# Arbitrage betting strategy to guarantee profit

Given the following bets you can make:

(1) A bet of $1$ wins $2$ if event A occurs

(2) a bet of $1$ wins $2$ if event B occurs

(3) a bet of $1$ wins $4$ if event C occurs.

In each case, you lose $1$ if the appropriate event does not win. Suppose that you know that event $A$ has a $90\%$ chance of occuring, event $B$ has a $9\%$ chance of occuring, and there is a $1\%$ chance of event $C$ occurs. Find a set of bets such that you are guaranteed to make money, regardless of the outcome.

My attempt: Based on my understanding, "guaranteed to make money" means profit is positive at the end. But we could easily see that if we bet $a$ dollars on event $A$ occurs, $b$ dollars on event $B$ occurs, and $c$ dollars on event C occurs, then our outcome is either $2a - b-c$ or $2b-a-c$ or $4c-a-b$. We need all these three $>0$ to guarantee we would make money. So if we solve for the intersection points of these 3 planes, we get: $a=b = 2c$.

My question: I got stuck on how to pick a good pair $(a,b,c)$ such that the overlapped regions of these $3$ planes give us the $3$ inequalities above. Plus, the information about the probability of each event occurring is not used at all, so I am skeptical whether my solution above is correct. Could someone please help me with this last step?

• Are the events $A,B,C$ disjoint? – carmichael561 Dec 2 '16 at 21:12
• Yes, they are disjoint. What I'm considering now is whether we could make a bet like: get $1$ if $A$ does NOT occur, and lose $1$ if $A$ does occur....Otherwise, there seems not to be an arbitrage here. – user177196 Dec 2 '16 at 21:13

You are correct that you want to find a bet so that your return in every possible situation is positve. If you put amounts (a, b, c) on each outcome then your possible profits, as you correctly deduced, are 2a - b - c, 2b - a - c and 4c - a - b.

One way of proceeding is to look for a solution where you make the same amount of money in each scenario, which means setting the three quantities above equal to each other. In this case you discover that you require a = b and 3a = 5c. You can satisfy these constraints by picking a = b = 5 and c = 3, in which case the possibilities are

• A wins - you make 10 - 5 - 3 = 2
• B wins - you make 10 - 5 - 3 = 2
• C wins - you make 12 - 5 - 5 = 2

In each case you have a profit of 2, so this is your arbitrage. Indeed, the real-world probabilities are completely irrelevant.

The key is the idea of implied probabilities. Since you lose 1 if A doesn't happen, but win 2 if it does happen, then the implied probability $p$ (defined to be the probability that would give you zero expected return on this bet) satisfies

$$(1-p)\times (-1) + p \times 2 = 0 \quad\Rightarrow\quad p = 1/3$$

Similarly the implied probability of B is 1/3 and the implied probability of C is 1/5. These sum to less than one (in bookmaking this is called an under-round) so there is the possibility of arbitrage. The arbitrage exists because the odds given are incoherent (they don't sum to one).

To take advantage, you make bets which are proportional to the inverse of the implied probability.

In the real world, a book keeper will always set odds that sum to more than one, meaning that any combination of bets which guarantees a payout, will always guarantee a negative payout (so there are no arbitrages).

• I think your last equation is wrong: it should be $2c=a$, not $3a=5c$. But your given pair actually works!! How amazing:) Thanks a ton. – user177196 Dec 2 '16 at 21:17
• You should check your algebra (in particular, start with 2a-b-c=2b-a-c=4c-a-b and add "a+b+c" to each equation to get 3a=3b=5c) – Chris Taylor Dec 2 '16 at 21:23