In 2009, Doug produced a Super Bowl Squares post, itself a revival of his old Sabernomics post eight years ago. In those posts, Doug derived the probability of winning a squares pool for each given square (or set of numbers). Unsurprisingly, he found that those lucky souls holding the ‘7/0’ squares were in good shape, while those left holding the ‘2/2’ ticket were screwed. You can download the Sports-Reference Super Bowl Squares app here, which is free, and should help you taunt your guests at your Super Bowl party.

Let’s say that this year, your Super Bowl squares pool allows you to either pick or trade squares: if that’s the case, this post is for you. I looked at every regular season and postseason game from the last ten years. The table below shows the likelihood of each score after each quarter, along with three final columns that show the expected value of a $100 prize pool under three different payout systems. The “10/” column shows the payout in a pool where 10% of the prize money is given out after each of the first three quarters and 70% after the end of the game; the next column is for pools that give out 12.5% of the pool after the first and third quarters, 25% at halftime, and 50% for the score at the end of the game. The final column is for pools that give out 25% of the pot after each quarter — since I think that is the most common pool structure, I’ve sorted the table by that column, but you can sort by any column you like.

Here’s how to read that. Take the top line for example. If you have one of the two ‘7/0’ squares (i.e., either Baltimore 7, SF 0, or SF 7, Baltimore 0), then your expected value is 12.2% of the first-quarter pot, 6.1% of the second-quarter pot, and so on. What’s interesting is that in the 10/10/10/70 system, this ticket is the second best ticket, but in the 25/25/25/25 structure, it’s nearly twice as valuable. That’s because ‘0/7’ is most likely to happen after the first quarter, so the value of your ticket will be tied not just to the numbers on the ticket but your prize structure.

The pool I’ve been at the last few years didn’t use the final digit of each team’s score, but the final digit of the sum of each team’s score. So a 17 would be an 8, a 22 would be a 4, and a 38 would be a 1. Here’s how that payout looks:

This changes things slightly, mostly by shrinking the differences between the numbers (a good thing, in my opinion; as Doug would say, the pool is less determined by the random assignment of squares and more determined by the random actions that happen as the game unfolds, and that’s how it ought to be.). Note that 0-0 was the top combination in the first table, but it is less than half as valuable in the “add the digits” scheme. In fact, since games almost never end in 0-0 (when combining the digits), that ticket is almost valueless if most of your pot goes to the score at the end of the game.

Finally, a bit of trivia if you want to make fun of the sucker that landed the 2-2 ticket. No matter which game you play, only two games (out of 2,669) in the last 10 years would have made the 2-2 ticket a winner. In 2004, the Bills beat Miami 42-32, a score matched by the Buccaneers over the Raiders this year. Meanwhile, in the ‘sum of the digits’ game, 2-2 would have only been a winner when the Chargers beat the Broncos 48-20 in 2006; two years later, the Eagles beat the Cardinals by the same score.