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.
Suppose  that
 The 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 (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.

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).

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.

The 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

Norman Fenton