Why Bookmakers Didn’t Find a Brexit More Likely – and why everyone thought they did

Bookies made a lot of money on the Brexit

As anyone knows by now, the British people voted for a Brexit, meaning that the United Kingdom will leave the European Union. The stock market didn’t particularly like this news, as indicated by the fact that the German DAX index opened 1.000 points, or nearly 10%, lower on the day the results became known.

It appears the majority of people expected the UK to stay in the EU. This expectation was likely fuelled by the fact that news agencies worldwide said that bookmakers (companies making their money by accepting bets) found it highly likely that Britain would stay in the EU. They pointed at the odds, or pay-offs, the bookies gave. The day before the referendum, the odds Betfair (a big bookmaker) gave were as follows: 1.16 for Bremain, 8 for Brexit. Meaning: if you would bet 1$ on a Brexit, and it happened, you would get 8$. For a Bremain you would get only 1.16$.

Such a big difference in pay-off would only occur if Betfair was very sure that Britain would stay in the EU, right? Otherwise they would lose a lot of money, right?

Wrong. This is nonsense. Betfair didn’t find it more likely that Britain would stay in the EU. It probably didn’t even care, because they would profit regardless of the outcome. Let me explain this via some easy calculations.

First I show the reasoning used by the public and news agencies, such as CNBC (which said that bookies found a Bremain at least twice as likely as a Brexit, and that this was because “voting for the unknown takes higher conviction than voting for the status quo”). Suppose odds of 1.16:8 as given by Betfair. Now suppose a person bets 1$ on Bremain. In case Bremain happens, Betfair has to pay him 1.16$. If not, they earn his 1$ bet. The same reasoning goes for a Brexit: the person earn 8$ in case of Brexit, Betfair earns 1$ in case of Bremain.

Now, what can Betfair expect to earn from this 1$ bet? Either way, it will get the 1$ bet. Now take p to be probability of Bremain. Then Betfair will have expected earnings of 1$ – 1.16$p in case of Bremain, and 1 – 8$(1-p) in case of Brexit. The bookie of course doesn’t want to lose money. Now the only value of p at which it doesn’t expect to lose money, regardless of a Bremain or Brexit occurring, is 0.87 (calculate this if you don’t believe me). For any other probability, the person betting the 1$ can expect to make money by betting on either Bremain or Brexit. In case of p = 0.6, for example, Betfair expects to lose 1 – 8$(0.6) = 3.8$ in case of a Brexit. Given that Betfair is not stupid, and doesn’t want to risk losing a lot of money, they must be damn sure of this 0.87 probability of a Bremain. Or so it would appear…

But this is not how it works.

You can also use the expected value calculation in another way, the way used by bookies. Contrary to my example, there is more than 1 person making a bet. You can you use this information. Let’s say that 100 people make a bet of 1$ each. Let’s say the bookie sets a pair of random odds of 4 to 9. Now assume 87 people bet for Bremain, and 13 for Brexit.

Suppose there is a Bremain. Then the bookie will earn 100$ (100 times the 1$ bets he receives) – 87 * 4$(pay-out per person) = -248$. In case of a Brexit they will earn 100$ – 13 * 9$ = -17$. Something is going wrong here: the bookie will lose money in case of either a Bremain or Brexit. Apparently the odds don’t match the proportion of people voting for Bremain and Brexit.  In case the bookie changes the odds to 4:6, he will make money only in case of a Brexit, but not in case of a Bremain. Now, the only odds that will make sure the bookie never loses or earns any money, given the 87 people betting Bremain and 13 Brexit, is 1.16 to 8. This is the break-even set of odds.

Proof: in case of Bremain, the bookie will earn 100$ – 1.16$*87 = + – 0$, and in case of Brexit 100$ – 13*8$ = +- 0$. So irrespective of what will happen, the bookie will never lose money.  Now, by only charging a fee for using its service, or a percentage of the amount bet, the bookie will always make money. The bookie can also change the odds slightly, so that – given the same 87 people voting for Bremain and 13 for Brexit – he can make money without charging fees: if he would change the odds to 1.12:8, for example, he is still sure not to lose money in case of a Brexit, but he will actually make money in case of a Bremain.

