23 January 2022
Football score puzzle
In American football, any drive can end with the following points – 0 (ball turned over), 2 (safety), 3 (field goal), 6 (touchdown with failed point after or failed two pointer), 7 (touchdown and successful point after) or 8 (touchdown and successful 2 point conversion).
Assuming no team scores more than 50 points in a playoff match (there has to be a winner), what are the different legitimate combinations of scores that a match can possibly end with? (e.g. a game can end 14-2 but can never have a 27-1 score).
This is the way I am thinking… Can you let me know if there is a flaw in the logic?
A team can score 0,2,3,6,7,8 in one drive. With two drives, it can score 4(2+2), 5(2+3). Which means in 2 drives, a team can score 0,2,3,4,5,6,7,8…
Which means it can score any number from 9-50 now – just add 7 to 2,3,4,5,6…. So a team can score 0,2,3,4…50 – 50 possible numbers.
First team can score 50 ways.
Second team can score 49 ways (to eliminate the same score as first which would result in a draw; in a playoff, you have to have a winner).
Since a score 7-13 is the same as 13-7 (we do not care which team won), we take half of the above and get to 1225.
The other approach would be – the winning team scores x, x>1 (winning team cannot score 0); losing team scores y, y
Fun one! Scoring 1 is actually possible in football, as crazy as that sounds, by the defensive team getting a safety on a PAT (point after touchdown attempt). Problem is, the winning team could not score 1 without the other team having scored, so the winning team could never score 1. By that logic, the winning team could score 2-50 (or 49 different score possibilities). The losing team by your logic could score 0-49, or another 49 possibilities. I think that leaves it with 1200 options? Then there’s a few caveats of a few scores that can never happen… the losing team scoring 1 can only happen if the winning team scores at least 6, so 2-1, 3-1, 4-1, 5-1 is out. 7-1 is also not possible, given that means that the winning team did in fact convert their extra point (and didn’t allow for the defensive team to get a safety on that play). All that to say, 1200 – 5 = 1195 would be my proposed answer, which is very open to critique 🙂
That is a very interesting angle – about team scoring 1. This was discussed in my crosspost in Facebook but you have a more nuanced way. I will tag you there.
From winning team having 49 different possibilities and losing team 49 possibilities – how did you get to 1200?
I am thinking first team can score 50 different ways (not counting 1). Second team can score 50 different way (not counting 1). That is 2500 combinations. But 50 of them are draws – 0-0, 2-2, 3-3 etc etc. So, we are down to 2450. But 7-13 and 13-7 is the same score (we do not care which team wins). So, we are down to 2450/2 = 1225 which is what we had posted.
But now, your logic haas to be added for the matches with one team scoring 1. (the other team cannot score 0,1,2,3,4,5 or 7. It can score 6 though. And 8 (2+6), 9 (3+6) etc etc. So, we have to add 44 possibilities. So it would be 1269. What do you think of this logic?
Good catch! I concur.
For kicks, here’s a twitter account that tracks the likelihood of a football game ending in a “scoragami”, or a score that is possible, but has never happened in actuality.
Good catch! I concur.
For kicks, here’s a twitter account that tracks the likelihood of a football game ending in a score that has never happened before!