Alright, I’ll bite. For purposes of this study, I’ll be using the Super Bowl scores for the last forty years.

Since no team has scored higher than 55 points, and no losing team has ever scored fewer than 3 points, and no total score has ever exceeded 75 points, we can set bounds on the “possible” scores. I ignore the lowest score ever because we’re also looking at all intermediate scores.

I count 5 possible Colts scores (7, 17, 27, 37, and 47) and 6 possible Bears scores (3, 13, 23, 33, 43, and 53), with a few high-scoring games that are too big. I count 6 scores that add to over 75, so you have 24 (thirty minus six) possible ways to score.

Some of the other boxes, for comparison:

1-1 only has 19 chances to score because you can’t score only one point

2-2 has 30 chances, but 11 of them rely on at least one safety (very unlikely), so let’s call that 19 as well.

3-3 has 26 possible scores: 36 possibilities and only 10 excluded for being too high.

4-4 is disastrous, because the only way to score four points is a double safety – this is so unlikely that we disregard the possibility. That leaves 25 possible scores, 10 of which are too high, for a total of 15 possible ways to score.

5-5 is also pretty bad because their single-digit score relies on safeties – also 15 chances to score.

6-6 is much better: 25 possibilities with only 3 scores that are too high. 22 total.

If you go through the rest of the grid, you get a total number of possibilities around 2,000*. If you have more than 21 possibilities, you’re doing better than most other boxes. If you have fewer than 21, then you paid too much for your box. The boxes with only 15 ways to score are particularly bad, since you paid for a full chance but only got about 75% of a chance. You appear to have a square with a 20% better than average chance of winning.

5 scores seems to be the minimum in a Super Bowl, and 16 seems like the max. $100 is the average payout per score except for the halftime score and the big payout at the end. There are going to be at least nine paying events, and up to twenty. Odds of you randomly hitting one event are (24/2000), so odds of it *not* being you for each score are 98.8%. The odds of missing nine events in a row are 89.7%, and the odds of missing twenty events in a row are 78.5%. So your expected value is somewhere above $11 (because you might hit two events, or the big payout at the end). An average square is worth slightly more than $9, and a bad square could be worth less than $7.

The bad news is that there are at least 5 squares on the board that have 30 (!) chances of coming up, and if your host knows what he’s doing, he has taken those in advance. Even if he only takes those 5 spots, he takes up 5% of the board but 7.5% of the potential scoring space (while leaving the “sucker” spots with only 15 chances in 2000 for the rest of you). If he takes 0-0, by the rules that you’ve set out, he gets $100 just for showing up! His expected value for any of those big boxes is something better than $13, which means that if he really gets **ten** boxes, he can go nuts and grab something like $120 in expected value *plus* the guaranteed $100 if he grabs the 0-0. He’s a greedy bastard.

All of this is dependent on me getting the right total number of possible scores (which I’m pretty sure I didn’t, but it’s close to 2,000), and assumes that scores are drawn randomly from all possible scores (which they aren’t). There’s probably some combinatorics math that can do this much more easily, but I don’t know it. Is **ultrafilter** in the house?