You might think: these odds work only in case 87 of 100 people vote for Bremain, and 13 for Brexit. In case the ratio changes, so will the bookie’s earnings. That’s true. Maybe more people will be lured into betting Brexit by the 8$ pay-off. You might get 20 of the next 100 people betting on Brexit, giving the bookie a negative pay-off of 100$ – 20*8$ = -60$ in case of Brexit. What to do then? Well, you just lower the pay-off: in this case a pay-off of 5$ for Brexit will do the trick. You can even leave the pay-off of Bremain at 1.16$, implying that you will make money in case of Bremain and not lose anything in case of Brexit. The point being: the odds are constantly adjusted, reflecting the ratio of bets at the time, so that the bookie is sure never to lose money. He is likely to use some margin of error in the odds so that, until a more accurate set of odds is reached, he will still not lose any money.

So we cannot say that the bookies thought a Bremain more likely than Brexit. Looking at the odds, we can only infer that much more money was bet on a Bremain than a Brexit, as shown by the implied probability of 0.87.

But what was the real probability that British people would vote for a Brexit? Until two days before the Brexit, polls were published, which showed the referendum to become a very close call. Some polls showed Leave to be in front with 52% to 48%, others (such as the Financial Times poll) showed Stay to be in front. Either way you look at it, judging by the polls, there was no reason to assume that a Bremain was much more likely than a Brexit. Given an expected value calculation, and assuming probabilities of 55% for Stay and 45% for Leave, betting for a Brexit would give you an expected: -1 + 0.45*8 = 3.60$ while betting on a Bremain would lose you an expected: -1 + 0.55*1.16 = 0.36$. Hence you should have always bet for a Brexit. Not only in retrospect (I didn’t do this; read the sidenote).

This is one of the clearest examples of a positive expected value calculation I have ever seen in practice. There is pretty much a 50-50 probability of something happening, and choice A provides you with a pay-off of 1.16, while choice B gives you 8. Then what do you choose?

Sidenote:
I must admit that I expected the British people to stay in the EU. I didn’t think it likely that the average British person would vote for an option that, to me so, obviously seems to decrease their wealth, and therefore their well-being. Leaving the EU forces the UK to start a lengthy process of negotiating new trade contracts with the EU. While the EU has many parties it can trade with, since it already has many contracts in place, the same will not hold for the UK once it is out of the EU. This implies that it has a bad bargaining position, since the need for the UK to trade with the EU is much larger than the other way around. This will likely lead to suboptimal trade contracts for the UK. Furthermore, given the slowdown in economic growth and downturn in the financial markets that were to be expected, the average Brit could have expected a decrease in his material well-being, either through a decrease in job certainty or a decrease of his pension money. Not to speak of the 15% more expensive holidays and imported computers. It was a valuable lesson to me: do not fare on possibly wrong assumptions while the data is so clear.

Why Trading Is Not a Game Of Chance

Roulette is not like trading

I recently had a discussion with a friend of mine. He said to me: ‘Be honest now. Trading is no different from playing roulette, right? You either win or you lose. It is just a matter of chance’.

What he really meant to say was that investing is a game of chance in a way similar to poker, rolling a dice or roulette. But although I agree with my friend that uncertainty is an intrinsic part of trading (or investing for that matter), it doesn’t follow that trading is a game of chance in the same way that poker, rolling a dice and roulette is. Not because trading requires certain skills that might allow you to beat the odds – for the same could be said of poker. It has to do with mathematics, and probability theory in particular.

I want to show this through the notion of expected value. The expected value of a random variable is its long-term average, or the value the variable takes on average per execution of the respective random process. Very briefly: in the case of rolling a dice, the expected value is 3.5 ((1 +2 + 3+ 4 + 5 + 6)/6), because in the long run you will get an average of 3.5 eyes per throw.

