1
$\begingroup$

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:

$$z=\large\frac{X-\mu}{\sigma}$$

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!

$\endgroup$
3
$\begingroup$

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):

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

$\endgroup$

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.