Why You Shouldn’t Buy Points for Super Bowl 2021.

Ryan Brill. January 2021.

Understanding the Question.

In 2021, we have been blessed with the Super Bowl matchup of our dreams, G.O.A.T. Brady vs. God Mahomes. The opening line on Bovada is Chiefs -3.5 (+100) and Bucs +3.5 (-120), as shown in the screenshot below.

I’ll be betting on the Chiefs because clearly they’re the better team, and I’ll take a 25 year-old Mahomes over a 43 year-old Brady. However, the -3.5 line gives me pause, because 3 points is the most common margin of victory in the NFL, and I don’t want to lose my bet via the Chiefs’ winning the Super Bowl by 3. So, Bovada provides some alternatives: I can “buy points”. In other words, Bovada has alternative lines, including the Chiefs -3 (-135) and the Chiefs -2.5 (-160), as indicated by the screenshot below.

So, is it worth it to pay the stiff juice of a \(-135\) or \(-160\) moneyline to avoid despair if the Chiefs win by a field goal?

Probability Model.

Let \(X_0\) be the random variable denoting the profit from betting \(\$B\) on the original line, Chiefs -3.5 (+100):

\[ X_0 = \begin{cases} +B & \text{if Chiefs win by } \geq 4 \\ -B & \text{else}. \\ \end{cases} \]

Let \(X_{1/2}\) denote the profit from betting \(\$B\) on the “buy \(1/2\) point” line, Chiefs -3 (-135):

\[ X_{1/2} = \begin{cases} +\frac{100}{135}B & \text{if Chiefs win by } \geq 4 \\ 0 & \text{if Chiefs win by exactly } 3 \\ -B & \text{else}. \\ \end{cases} \]

Let \(X_1\) denote the profit from betting \(\$B\) on the “buy \(1\) point” line, Chiefs -2.5 (-160):

\[ X_1 = \begin{cases} +\frac{100}{160}B & \text{if Chiefs win by } \geq 3 \\ -B & \text{else}. \\ \end{cases} \]

To make the model more explicit, we introduce 2 parameters. Let \(p\) denote the probability that the Chiefs win by \(\geq 4\) points, and let \(q\) denote the probability that the Chiefs win by exactly \(3\) points. Then

\[ X_0 = \begin{cases} +B & \text{with probability } p \\ -B & \text{with probability } 1-p \end{cases} \]

\[ X_{1/2} = \begin{cases} +\frac{100}{135}B & \text{with probability } p \\ 0 & \text{with probability } q \\ -B & \text{with probability } 1-p-q \\ \end{cases} \]

\[ X_1 = \begin{cases} +\frac{100}{160}B & \text{with probability } p+q \\ -B & \text{with probability } 1-p-q. \\ \end{cases} \]

The most profitable strategy has the highest expected profit. So, we will compare \(\mathbb{E}X_0,\mathbb{E}X_{1/2},\text{ and }\mathbb{E}X_1\), and choose the strategy with the highest such value. Well,

\[ \begin{aligned} \mathbb{E}X_0 &= B(p) - B(1-p) \\ &= B(2p-1)\\ \\ \mathbb{E}X_{1/2} &= B(\frac{100}{135}p) - B(1-p-q) \\ &= B(\frac{47}{27}p + q - 1) \\ \\ \mathbb{E}X_1 &= B(\frac{100}{160}(p+q)) - B(1-p-q) \\ &= B(\frac{13}{8}p+\frac{13}{8}q-1). \end{aligned} \]

Now, we need only calculate \(p\) and \(q\).

Estimating \(p\) and \(q\).

Given that the Chiefs are a \(3.5\) point favorite, \(p\) is the probability that the Chiefs win by \(\geq 4\) points, and \(q\) is the probability that the Chiefs win by exactly \(3\) points. Our task is to estimate \(p\) and \(q\).

