0
$\begingroup$

I am trying to confirm the probability of being a weekly winner in an NFL football pool at least once during a given NFL season.

The pool is set up as the following:

Each player picks five teams to beat the spread each week. It is a standard NFL season, meaning that matchups, players and injuries change each week.
Their score is determined by the total points that their five teams outscore the spread, so their score could be positive or negative. The highest positive score wins the week. Nothing prevents multiple players (or all players for that matter) from selecting the same team or teams.

I believe that the above have nothing to do with the problem, but the pertinent facts and assumptions are below:

Assume that there are 90 players, and they all play each week. There are seventeen weeks in the NFL season, so there are seventeen independent contests (there is a cumulative winner, but that is irrelevant to this question). Winning (or losing) in a given week has no impact on the next week or any of the subsequent weeks. Each player starts with a “clean slate” for each week. Each player has an equal chance of winning each week. In other words, assume that skill does not play into the weekly outcome. For the sake of this problem, assume that there are no ties (or there is some form of tiebreaker that makes sure there is only one winner each week).

The question is: What is the probability for a given player to win a week at least once during one NFL season?

$\endgroup$
  • $\begingroup$ If people are using knowledge of the teams rather than random selection there is no way to answer the question. $\endgroup$ – Michael R. Chernick Mar 22 at 4:23
0
$\begingroup$

It seems to me that the key information is that you are considering 90 identical players.

This means that we can assume that the knowledge of each of the 90 players is similar. If this is the case, then the probability of any of them to win in a given week is $1/90$. Thus, the probability of each player not winning in a given week will be $89/90$.

Then, if we assume that the number of players is unchanged and that the weeks are independent, the probability of a given player not winning in the 17 weeks is $(89/90)^{17}$.

Therefore, the probability of that given player winning at least once is the probability of the complementary event $= 1 - (89/90)^{17} \approx 17.3\%$.

Note that this probability decreases with the number of players.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.