Are the bookies being fair on 'double' bets?
There is a very
popular fallacy among sports betters that, if you want to bet on TWO
results, then the ‘odds’ you get should simply reflect the
multiplication of the two separate probabilities.
Suppose, for example, that you want to bet on horse X winning race one AND horse Y winning race two.
The probability of both X and Y winning is therefore assumed to be 0.2 times 0.5 which is 0.1.
- the probability of horse X winning race one is 0.2
- the probability of horse Y winning race two is 0.5.
Using the ‘odds’ way of representing probabilities (and
ignoring the bookies’ mark-up of the ‘odds’) then the
separate bets would be 4 to 1 and evens 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 ten back.
This seems like a good deal. But in many cases you are being cheated.
The only time you can simply multiply the probabilities together to
calculate the correct probability of the ‘joint’ event is
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.
Think of the following extreme example.
Then the probability
of both Spurs winning AND Arsenal winning is 0.08, 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
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.
- The probability of Spurs winning a particular football match is 0.4
- The probability of Arsenal winning a particular football match is 0.2.
OK, so that was an extreme
case where it was easy to realise that multiplying the probabilities
together wasn’t correct. But 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.
For 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 the
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:
It 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 Spurs win GIVEN that Tevez scores is 0.2
- Probability Spurs win GIVEN that Tevez does not score is 0.55
The real 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.
above argument has introduced you discretely to the idea of conditional
probability. For a more extensive introduction to this subject try the Tutorial on Probability and Bayesian networks.
Return to Main Page Making Sense of Probability: Fallacies, Myths and Puzzles