## 2018-06-30

### manipulating betting odds into probabilities and other odds

A colleague and I wanted to enter into a bet on a group-stage game of the World Cup of Football. We wanted our bet to have the following structure: He pays me something if Team A wins, I pay him something if Team B wins, and we push (that is, neither of us pays) if it is a draw between Team A and Team B. We are friends, so we wanted to give reasonable odds on this bet. How to calculate them?

Online, we saw the following odds posted on a reputable sportsbook (yes, such things exist) in European style:

 Team A wins 1.625 Team B wins 6.71 tie 3.91

Is it possible to convert these data into what we want to know? Of course the answer is yes, but we need to make assumptions. Before we continue, I should note that these odds would be written in American style as follows:

 Team A wins -160 Team B wins +571 tie +291

Because that may be confusing, let's just check in on what these odds mean. Looking at the European odds, what they mean is that if I bet \$100 that Team A wins, I will be paid \$162.50 if they do indeed win. That is, I will be paid back my \$100 plus \$62.50 in winnings. If I bet \$100 that there will be a draw, then I will be paid \$391.00 (my \$100 plus \$291 in winnings) if they do indeed draw. Looking at the American odds, what they mean is that if I bet \$160 that Team A will win, I will be paid \$100 in winnings plus my original bet back, or a total of \$260, if they do indeed win. And they mean that if I bet \$100 that Team B wins, I will be paid \$571 plus my original bet back, or a total of \$671, if they do indeed win. The +/- sign at the beginning of those odds makes a big difference! So in this sense, to a scientist, usually the European-style odds writing makes more sense.

Strictly, what's written in European-style odds is not the odds, but the odds-plus-one. If the book was not out to make money—If the book was just trying to break even—then these odds would be the book's approximation to the inverse of the probabilities of the outcomes. If you want to see why, you can think about placing a Dutch bet: Putting money on each outcome in proportion to the probability. If you do that, and the odds are just right, then you will get the same payout no matter what the outcome. That's an exercise to the reader! But if we interpret these European odds as inverse probabilities, then the implied probabilities of the various outcomes would be:

 Team A wins 0.615 Team B wins 0.149 draw 0.256

Which you can get just by inverting the numbers in the first chart. The observant reader will notice that these three numbers don't add up to one! They add up to a bit more than one. Why? Because the book is making money, and so they pay out odds slightly lower than what's fair, which means that they have an edge over the bettors, on average.

Now if we make the assumption that the book is relatively capable (or that the betting public is pushing the book's odds to something sensible), then we can assume that these are close to the correct probabilities for the three outcomes. That's a great start! But remember that we want odds for a win bet, with the draw as a push. The idea is, we want to know: What's the probability that Team A wins given that there isn't a draw. For that we condition on there not being a draw and re-normalize that probabilities. Probabilities for these outcomes are:

 Team A wins (conditioned on no draw) 0.805 Team B wins (conditioned on no draw) 0.195

I got these by re-normalizing the two outcomes my colleague and I care about to get unity. Now we can convert to European-style odds by inverting again, and the odds we get are:

 Team A wins (conditioned on no draw) 1.242 Team B wins (conditioned on no draw) 5.128

That is I should bet \$100, and my colleague \$24.20, and whoever wins gets it all. If there is a draw, we take our money back. Or at larger stakes, I should bet \$412.80, my colleague \$100, and so on.