Certain requirements have to be met in order to to be able to calculate the expected value of a random variable. First of all, one should be able to fix the sample space of the process, or ‘the set of all possible results’ of the random process. In case of roulette, this set is unambiguous: number 1, 2,….., 38, because the ball can fall on one (and only one) of these numbers. Now, knowing this sample space plus the probability of the ball falling on any of these numbers plus the pay-offs of the ball falling on any of these numbers, you can determine whether you should take a bet or not. For example: let’s say you get 100$ per every 1$ you bet on the ball falling on number x. Then the expected value of betting 1$ on number x is (1/38 times 100$) + (38/38 times -1) = 1.63, meaning that in the long run you will make an average of 1.63$ per round per 1$ you bet when following this strategy. Since you will make money on average, you should pursue this strategy.

All well and good, but what happens when we take the market, instead of a roulette wheel, to be the random process we focus at? Let’s help ourselves a bit, and focus on a very restricted part of the market: the Brexit-debate. We can take the relevant possible results of the Brexit-debate as our sample space, the value of the DAX-index to be our pay-off, and the probability simply the probability of each possible result happening.

Now we come to face a couple of great difficulties.

Problem 1: sample space and the unknown unknowns
There appear to be only two possible results of the debate – Brexit, Bremain. Now we can define a random variable X such that any outcome of the random process is mapped to a real value. We choose the DAX-index to be our real value. You can for example say that in case of a Brexit, the value of the DAX-index will be 9000, and in case of a Bremain 10.400. Assuming that you can define a probability function on this variable, you can calculate the expected value of a trade.

But are these really the only two relevant results when it comes down to the Brexit-debate? No, it appears. There could be an explosion in a chemical factory in Germany that coincides with the Brexit or Bremain, but that significantly alters the course of the DAX-index. Maybe a politician will be murdered in case of a Bremain, and the DAX-index will collapse, even though the UK stays in the EU. There is an infinite list of possible events, not all of which can even be conceived: the black swan events, or the unknown unknowns. Since these results are by definition unknown, but nevertheless possible to happen and relevant for the DAX-index, your calculation will necessarily lack all relevant information, thereby giving an incomplete sample space. Such a thing can never be the case in roulette.

Problem 2: pay-offs
Furthermore, it is unclear what the DAX-index will do in case of either a Brexit or Bremain.This can be seen from the many different predictions made by various analysts. No-one knows exactly what the result of a Brexit or Bremain will be. Hence it is impossible to put a value on each of these results. So the second component of expected value, the pay-offs, is doubt-worthy too. This too can never be the case in roulette.

Problem 3: probability
But there is another, at least as stressing, issue. For expected value to be calculated, every event in the sample space must be assigned a probability.  While this is relatively non-controversial in the case of roulette (probability of either 1,2,…, 38 is 1/38), how do you come to know the probability of an event such as a Brexit, an interest rate hike, or any event that has never happened before?

It seems impossible to apply the so-called frequentist interpretation of probability, in which you conduct experiments and measure how often an event occurs relative to the total number of experiments. First of all because it is impossible to conduct experiments of this sort in the market. And second of all: even if you somehow manage to calculate how often an event occurred in the past and divide that by the total number of experiments done, you will necessarily get a 0% probability for events such as a Brexit, which have never occurred before. This seems absurd.

Using a subjectivist interpretation of probability will not help you much further. You can of course assign a probability to Brexit or Bremain by judging the available evidence, but the question is: how should you judge the available evidence given that you have no information about what happened in the past given the same set of available evidence, for this set of available evidence is surely to differ from any set in the past (Bayesian probability). One thing is for sure: certainly no Brexit has ever happened, under no set of available evidence, so again, the probability of a Brexit should be 0% (in light of any set of available evidence, in case you apply the rules of Bayesian statistics rightfully), which seems absurd.

This shows why probability theory and trading are no happy marriage, and why trading is not a game of chance like poker, rolling a dice or roulette.