### 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.

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)

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.

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.

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.

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.

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.

## 28 Comments:

Dear Acey,

That's pretty neat. Now if you could model one for the coming 2006 World Cup and explain it in jargon-free language using your [C_#] skills... you're set for life...

CM rubs his hands in anticipation of getting his grubby hands on the magic crystal ball...!

Obviously, unlike the EPL you're going to have to input the following variables:

1. Develop a random fluke into your model to take into account of the black horse and wildcards.

2. The assumptions of the odds of win/losses needs to be revised. Historical data needs to be revised to take into account current team strengths.

3. Take into account the various team's performance in the German climate...

Oh...so many variables and so little time...

So, how about it, o' Great Seer...??? What is your prediction...

CM

I'm actually not too keen on international football. I find it rather bland when compared to club football.

But having said that, I'm (always) hoping for a first time winner in the World Cups. It's invigorating to see a nation's celebrations when they win it for the first time. So in that sense, I'm hoping the perennial underachieving countries of Spain and Holland will get it.

If I had started work by then, chances are I won't be watching much of it.

Well, you can always join the legions of blurry eyed workers in the morning :) if you do watch

As for world cup vs. leagues, I tend to prefer the World Cup. Once every four years vs. the yearly drill of league cup...

Oh well, to each his own.

BTW - I have a football question.

When a person scores three goals it's called a hattrick... Why not a tri-trick?

And what do you call it when it scores 4/5 goals?

4 = quattrick?

5 = pentrick?????

LOL - PS. Usually, I eagerly scan newspapers for the above terms but I get the lame-o, 4 goals/5 goals etc....

CM

I've heard of the explanation for the "hattrick" term before. On Singapore radio, UFM100.3. There's a proper story behind its origins. Unfortunately, I've forgotten the story.

woohooo..can use your results to bet :P

ginger_ale

Good luck. I am not responsible for any losses and am 100% responsible for any winnings. :P

hahaha.. do you always have to work out the numbers to prove everything, even the obvious? Very good, very good. You have the makings of a great actuary!

p.s.: Did you have a 'checker' for your results?

Just for fun and laughter mah. You see, you laughed, didn't you?

What's a checker?

Acey,

I think Jade means an auditor...i.e. someone who checks your assumption to see whether they make sense....

For instance, does your Binomial distribution for the remaining matches of Chelsea take into account the current dip in their form.

If not, the probabilty of Man-U winning should be upped say by another 5%?

35%/65% odds instead???

Hehe :))

CM

So des ne.

No $$$ to hire one. :P

Ac,

What does "so des ne" mean?

Once you market you "actuarial model" for EPL, why you'll be swimming in so much dough...

You don't need to worry about any $$$ issues then.

CM

So des ne = Oh I see in Japanese. We use it a lot in Singapore.

Yes Acey, I did laugh.. actually, I wanted to do a model for soccer betting some time ago too.. to calculate the expected winnings or losing of my betting strategy per weekend, but no time lah..

Yes CM is right. A 'checker' is someone who checks your assumptions and model. Maybe CM can be you checker and both of you come up with an excel programme to calculate expected winnings, where parameters are inputted by the user? :)

I wait eagerly for your work! :P

Write program in Excel? Where got time? My life so short. :P

Ac,

You must think of the potential returns...

Just imagine the money you would make off the programme... After a couple of fine tuning, you could start marketing it to the bookies and make loads of money of it...

And charge them huge fees to tweak it every year ....

CM

:))

Money does not possess much attraction to me.

Just make enough $$$ to retire from work then :)

I'm not so sure retirement holds that much of an attraction for me either.

okay,

some money for your niece's future education then...

:))

just joking about the model-lar...it may give ppl false security and encourage them to gamble more...which is never a good thing IMHO.

btw - Ac have your copyrighted your model?

CM

Nope.

Ah-ha,

I sense an opportunity brewing here... :)

Ac, jokes aside...err based on your experience, do you think the bookies actually have a model similar to yours?

After all, how would they determine what odds to give? Any idea whether any actuaries employed by these companies?

Just curious.

CM

As far as I know, they do not use models. It's more an educated guess based on previous experience. Probably padded with lots of safety margins.

Just my guess.

And I have not heard of them hiring actuaries yet.

Guesswork huh, that's pretty unscientific...

I find it strange that actuaries are not involved...After all, these things are right up the profession's alley [so to speak]

What a huge unexplored growth area :)...

That's assuming bookies don't do self help and try to *fix* the results in advance :((

It's just my guess.

Well if you ever do find out any actuaries involved in it, drop me a mail.

Would be very interesting to see what they actually do....

CM

Wilco

thats pretty cool..but I do think united will win, well i hope so!

be sure to pass by my space and leave a message, I will try and come by again!

Thanks, sporty muslimah, for dropping by.

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