Reply to comment

Outer space: Racing certainties

Submitted by plusadmin on January 1, 2005
in
January 2005

Horse races

A few months ago I saw a TV crime drama that involved a plan to defraud bookmakers by nobbling the favourite for a race. The drama centred around other events and the basis for the betting fraud was never explained. What might have been going on?

Suppose that you have a race where there are published odds on the competitors of $a_{1}$ to 1, $a_{2}$ to 1, $a_{3}$ to 1, and so on for any number, $N$, of runners. For example, if the odds are 5 to 4 then we express that as an $a_{i}$ of 5/4 to 1. Now if we lay bets on all of the $N$ runners in proportion to the odds so that we bet a fraction $1/(a_{i}+1)$ of the total stake money on the runner with odds of $a_{i}$ to 1 then we will always show a profit so long as the odds satisfy the inequality

  \[ Q=\sum _{i=1}^{N}\frac{1}{a_{i}+1}<1, \]    
and if $Q$ is less than $1$ then our winnings will be at least equal to
  \[ W=\left[ \frac{\ 1}{Q}-1\right] \times \mbox{our total stake}. \]    

More horse races
Let's look at some examples. Suppose there are four runners and the odds for each are 6 to 1, 7 to 2, 2 to 1 and 8 to 1. Then we have $a_{1}=6,a_{2}=7/2,a_{3}=2$ and $a_{4}=8$ and
  \[ Q=\frac{1}{7}+\frac{2}{9}+\frac{1}{3}+\frac{1}{9}=\frac{51}{63}<1 \]    
and so by betting our stake money with 1/7 on runner 1, 2/9 on runner 2, 1/3 on runner 3, and 1/9 on runner 4 we will win at least 51/63 of the money we staked (and of course we get our stake money back as well).

However, suppose that in the next race the odds on the four runners are 3 to 1, 7 to 1, 3 to 2 and 1 to 1 (ie evens). Now we see that we have

  \[ Q=\frac{1}{4}+\frac{1}{8}+\frac{2}{5}+\frac{1}{2}=\frac{51}{40}>1 \]    
and there is no way that we can guarantee a positive return. Generally, we can see that if there is a large field of runners ($N$ is big) there is is likely to be a better chance of $Q$ being greater than 1. But large $N$ doesn't guarantee $Q>1$. Pick each of the odds by the formula $a_{i}=i(i+2)-1$ and you can get $Q=3/4$ and a healthy 30% return even when $N$ is infinite!

But let's return to the TV programme. How is the situation changed if we know ahead of the race that the favourite in our $Q>1$ example will not be a contender because he has been doped?

Reply

  • Web page addresses and e-mail addresses turn into links automatically.
  • Allowed HTML tags: <a> <em> <strong> <cite> <code> <ul> <ol> <li> <dl> <dt> <dd>
  • Lines and paragraphs break automatically.

More information about formatting options

By submitting this form, you accept the Mollom privacy policy.