It’s one of those football last-digit-score pools… setup:

A 10x10 grid, numbered 0-9 on each axis. On one axis you have one team, on the other axis you have the other team.

squares are meted out randomly, there are 100 unique players.

your “square” wins if the ending digits of a score of a game are equal to that square. so, for example if you have square (2,7) then you win if the team score is, say: Team A 12, Team B, 27

this is done 4 times, once at the end of each quarter. thus, the entire pool is divided by 4 (payouts are the same per quarter). you can win multiple times

entrance is $100 per square purchased, so the entire pot is 10,000. Divided 4 ways, you have a prize potential of 2,500 per round.
how do you calculate the expected payout? is this even possible without knowing the probability of each score permutation? (i.e. a 0-0 game is less probable than a game that ends in 6, say). if it’s not possible to calculate e.v. because of the dynamic of the game, just assume that the square is also picked randomly, 4 times.

I would think the first step is to determine what the distribution looks like for the final digit in football scores. My gut instinct is to think that certain numbers are more likely than others, given the scoring increments common in football. For example, “7” might come up more frequently than “2”, just because 7 is the score after a successful touchdown and PAT, and then any equal number of additional touchdowns and field goals will preserve this final digit. Compare to “2”, where you would need a safety (starting at 0) or 4 field goals without a touchdown, which are more unlikely to happen. I realize there are other ways to reach these scores, but as a quick analysis, I think the point is clear that certain scores are more likely than others.

Lets say that we have a statistical analysis of all football games played and know the probability distribution of the final score digit. Assuming that each team’s score digit is independent of the other (not necessarily true, but helps the analysis), then you could punch in the probabilities and your expected value should be pretty straightforward for any given pair of numbers.

I suppose you’d have to know the likelihood of each number happening. Willing to bet such numbers are obtainable (heck, you could input the data and find out yourself). Some weird things do happen. For instance:

Regardless though I do not think you can calculate this because cell assignments are random (at least every one of these I have played). That is, people pick squares ahead of time with no numbers assigned on the X,Y axis. This is specifically to avoid people picking high occurrence numbers and the last squares never getting bought because they suck.

Right before the game some random method is used to place numbers on the axis and you get what you get.

I suppose you could calculate your odds against the historical record (if that even means anything) but at that point but you are stuck anyway.

let’s just say it’s a straight game where the winning numbers are pulled from two separate piles… and the square is assigned at random

would the ev be:

1% chance at 2500 = 25.00 ev

but you have 4 chances to win (let’s assume here that the numbers are totally random, and not related to prior drawn numbers (which isn’t anything like the actual game))… so would the ev be $100 (i.e. equal to your contribution?)

doesn’t sound right, but my expected value ability is a bit rusty

If we assume that every square has equal probability of winning, and every square is played, then the expected value is -0-, because all the bets are distributed as winnings. (This is better odds than a Las Vegas roulette wheel, which has a negative expected value.)

If we assume that different squares have a different probability of winning, then you have to calculate the expected value for a given square, not for playing the game in general. The best way to do that would be to research all football scores for games similar to the one you’re betting on and count 'em up.

ETA:

The expected value is your net, so it would be your figure of $100 minus the initial bet of $100, or zero.

Well as it is with all things, with more information, comes a better model.

If you’re completely in the dark in choosing the numbers corresponding to your square as Whack-a-mole claims, then your odds are quite easy to figure out - 1/100 for each square.

If you can choose the number pairing but don’t have the historical data, or as the OP said “probability of the score permutations” you can still make an educated guess. Point totals can be summing combinations of 2, 3, 6, 7, and 8 upwards to an arbitrary amount… say 13 (an insanely high score 1 quarter of play) and modulo the resultes by 10 to get your 1’s digits and bet according to that distribution.

Of course with the most information - score by quarter for SB games, or score by quarter for all games, whatever, you can go through and figure out the probability that way. Sounds like the most fair way is the blind assignment of numbers… which is also the easiest way to calculate expected value. I too have seen it played this way. It doesn’t allow for arbitrage.

When I’ve seen football pools like this, it was in a way similar to a lottery, with little skill involved. There are 100 squares, you can buy one or more squares, and after all the squares have been purchased, only then are the numbers arranged in a random sequence on the top, and another (different) random sequence on the left. So, ahead of time, no one knows if they will get square (7,2) or (2, 7) or (0,0).

This doesn’t answer the question of which square is more valuable than another square, of course.

ETA: never mind, Whack-a-Mole had said all this first, but since I typed it in I’ll let it stand.

When a game like this pays out to winners 100% of the money bet, the expected value is (pretty much by definition) equal to the amount bet - IOW, on average you get back 1 for each you put in.

ntucker notes what most people would expect: the chance of each square being a winner is not equal. If they’re randomly assigned, no problem. But it’s a common scam for a game like this to have the high-probability squares assigned to those who organized the game (and perhaps their friends) leaving the suckers to supply much of the money in return for a lower chance of winning.

showing similar patterns to yours: lots of scores ending in 0, 3, 4, 7; few ending in 2, 5, 8, 9. (x7-x0 and x3-x0 endings are the most common.)

Interestingly, the most common x7 scores are 17, then 27, then 7; and the most common x0 scores are 20, 10, 30, 0. So these are not simply explained as scoreless or one-score games.

This is correct. As others have pointed out, for “friendly” pots like this where this is no house rake, the EV comes out to 0. Using general nomenclature: EV = (probability(x) * prize) - cost

Also note that it’s rare for all four quarters to pay off identically. Typically, the final score pays the most (about half the pot give or take), the 1st and 3rd Quarter scores pay the least (~15% each), and the halftime score pays somewhere in between.

As long as the total amount invested = total amount distributed (no house cut of vig) and every player have equal chance to win (randomly chosen squares) I believe it will always be an even money game in the long run.