Many sportsbook promos take the form of a “risk-free bet”. In spite of what the name might suggest, these bets are not truly risk-free (at least not in isolation). However, through a process known as matched betting, they can be combined with appropriate hedges to lock in a risk-free payout. In this writeup, I explain what a risk-free bet is, and how one can derive the optimal matched betting strategy.
To adapt an old quip “Risk-free bets — aren’t.” Risk-free bets are nonetheless called risk-free because the sportsbook give you some compensation if you lose the bet. In particular, for a risk-free bet wager of $X$ dollars, if the wager wins it is treated as a normal win (i.e. you pocket the winnings prescribed by the given odds) but if it loses, you get what is known as a “free bet” with value $X$.
Free bets are like normal bets with two important differences: first, you have to wager the $X$ free bet (you cannot withdraw the cash). Second, if you win a free bet, you do not keep the principal (the $X$ dollars); only the winnings. As a result, wagering the free bet on an extremely likely outcome will generate low returns, since the winnings will be small and the principal is not returned to the bettor.
It is useful to ask what the value is of a “risk-free bet” promo. Loosely speaking we have the following structure
The individual starts by wagering $X$ on a risk-free bet. If he wins he ends at the top node, collecting the principal and winnings. If he loses the risk free bet, he earns a free bet of size $X$. This in turn is wagered again: if the free bet loses, the bettor has zero in his account (bottom node). If the free bet wins, the bettor collects the profit from the free bet, but not the principal.
We are interested in valuing the risk-free bet. To do so, let’s begin at the outside node, Free Bet, and solve the subtree. At the free bet node, we will wager the free bet $X$ on some event with probability $q$, and hedge in another book on the other side of the bet with amount $Y$. Our choice variables are thus the bet probaility, $q$, and the hedge amount $Y$. We let $v$ denote the vig — the fee the bookmaker collects by shading down the odds (equivalently, shading up the odds-implied probabilities). We assume the vig is symmetrically added to both sides of the event, i.e. each odds-implied probability is shifted up by $v/2$.
Because we are interested in minimizing risk, our problem at the free bet node is to maximize the minimum payoff:
$$ \begin{equation} \max_{q,Y} \min\left\{ \underbrace{X\left(\frac{1}{q+v/2}-1\right)-Y}{\text{Win Free Bet}},\underbrace{Y\left(\frac{1}{1-q+v/2}-1\right)}{\text{Win Hedge}}\right\} \end{equation} $$
The first element of the minimand is the payoff from winning the free bet leg: $X$ dollars wagered earn the vig-adjusted payout $1/(q+v/2)$ less the principal ($-1$) and less the full value of the hedge $Y$ (which must have lost for the free bet to have won). The second element of the minimand is the payoff from winning the hedge. $Y$ dollars wagered earn the payoffs associated with the residual vig-adjusted probability $1/(1-q+v/2)$ less the initial investment ($-1$). Note that these two events are exclusive since the hedge, by construction, is taken on the complement outcome. This is captured by the fact that the probability of $Y$ winning ($1-q$) and the probability of $X$ winning ($q$) sum to 1.
The two conditions for optimality in the maximin problem show in equation (1) are first, that the two inner terms are equal:
$$ \begin{align}X\left(\frac{1}{q+v/2}-1\right)-Y \nonumber\ &= Y\left(\frac{1}{1-q+v/2}-1\right) \\ Y&=X\left(\frac{\left(1-q+v/2\right)}{q+v/2}-\left(1-q+v/2\right)\right) \end{align} $$
and second, that we maximize either condition with respect to $q$. To do this, we substitute in the expression for $Y$, and set the derivative with respect to $q$ equal to 0: