is a very
popular fallacy among sports betters that, if you want to bet on TWO
results, then the ‘odds’ you get should simply
multiplication of the two separate probabilities.
for example, that you want to bet on horse X winning race one AND horse
Y winning race two.
the probability of horse X
winning race one is 0.2
the probability of horse Y
winning race two is 0.5.
probability of both X and Y winning is therefore assumed to be 0.2
times 0.5 which is 0.1.
Using the ‘odds’ way of representing probabilities
ignoring the bookies’ mark-up of the
‘odds’) then the
separate bets would be 4 to 1
respectively and the ‘double’ bet would be 9 to 1.
Whichever way you look at it, if you bet one pound then you will get
like a sensible way to calculate the odds of the 'double' event.
But, the only time you can simply multiply the probabilities together to
calculate the correct probability of the ‘joint’
if both events are independent.
This means the outcome of the second event in no way depends on the
outcome of the first. For the horse races (assuming all the horses in
the two races are different) this might be a reasonable assumption. But
for many sporting doubles it will not be.
the following extreme example.
The probability of Spurs drawing
a particular football match is 0.2 (odds of 4 to 1)
The probability of Arsenal drawing a particular football match is 0.2. (odds of 4 to 1)
the probability of both Spurs drawing AND Arsenal drawing is 0.04, is
it not? Well it might be if you knew that we were talking about TWO
different matches, but it certainly is NOT if the match happens to be
the same, i.e. Spurs versus Arsenal. In this case the probability of
both Spurs drawing AND Arsenal drawing is 0.2. So, if a bookie was
generous enough to offer you anything better than 4 to 1 you would be daft NOT to have a flutter.
In fact, in most cases it turns out bookies calculate double odds in a
way that benefits them by capitilising on fans wrongly assuming events
are independent. Consider the following example
The probability of Spurs winning
a particular football match is 0.4
The probability of Arsenal
winning a particular football match is 0.2.
of both Spurs winning AND Arsenal winning is 0.08, is it not? Well, as above, it
might be if you knew that we were talking about TWO different matches, but again it
certainly is NOT if the match happens to be the same,
i.e. Spurs versus Arsenal. In this case the probability of both Spurs
winning AND Arsenal winning is zero. So even if the bookies offered you
1,000,000 to 1 you would be pretty daft to have a flutter.
It turns out that every day punters are seduced by
apparently attractive odds on double bets which are actually nothing
less than a rip-off. The classic example of such bets in football is
the “Joe Bloggs scores AND Team X wins” variety.
example, in March 2007
Spurs were playing West Ham. Based on odds offered by bookies the
probability of Spurs winning was just over 0.5. The probability
of the West Ham player Tevez scoring was 0.1 (he
had never previously scored a goal in English football).
A good friend of mine
confidently predicted that Tevez would score but that Spurs would win.
Multiplying the two probabilities together, the bookies would have
given my friend odds representing a probability of 0.05 for the double
bet. That looks good -- 20 pounds won for each pound bet. But he would
have been massively cheated by this. The real probability of
double (a prediction which did in fact come true) was about 0.02, and
so the odds he should have got were about 50 to 1.
It is all
because the two
events are definitely NOT independent. If we know that Tevez
scores, then the probability that Spurs win is significantly reduced.
In fact the following assumptions are reasonable:
Probability Spurs win GIVEN that
Tevez scores is 0.2
Probability Spurs win GIVEN that
Tevez does not score is 0.55
turns out that
with these probabilities the overall probability of Spurs winning is
0.515 which matches the prior odds (to see this you can run this model in AgenaRisk).
probability that we
need to calculate is the probability that Tevez scores AND that Spurs
win given that Tevez scores. This probability is equal to 0.1 times
0.2, i.e. 0.02. The
above argument has introduced you discretely to the idea of conditional
note: Much of the material here has now been incorporated (and
improved) into our new book "Risk Assessment and Decision Analysis
with Bayesian Networks" by Norman Fenton and Martin Neil. The book
includes many more worked examples like these. The website for the book
is here while the book's associated blog is here. Also, we now have a regularly updated blog addressing similar themes: Probability and Risk. footer Norman