What is the probability of odd versus even in a football game?

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Discussion Overview

The discussion revolves around the probability of the sum of scores in a football game being odd versus even. Participants explore the mathematical implications of scoring in football, considering the types of scores and their frequencies. The conversation includes theoretical reasoning and suggestions for empirical analysis.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that the probability is not a simple 50/50 due to the specific scoring options in football (2, 3, 6, 7, 8) and proposes that even scores may have an advantage because safeties are rarely scored.
  • Another participant argues that the outcome is influenced primarily by the odd-numbered scores, and that the even scores do not change the parity unless they are the only scoring events in a game.
  • A different participant expresses a desire for a basic formula to determine the probability and questions whether it can be proven that the chance is not 50/50.
  • One participant proposes using Poisson distributions to model the scoring events and suggests that knowing the average number of touchdowns and field goals could help calculate the probabilities of even versus odd sums.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the probability of odd versus even scores is 50/50. There are competing views on the influence of different scoring types and the need for empirical data versus theoretical models.

Contextual Notes

Limitations include the dependence on the assumption of scoring distributions and the need for specific average scoring data to perform calculations.

sophmore
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Who can figure this one out?

The topic of discussion amongst my friends and I:

What is the probability of the sum of a football game landing on odd versus even.

My friends say that it is a simple 50/50 coin toss. My rebuttal is that while in football the only integers that can be achieved by one score are: 2 3 6 7 or 8 thus limiting the possibility
of a 50/50 ratio.

2 = safety
3 = field goal
6 = touchdown w/ missed PAT, or missed 2pt. conversion
7 = touchdown w/ made PAT
8 = touchdown w/ made 2pt. conversion

My Hypothesis: Since safeties are rarely scored in a football game, I believe that even has the advantage versus odd.

Now I have read some responses of unrelated topics, and there seems to be a high number of intelligent people subscribed to this forum. Everyone likes to 1-up the next guy, which is good thus creating a large field of answers.

Please approach this question with an open mind, and really give it some thought as I have not come to a conclusion on my own.
 
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you can skip the potentially difficult mathematical proof and go to nfl.com and start analyzing actual data.

Since safetys, touchdowns w/ no extra pts, and touchdowns w/ 2 points all are even scores, then they don't change the outcome unless they are the only scoring event in a game (an odd number plus 2,6 or 8 is still an odd number, and even number is still an even number) the only contributors are the odd number scores since it really depends on if there are an odd number of odd scores or not (i.e. touchdown w/ ep and fieldgoals, which are also the most common) so that makes it easieir to think about.

If I were you, id find an estimate of the probability of even vs odd using actual NFL scores, then make a a bunch of bets with your friends that gives you the slight edge (if in fact there is a difference between the two)
 


As i appreciate your response, I don't think it helps me much. I figured that if I analyzed actual NFL data that I would come up with a rough estimate. I just don't want to have to do that, so I asked if anyone knew a basic formula to figure it out. Can it be proved that it's not a 50/50% chance? I know this isin't a common question in this forum, but I ask if anyone can answer it.
 


Well, if we assumed that safeties, touchdowns (all 3 kinds separately), and field goals were 'rare events' with independent Poisson distributions, we could find the probabilities of even/odd. Actually we'd need only the chance of the 3 and 7 point ones.

If you can give us the *average* number of touchdown+PAT combinations in a game, plus the *average* number of field goals, we can calculate the probabilities for you.
 

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