = Probability(Chelsea wins 4) + Probability(Chelsea wins 3) + Probability(Chelsea wins 2) + Probability(Chelsea wins 1) + Probability(Chelsea wins none)

OK, *Phew*, now putting all that together (you'll have to trust my binomial calculations), we obtain

Probability(Man Utd wins the league) = 0.3 x 0.113 + 0.7 x 0.003 = 0.037

In other words, 3.7% chance or an odds of about 1 in 27.

Conclusion

So like everyone has suspected or assumed, we can basically kiss our chances goodbye. Till next season then.

Akan Datang: Diary, blog

18 days to go.