# Implied odds behind lay betting

In a precedent post,  I introduced the concept of back and lay bets on a given event, for example the fact that it will rain on a given day.  We also said that we are given some odds $o$, which express how much we can earn if we guess the right outcome between the event happening ($E$) or not ($\bar{E}$). Usually, people betting are very familiar with the “common” back bet and the odds $o$, which generate a net cash flow of $x (o-1)$ when we guess right and $-x$ when we guess wrong on an invested amount of $x$. However, I always found lay bets less intuitive and I prefer to see them as betting on the event not to happen.  This post about seeing a lay bet on $E$ as a back bet on $\bar{E}$, for example backing that it will not rain on a given day. The difference is explained in the table below:

$E$ $\bar{E}$
Laying $E$ $-x(o-1)$ $x$
Backing $\bar{E}$ $-y$ $y(\bar{o}-1)$

We see a new term $\bar{0}$ appearing, which are the odds of backing the event not to happen. Because both lines actually imply betting on the same thing (the event not to happen), the cash flows in case of $E$ or $\bar{E}$ must be the same. Therefore, we get the following system of equations:

\begin{align} -x(o-1) &= -y\\ x &= y (\bar{o}-1) \end{align}

The fist equation tell us that $y=x(o-1)$ and we can substitute in the second equation to solve for $\bar{o}$:
\begin{align} x &= y (\bar{o}-1)\\ x &= x(o-1) (\bar{o}-1)\\ \frac{x}{x(o-1)} &= \bar{o}-1\\ \bar{o} &= 1+\frac{1}{o-1}\\ \bar{o} &= \frac{o}{o-1}\\ \end{align}

Now, we can take bets on the event not to happen much more intuitively. Let’s go back to our example and let’s say that the odds of the bet on having rain today is $o = 5$. If we want to bet on the fact that it will not rain today, we need to place a lay bet, but what are our odds $\bar{o}$? Using the formula we derived above, we can easily compute them:

$$\bar{o} = \frac{o}{o-1} = \frac{5}{5-1} = 1.25$$

Let’s say that we are happy with those odds, and we’d like to place a bet of $y=2$ at $\bar{o}=1.25$. Well, if we look at your system of equation, we see that $x = y (\bar{o}-1)$, so we need to bet:

$$x = y (\bar{o}-1) = 2 (1.25-1)=0.5$$

Let’s check:

$E$ $\bar{E}$
Laying $E$ $-x(o-1) = -0.5 (5-1) = -2$ $x = 0.5$

This is correct! An easy way of computing $x$ is to keep in mind that $x = y (\bar{o}-1)$ is simply the amount you expect to win by betting on the event not to happen.