I found a dataset containing the opening and closing pointspreads from every NFL game during the last 10 years, ending with NFL Week 14, 2020. Here is a peek at the head of the dataset, showing the some of the Week 14 results.

        date away.team home.team away.score home.score home.line.open home.line.close
1 2020-12-14    Ravens    Browns         47         42            1.0             3.0
2 2020-12-13  Steelers     Bills         15         26            2.0            -2.0
3 2020-12-13   Falcons  Chargers         17         20            2.5            -1.0
4 2020-12-13    Saints    Eagles         21         24            7.0             7.5

Now, we narrow the dataset to include only the games that had an opening or closing pointspread of \(3.5\). Then, we modify the dataset to include 2 new columns: fave.covers, which is TRUE iff the favorite covered the \(-3.5\) spread, and exactly.3, which is TRUE iff the final score ended in the favorite winning by exactly 3 points. Here is a peek at the head of the updated dataset, omiting the column home.line.close to save space for aesthetics.

          date away.team  home.team away.score home.score home.line.open fave.covers exactly.3
1   2020-12-13   Broncos   Panthers         32         27           -3.5       FALSE     FALSE
286 2020-12-06     Colts     Texans         26         20            2.5        TRUE     FALSE
2   2020-12-06    Browns     Titans         41         35           -4.0       FALSE     FALSE
3   2020-12-02    Ravens   Steelers         14         19           -3.5        TRUE     FALSE
287 2020-11-29    Chiefs Buccaneers         27         24            3.5       FALSE      TRUE
4   2020-11-23      Rams Buccaneers         27         24           -3.5       FALSE     FALSE

Interestingly, as shown above, the Chiefs and Bucs met earlier in the season, on November 29, 2020. The Chiefs were also \(3.5\) point favorites then, and they won by exactly 3 points! Spooky!

Now, we estimate \(p\), the probability that a \(-3.5\) point favorite covers the spread, and \(q\), the probability that a \(-3.5\) point favorite wins by exactly 3 points, by the empirical proportions from our dataset. Hence

\[p = 0.48\]

and

\[q = 0.08.\]

Finally, note that our dataset of games having a pointspread of \(3.5\) contains \(463\) games, which is a nice sample size, enough to give us some confidence in our estimates for \(p\) and \(q\).

Conclusion.

As calculated previously,

\[ \begin{cases} \mathbb{E}X_0 = B(2p-1)\\ \mathbb{E}X_{1/2} = B(\frac{47}{27}p + q - 1) \\ \mathbb{E}X_1 = B(\frac{13}{8}p+\frac{13}{8}q-1). \end{cases} \]

Using \(p = 0.48\) and \(q = 0.08\), we get

\[ \begin{cases} \mathbb{E}X_0 = -0.04B \\ \mathbb{E}X_{1/2} = -0.08B \\ \mathbb{E}X_1 = -0.09B. \end{cases} \]

To make this calculation more concrete, we choose a value for \(B\). Since Ryan Brill and his buddy Nick Miller made \(\$ 50\) worth of NFL bets each week on the 2020 edition of the Bet, Sweat, and Forget Podcast, we shall use \(B =50\), yielding

\[ \begin{cases} \mathbb{E}X_0 = -2 \\ \mathbb{E}X_{1/2} = -4.22 \\ \mathbb{E}X_1 = -4.5. \end{cases} \]

So, betting the original line is the most profitable strategy, because its expected profit is the least negative. In fact, betting \(\$50\) and not buying points saves you at least \(\$ 2.22\) on average.

Hence it is more profitable on average not to buy points, no matter how tempting it may seem!

More General Moneylines.

More generally, we suppose that the original pointspread has a moneyline of \(-M_0\) for the favorite, the “buy a half point” spread has a moneyline of \(-M_{1/2}\), and the “buy a point” spread has a moneyline of \(-M_1\). Suppose we bet \(\$ B\). Let \(p\) be the probability that the favorite covers in the original bet, and let \(q\) be the probability that the final game score is exactly the pointspread in the “buy a half point” bet. Then the profits \(X_0, X_{1/2}, \text{ and } X_1\) of betting on the respective pointspreads are

\[ X_0 = \begin{cases} +B\frac{100}{M_0} & \text{with probability } p \\ -B & \text{with probability } 1-p \end{cases} \]

\[ X_{1/2} = \begin{cases} +\frac{100}{M_{1/2}}B & \text{with probability } p \\ 0 & \text{with probability } q \\ -B & \text{with probability } 1-p-q \\ \end{cases} \]

\[ X_1 = \begin{cases} +\frac{100}{M_1}B & \text{with probability } p+q \\ -B & \text{with probability } 1-p-q. \\ \end{cases} \]

The expected profits are thus

\[ \begin{cases} \mathbb{E}X_0 = B\frac{100}{M_0} p - B (1-p) \\ \mathbb{E}X_{1/2} = B\frac{100}{M_{1/2}} p -B (1-p-q)\\ \mathbb{E}X_1 = B\frac{100}{M_1} (p+q) -B (1-p-q). \\ \end{cases} \]

The most profitable strategy has the highest expected profit.