Sports Book Odds, How to Make Smart Bets, & Why You’ll Still (Probably) End Up Losing

Joe Lucero —  October 29, 2012 — 15 Comments

On October 13, 2012, one of the major sports books in Las Vegas said that there was a 108.8% chance of one of the four teams left in the baseball postseason would win the World Series. Of course it didn’t actually say there was a 108.8% chance of this happening, but the odds that they released to bettors did and helped ensure that over the long run, Vegas wins and we, as a whole, lose.

If you haven’t already, check out Part 1 for a review of AND vs OR probability. Now let’s imagine that instead of betting on outcomes,gmat odds like we did in the previous article, you’ve wised up and decided to open your own sports book, gMATH. You decide to start simple and offer bettors a chance to bet on which number, 1-4, randomly rolls out of a bingo cage. You realize that the probability of each number being selected is 25%, but you need a way to translate this for paying bettors. In a scenario where four different people each put down $1 on each of the four numbers, one person would win $3 ($4 total – $1 they bet). So you place the very first odds at gMATH’s number guessing game at 3 to 1.

In the long run, gMATH’s inaugural betting event may attract a clientele of people who enjoy watching ping pong balls with painted numbers roll around, but it won’t be bringing you the fortunes that you passed up on business school for. You realize that you need a new betting game that will attract more than just the bingo-loving crowd, involves a small amount of luck, and allow you to make a profit no matter which team wins. As there are exactly four teams left in the postseason, you decide that baseball would make a perfect switch.

The key difference in a game of pure chance (like guessing numbers or flipping a coin) and sport is that everyone THINKS they know something about sports. Allison thinks that the Tigers have the best pitchers left in the postseason. Steve thinks that the Cardinals have a lineup that matches up well against anyone. Tom thinks that the Yankees have great uniforms. So each of these bettors are willing to put money down on an event without making sure that the odds are fair. On the morning of October 13th, when exactly four teams were getting ready to begin the American & National League Championship Series, the odds on each team winning the World Series were:

Tigers: 2.5 to 1

Yankees: 2.5 to 1

Giants: 2.75 to 1

Cardinals: 3 to 1

If you tried to adjust the gMATH’s number guessing game’s odds to this, people wouldn’t put a dime down on your game because the odds don’t add up. Recall that in our guessing game, we gave 3 to 1 odds for each number because each one had a 1/4 = 25% probability of being called. We can then create our sports book’s stated likelihood of any event occurring based on the x to y betting odds by using the equation y/(x+y). Numbers on a roulette table payout 35 to 1 because 1 number will be selected out of the 1 + 35 = 36 numbers. But because there are actually 37 or 38 numbers (0 on all tables and 00 on some), the bettor is playing a game where they have a 1/37 or 1/38 probability of actually winning, but the payouts are only fair for an event that would happen 1 out of every 36 times. The same thing happens when sports books make future odds like the ones for the World Series. Using our equation y/(x+y) on the Tigers 2.5 to 1 odds to win the World Series, Vegas is paying out as though the Tigers have ≈ 28.6% probability of winning it all. As for the rest of the teams:

Tigers: 2.5 to 1 ≈ 28.57%

Yankees: 2.5 to 1 ≈ 28.57%

Giants: 2.75 to 1 ≈ 26.67%

Cardinals: 3 to 1 = 25%

Since either the Tigers OR the Yankees OR the Giants OR the Cardinals must win the World Series, we add up the four probabilities to show that the betting odds given for one of these four teams to win the World Series is ≈ 108.8%. Over the long run, if people were to bet as predicted (slightly more overall money placed on favorites), Vegas will be able to pay out all successful bets and still keep ≈ 7-11% of all bets as profit. This is why Vegas has gondola rides and your house doesn’t.

When the odds are listed as above, it becomes much more obvious what an advantage the house does, so it does a few things to make them less obvious to the average gambler. Instead of 2.75 to 1 odds, Vegas states this as 11 to 4 odds (which also explains why we all have to relearn how to use decimals when studying for the GMAT) or moneyline odds like +275. Just like on the GMAT, Vegas gives you a confusing term and when you can translate it into something you understand better (such as percentages), it’s easier to understand where there is and is not value for a certain bet. But no matter what form these odds are listed as, you can be sure that Vegas isn’t worried about who emerges victorious because the real winner will be Vegas.

