New NFL Data
This week, I update my analysis of some common NFL bets through the 2022 season. If this topic doesn’t interest you, remember to jump to the end for this week’s new puzzle.
I recently obtained NFL for the 2015 through 2022 seasons. My web site and book previously only went through the 2017 season. After analyzing 1,889 games, I present the following analysis. Comments on expected value are based on laying 11 to win 10.
Home vs. Visiting Team Against the Spread
In the period studied, the home team beat the spread in 880 games, the visiting team in 953, and game fell exactly on the spread in 56. Of bets resolved, the home team won 48.0% and the visiting team 52.0%. Flat betting would have resulted in a loss of 8.3% on home teams and 0.7% on visiting teams.
This difference was rather surprising. So, I did a little digging. I find on average the home team scored 1.64 more points only. The “rule of thumb” I’ve heard quoted many times is the home field advantage is worth 3 points. I speculate the betting public is overvaluing the home field advantage, creating value the other way.
Underdog vs. Favorite Against the Spread
In the eight seasons studied with a non-zero point spread, the underdog beat the spread 898 times, the favorite covered 879 times and the game fell exactly on the spread 56 times. Not counting the pushes, the underdog won50.5% of the time and the favorite covered 49.5%. That would equate to a loss of 3.5% on underdogs and 5.6% on favorites.
I have always been a champion of betting underdogs. This still holds true, but the 1% margin is less than I was expecting.
Over vs. Under against the total.
In the eight seasons, 963 games the under won 963 times, the over 909 and the line was exactly right 17 times. The under won 51.4% of bets resolved. The expected loss was 1.8% on the under and 7.3% on the over.
Money Line
For money line bets I looked at both sides when the underdog paid at least even money. So, money lines like -115/-105 were not counted. All bets were one unit, whether betting on the underdog or favorite. That said, the overall loss was 0.9% on underdogs and 5.6% on favorites.
September 19, 2024 Puzzle Question
An evil warden gets together 100 prisoners and gives each a unique number from 1 to 100.
Inside another room are 100 numbered boxes. The warden takes pieces of paper, numbered 1 to 100, and randomly places them in the boxes, one piece per box.
The next day, the prisoners will be allowed into the boxes room one at a time. Each prisoner may open 50 boxes. If a prisoner finds his own number (for example prisoner 23 finds the box containing the numbers 23) then he will be "successful" and may leave early if he finds it before the 50th opening. Exits are made through a separate door than the entrance.
If all 100 prisoners are successful, they will all be set free. However, if one or more are unsuccessful, then they are all immediately put to death.
The prisoners are allowed a day together to strategize. Once the first prisoner enters the boxes room, no further communication is allowed. Examples of communication include, but are not limited to, moving the papers around and leaving the lids open. If any communication is detected, all prisoners will be put to death immediately and painfully.
What strategy will maximize their probability of being set free and what is that probability?
September 19, 2024 Puzzle Answer
I admit I asked this puzzle in the Wizard column #369. However, I’m not happy with my answer. I am going to attempt to give a simpler explanation here.
First, recognize that the 100 boxes will consist of some number of closed loops. What is a closed loop? It is series of boxes that lead back to the original box. For example, if box 17 leads to box 79, box 79 leads to box 5, and box 5 leads to box 17, then those three boxes form a closed loop.
The strategy of each prisoner will be to open the box corresponding to his own number. He will read the paper inside and then open the box on that paper. If it were not for the limit of 50 openings, the prisoner would eventually open the box that contains his number. This is because by choosing the box with his number, he is at least on the closed loop that contains his number.
However, this is no guarantee of success. There is a good chance there will be a closed loop of size 51 or more. If that is the case, none of the prisoners in that closed loop will have enough openings to find their number.
Next, let’s find the number of ways there is a closed loop of size 100. For this first box, there are 99 possible numbers inside that don’t match the box number, which would lead to a closed loop of 1. For the second box, there are 98 numbers that don’t match the number of the first or second box, which would lead to a closed loop of 2. For the third box, there are 97 numbers that don’t match the number of the first three boxes, which would lead to a closed loop of 3. Extending this logic, there are 99*98*97 * … * 1 = 99! Ways to have a closed loop of 100. There are 100! ways to order the 100 papers. The probability of a closed loop of 100 is the number of successful combinations divided by the number of all combinations. This is 99!/100! = 1/100.
Next, let’s find the number of ways to have a closed loop of 99. There are 100 possibilities for the other box that leads to itself, forming a closed loop of 1. Then, how many orders are there for the other 99 boxes to form a closed loop of 99? By the logic above for 100 boxes, the number of permutations for a closed loop of 99 is 98!. The number of combinations for a closed loop 99 and a closed loop of 1 is 100*98!. Dividing that by 100!, the total number of combinations, we get 1/99.
Next, let’s find the number of ways to have a closed loop of 98. There are permut(100,2)=100!/98! = 9900 possibilities to pick two boxes out of 100 that are not part of the closed loop of 98, with regard to order. Then, how many orders are there for the other 98 boxes to form a closed loop of 98? By the logic above for 99 and 100 boxes, the number of permutations for a closed loop of 98 is 97!. The number of permutations for a closed loop 98 and all the ways to order the other 2 is 100*99*97!. The probability of a closed loop of 98 is thus 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 the case of this problem, the probability of a closed loop of 51 to 100 was 1/51 + 1/52 + … + 1/100. This could be expressed as:
(1/1 + 1/2 + 1/3 + … + 1/100) – (1/1 + 1/2 + 1/3 + … + 1/50)
Using the approximation formula above, this is approximately…
(ln(100) + c) – (ln(50) + c) = ln(100) – ln(50) = 4.605170 – 3.912023 = 0.693417. The alternative is the probability of success, of 0.306853 = 30.69%. Recall the actual probability was 31.18%. So, the approximation is off by 0.50%.
September 26, 2024 Puzzle Question
You have a flashlight and eight batteries. The battery requires two good batteries to work. Four of the eight batteries work and four do not. As usual, you can't tell the good from the bad ones by appearance. How can you turn on the flashlight with a maximum of seven attempts?