New NFL Data

This week, I have revised my evaluation of frequent NFL wagers based on the 2022 season. If this subject doesn’t pique your interest, feel free to skip to the conclusion for this week’s new brainteaser.

I have recently gathered NFL data spanning from 2015 to 2022. Previously, my website and publication covered only up to the 2017 season. After thoroughly analyzing a total of 1,889 games, I'm excited to share my findings. The insights related to expected value are based on wagering 11 to earn 10.

Home Teams vs. Visiting Teams Coverage

Within the analyzed timeframe, home teams successfully beat the spread in 880 instances, while visiting teams managed to do so in 953 cases, and 56 games ended exactly on the line. Among completed bets, home teams had a success rate of 48.0% compared to 52.0% for visitors. Employing a flat betting strategy would have resulted in an overall loss of 8.3% when betting on home teams, and a 0.7% loss on visiting teams.

This disparity caught my attention, prompting further investigation. I discovered that, on average, home teams scored only 1.64 points more. The commonly mentioned guideline suggests that home field advantage should equate to 3 points. I suspect that bettors are overestimating this advantage, which may inadvertently create favorable conditions for the opposite side.

Underdogs vs. Favorites Coverage

Throughout the eight seasons analyzed with a distinct point spread, underdogs managed to cover 898 times while favorites did so 879 times, and 56 games landed precisely on the spread. Excluding the ties, underdogs succeeded 50.5% of the time, compared to 49.5% for favorites. This indicates potential losses of 3.5% for underdog bets and 5.6% for favorites.

I've always advocated for betting on underdogs, and that belief remains strong, but the 1% discrepancy is lower than I anticipated.

Over vs. Under against the total.

Over the span of eight seasons, the total number of games where the under hit was 963, while the over hit 909 times, and the line was spot on only 17 times. The under covered 51.4% of the resolved bets, leading to expected losses of 1.8% on unders and 7.3% on overs.

Money Line

Regarding money line wagers, I analyzed both sides where underdogs provided at least even money. Thus, money lines such as -115/-105 were excluded. Each bet was considered one unit, whether placed on underdogs or favorites. In total, the net loss was 0.9% on underdog bets and 5.6% on favorites.

September 19, 2024 Puzzle Question

A nefarious jailer gathers 100 inmates, assigning each a unique identification number from 1 to 100.

In a separate room, there are 100 boxes, each marked with a number. The warden takes slips of paper labeled from 1 to 100 and randomly inserts them into the boxes, ensuring one slip per box.

On the following day, each prisoner will enter the box room one at a time, with the opportunity to open up to 50 boxes. If a prisoner successfully locates his own number (for example, prisoner number 23 finds the slip in box number 23), he will be deemed \"successful,\" allowing him to exit early if he finds it before opening the fiftieth box. Exits are facilitated through a different door than the entry.

If all 100 inmates find their numbers, they will gain their freedom. Conversely, if even one fails, the entire group faces immediate execution.

Before the first inmate enters the box room, the prisoners are granted a day to strategize together. Once one person steps inside, no further communication is permitted. This includes, but is not limited to, rearranging the papers or leaving boxes ajar. Any sign of communication would result in the entire group being executed swiftly and mercilessly.

What approach will enhance their chances of freedom, and what is that probability?

September 19, 2024 Puzzle Answer

I admit I asked this puzzle in the Wizard column #369 Despite my previous explanation, I still feel it can be clearer. I aim to provide a more straightforward clarification here.

First, it's essential to understand that the 100 boxes will form various closed loops. So, what exactly is a closed loop? It's essentially a sequence of boxes that eventually leads back to the original box. For instance, if box 17 contains the number for box 79, which in turn leads to box 5, and box 5 redirects back to box 17, those three boxes create a closed loop.

Each inmate should start by opening the box associated with their assigned number. They'll take note of the number they find and subsequently open the box corresponding to that number. If there weren’t a cap of 50 openings, a prisoner would eventually uncover their number due to being part of the closed loop containing their number.

That said, success isn’t a certainty. There’s a significant possibility that a closed loop might encompass 51 or more boxes. Should that occur, none of the inmates within that loop would have sufficient opportunities to find their number.

Next, we need to analyze how many configurations can lead to a closed loop comprising 100 boxes. For the very first box, there are 99 different possible numbers that can lead to a loop of 1. Moving on to the second box, there are 98 numbers that can be allocated so they don’t match either of the first two boxes, allowing for a loop of 2. Following that logic, the third box has 97 suitable numbers, which would continue to form a loop of 3. Extending this reasoning, the formula for obtaining a closed loop of 100 is 99*98*97 * … * 1 = 99!. Since arrangements for 100 papers total 100!, the probability of closing a loop that encompasses all 100 boxes is calculated by dividing the successful configurations by the total arrangements, yielding 99!/100! = 1/100.

Now, let’s determine how many configurations lead to a closed loop of 99. There are 100 possible choices for one box that leads to itself, thus creating a loop of 1. Regarding how to arrange the remaining 99 boxes to form a closed loop of 99, based on the previous logic for 100 boxes, we find that permutations of a closed loop of 99 equal 98!. The combination of a 99-loop with a 1-loop would then total 100*98!. Dividing this by the total arrangements of 100! results in a probability of 1/99.

Shifting focus to find a closed loop of 98, we apply the formula for selecting two boxes from 100 that are not linked in the 98-loop, yielding permut(100,2)=100!/98! = 9900 options. Next, we again apply the earlier logic for the count of arrangements forming a closed loop of 98, yielding 97!. The configurations for a closed loop of 98 combined with arranging the other two amounts to 100*99*97!. The probability for a closed loop of 98 turns out to be 100*99*97!/100! = 1/98.

Extending this logic down to a closed loop of 51, the probability of a closed loop of 51 to 99 is 1/51 + 1/52 + 1/53 + … + 1/100 = 68.82%. The alternative is success, that there are no closed loops of 51 or more, meaning every prisoner finds his number. That probability is 31.18%.

A quick way to get a somewhat crude approximation is to use Euler’s constant (not to be confused with Euler’s number). Let c = Euler’s constant = 0.577216. The associated formula says:

1/1 + 1/2 + 1/3 + … + 1/n = ln(n) + c.

In this particular scenario, the likelihood of achieving a closed loop that spans from 51 to 100 sums as 1/51 + 1/52 + … + 1/100. This can be expressed as:

(1/1 + 1/2 + 1/3 + … + 1/100) – (1/1 + 1/2 + 1/3 + … + 1/50)

Following the approximation formula delineated previously, this comes to about…

(ln(100) + c) – (ln(50) + c) = ln(100) – ln(50) = 4.605170 – 3.912023 = 0.693417. Alternatively, the probability of succeeding is roughly 0.306853 or 30.69%. Keep in mind, the actual probability calculated was 31.18%, indicating the approximation is slightly off by 0.50%.

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September 26, 2024 Puzzle Question

You possess a flashlight along with eight batteries, two of which are required to power the flashlight. Out of the eight batteries, four are functional while four are defective. As it often is, the visual distinction between good and bad batteries is not possible. How can you activate the flashlight within a maximum of seven tries?