The interesting thing about betting on sports is that the more uncertainty there is about the outcome, the more Vegas wins. At the beginning of the World Series (2 teams), Championship Series (4 teams), Division Series (8 teams), and before the season (30 teams), the probabilities that corresponded with the betting odds of all teams winning the World Series was approximately 103%, 109%, 113%, and 131%! Vegas (like everyone else) has no idea who will win the World Series before the season starts. But if the payouts are low enough, Vegas won’t have to worry.

This is both a blessing & a curse. Ask any seasoned gambler and you won’t find them betting on what they think WILL happen, but on when they think the odds are better than advertised. If you asked me to bet on whether I thought a dice would roll a six, I would bet not, but if you gave me 100 to 1 odds that the dice would roll a six, I would take the bet because even if I wasn’t likely to win, the payout was worth the risk. And the more times that I took that bet, the more I would eventually win. Before the season, the Baltimore Orioles were 150 to 1 (≈ 0.66%) odds to win the World Series. No one thought that the Orioles would be the best team in baseball and they were right. But if I thought that the Orioles had a 1% chance of winning the World Series, I would bet on the Orioles because if they won 1 time out of every 100 World Series at those odds, I would make $150 on 100, $1 bets. On the other hand, the Philadelphia Phillies were the most popular pre-season team in baseball and if I had to bet my life on who I thought would win in October, they would have been my pick. But at 11 to 2 odds (≈ 15.38%), I would doubt that the Phillies would win 1 out of every 6.5 World Series.

Which leads us to a classic GMAT-style of question. Post your guesses in the comments section and the correct answer (and solution) will be posted there next week:

Exactly two candidates, Donkey & Elephant, are competing for a job. If the Donkey has d to 1 odds to win and Elephant has e to 1 odds to win, assuming that the combined probability of the two winning is 100%, what is d in terms of e?

 

(A)  1-e

(B)  1/e

(C)  (e+1)/(e-1)

(D)  e/(e-1)

(E)  1/(e+1)

Joe Lucero

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Joe Lucero has both a Biology degree and a Master of Education from the University of Notre Dame. He also has a 780 on his GMAT. In the fall, you will find Joe in a much better mood during weeks after the Fighting Irish win their football game. During the rest of the year, you will find him looking for new places to travel, reading almost anything non-fiction, crossfitting, and trying to solve every challenge problem in the Manhattan GMAT Student Center.

15 responses to Sports Book Odds, How to Make Smart Bets, & Why You’ll Still (Probably) End Up Losing

  1. Correct Answer: B

    The easiest way to solve this problem is to plug in numbers d=1, e=1. Flipping a coin should be 1 to 1 odds, since you would win $1 for every $1 bet. Only B gives you the right answer.

    Algebraically, you can say that the combined probabilities for the two events happening should add up to 100% or 1, using the x to y odds = y/(x+y) formula explained above:

    (d to 1) + (e to 1) = 1
    1 / (d+1) + 1 / (e+1) = 1
    [(e+1) / (d+1)(e+1)] + [(d+1) / (d+1)(e+1)] = 1
    (e+d+2) / (d+1)(e+1) = 1
    e+d+2 = (d+1)(e+1)
    e+d+2 = de+e+d+1
    1 = ed
    d = 1/e

  2. Hey Joe,

    Entertaining (and enlightening) article. I initially used the algebraic approach for the example problem, but your explanation using plug-in numbers was more satisfying.

    I have a question, though. At first, I thought that one would have to test multiple numbers to be sure of the answer; however, after thinking about the problem and the answer choices logically, reason told me that if d can be written in terms of e then any combination of numbers (that add up to a 100% probability) would satisfy the equation. That is why it would make sense to choose a simple number like 1 to plug-in.

    Is my reasoning applicable?

  3. Correct. These types of Variable In (Answer) Choices, or VIC problems, require you to select the formula A-E that works for all numbers. You can either prove that you can algebraically prove the formula or you can find numbers that fit the scenario and eliminate the formulas that don’t work with the scenario you’ve selected.

    Most of the time, choosing “easy” numbers (i.e. d=1, e=1) will not allow you to eliminate all 4 incorrect answer choices, but in some cases it will. But if I can choose numbers that quickly eliminate even 2-3 answer choices, I have a lot less math I need to solve when I plug in another set of numbers (i.e. d=2, e=1/2).

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