A friend of mine came to me a few hours ago wondering what the probability of him making the playoffs were in our fantasy football league. I originally thought it wouldn't be too hard to figure out, and perhaps it's not and I am simply missing a few details.

The league has $10$ teams and $6$ teams make the playoffs. Assume each team is equally likely to win each week. There are $13$ weeks in our season and the expected number of wins to make the playoffs we rounded up to $7$ (there are no ties allowed or fractional points for wins/losses).

If we look at this from the start of the season then the answer is intuitively $60\%$, right?

$$\large\frac{{1\choose 1} {9\choose 5}}{10\choose 6}=.6$$

Now, if we wanted to find the probability from any point in the season how could we do it? We attempted looking at it as a normal distribution. Currently, we have completed four weeks in the season so the average number of wins is $2$ and the standard deviation is $.8165$.

Let's say you have a $3$-$1$ record and you wanted to determine $P(X\ge 7)$ where $X$ is the number of wins you have at the end of the season. How would you find that? When I've tried determining it with $z$-scores I get probabilities that don't seem to make sense (i.e $.9999$). Using:


I've tried $X=7$ and $X=4$ since that's how many wins are needed. My probabilities are always too high and this isn't even taking into account adding in $P(8)+P(9)..$ and so forth. I'm thinking it may not be appropriate to assume it's normally distributed.

How can we generalize it for any point in the season/any number of teams/games played?

Any help would be great!


If you have a $3-1$ record, $9$ games left to play, and need $4$ or more additional wins to make the playoffs, this is the same as "if I flip $9$ coins, what are the chances at least $4$ of them come up heads?"

To compute this, you count the number of ways to win $4,5,6,7,8$ or $9$ games, and divide by the total number of possibilities:

$$ \dfrac{\sum_{k=4}^9 {9\choose{k}}}{2^9} $$

or equivalently (and slightly easier to calculate):

$$ 1 - \dfrac{\sum_{k=1}^3 {9\choose{k}}}{2^9} = \frac{383}{512} \simeq 75\% $$


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