Good evening,
It’s been a very long time since I last posted something on this blog because I have been very busy at work but I thought I will try to be more active from now on.
Today, I’d like to share some basic maths about exchange sports betting, and actually we could generalize this to exchange betting in general.
Exchange betting has emerged over the last decade on the internet. Companies such as Betfair act as intermediary between individuals (or actors) willing to bet on a specific event.
From a general point of view, actors bet on an event $E$ to happen or not to happen.
 When you want to bet on the fact that $E$ does happen, we say that you back $E$
 When you want to bet on the fact that $E$ does not happen (denoted $\bar{E}$), we say that you lay $E$.
I chose to discuss exchange betting in this post because you have the opportunity to directly bet both on $E$ and $\bar{E}$. When you bet against a bookmaker, usually he will allow you only on an event to happen. For example, on a football game between Team A and Team B, the bookmaker will allow you to bet on Team A to win, Draw or Team B to win, but he will not let you bet directly on Team A not to win. This does not mean that you cannot recreate this bet synthetically, but I wanted to keep things as simple as possible on this post.
Let’s keep the example of the football game, and let’s say that the event of Team A to win is denoted by $A$. When you place a bet, you do it at given odds, which we will denote by $o$. For example, let’s say that $o=2$. Here I am using the European convention, which means that if you bet an initial amount $x$ on $A$,
 If $A$ happens, the payoff is $x \cdot (o1)=x \cdot (21)=x$.
 If $A$ does not happen, the payoff is $x$.
Now let’s say that the odds do not change and that you lay an initial amount of $x$ on $A$:
 If $A$ happens, the payoff is $x \cdot (o1)=x$.
 If $A$ does not happen, the payoff is $x$.
Let’s draw a table to see what happens now in both situations, using generic odds $o$:

$A$ happens 
$A$ does not happen 
Back $x$ at $o$ 
$x(o1)$ 
$x$ 
Lay $x$ at $o$ 
$x(o1)$ 
$x$ 
Net payoff 
$0$ 
$0$ 
As expected, if you back and lay $A$ for the same odds $o$, then your payoff will be $0$ in both cases.
Let’s now consider a case where the odds change in time. Originally, you back $A$ at $o_1$, and then the odds change to $o_2$. What I would like to show on this post is that you can again guarantee a similar payoff in both cases ($A$ happens or not) by laying some amount $y$ on $A$ at $o_2$.
Let’s draw the table again:

$A$ happens 
$A$ does not happen 
Back $x$ at $o_1$ 
$x(o_11)$ 
$x$ 
Lay $y$ at $o_2$ 
$y(o_21)$ 
$y$ 
Net payoff 
$x(o_11)y(o_21)$ 
$yx$ 
Now, we want the net payoff to be equal in both cases:
$$x(o_11)y(o_21) = yx$$
$$x(o_11) + x = y + y(o_21)$$
$$x(o_11+1) = y(o_21+1)$$
$$x o_1 = y o_2$$
$$y = x \frac{o_1}{ o_2}$$
By solving this equation, we know that if we bet $y = x \frac{o_1}{ o_2}$, we will have the same payoff in both cases. We can also compute the value of this payoff:
$$\text{Payoff} = yx = x \frac{o_1}{ o_2} – x = x \left(\frac{o_1}{ o_2} – 1 \right)$$
So we see here that if $\frac{o_1}{ o_2} – 1 > 0$, we will have a positive payoff in any case. In particular, if $\frac{o_1}{ o_2}>1$, we make money, if $\frac{o_1}{ o_2} < 1$, we lose money, and if $\frac{o_1}{ o_2}=1$, we make $0$ (which is the case $o_1=o_2=o$ we saw previously).
In conclusion, we make profit if and only if $o_2<o_1$. This makes a lot of sense; let’s say $o_1=3$ and $o_2=2$, the odds became lower because the market acknowledged that $A$ is more likely to happen because it’s now willing to pay less if $A$ happens than before. This means that you were right, and hence that you have made money.
Now, how does it work if you want to take the opposite position originally, namely you lay an initial amount $x$ on $A$ at $o_1$, and then the odds change to $o_2$. How can you close the position by backing $A$ with an amount $y$?

$A$ happens 
$A$ does not happen 
Lay $x$ at $o_1$ 
$x(o_11)$ 
$x$ 
Back $y$ at $o_2$ 
$y(o_21)$ 
$y 
Net payoff 
$y(o_21)x(o_11)$ 
$xy$ 
We then need:
$$y(o_21)x(o_11)=xy$$
$$y + y(o_21)=x + x(o_11)$$
$$y o_2 = x o_1 $$
$$y = x \frac{o_1}{o_2} $$
Then, the payoff is:
$$\text{Payoff} = xy = x – x \frac{o_1}{o_2} = x \left( 1 – \frac{o_1}{o_2} \right) $$
In this case, we see that we make a positive payoff if $\frac{o_1}{o_2} < 1$, that is, if $o_1 < o_2$. Again, this makes sense intuitively; if you lay $A$ at $o_1=2$ and then $o_2 = 3$, then the market believes that $A$ is now less likely to happen than before, which means that you were originally right to lay $A$ and hence deserve a payoff.
Let’s do a final sanity check on what we just said. First, we know that odds can take any value in $[1, \infty]$. When $o=1$, then $E$ will happen with probability $1$. When $o = \infty$, then $E$ will never happen (probability is $0$).
Assume you had a bet at $o_1$ and then $o_2 = 1$, which means that $E$ will happen. Then, if you backed $E$ you can get a payoff of $x \left( \frac{o_1}{1} – 1 \right) = x (o_1 1 )$. If you laid $E$, then the payoff is $x \left( 1 – \frac{o_1}{1} \right) = x (o_1 – 1)$. These match the payoff we defined orginally for back and lay bets on $E$.
Assume now $o_2 = \infty$, then if you backed $E$ the payoff is $x ( 01) = x$ and if you laid $E$ the payoff is $x (10)=x$. This also matches the original definition and we have shown here that the approach above is sound.
Please keep in mind that all the above assumes there is no transaction costs to take into account, which is not exactly correct and which I will address in a later post, but the idea is there.
That’s all for now, hope you enjoyed that and I’ll be back with more.