# Poor Luck in Fantasy Football

This year in fantasy football has been frustrating to say the least. I have the best team in the league by far when looking at overall points, yet I sit at week 10 with a record of 6-4 and barely holding on to second place. All four of my losses have been because I have gone up against a team that ends up scoring the most points in the league that week.

If there are n=10 teams in the league then (just theoretically speaking) there is a 1/10 chance the team I play will score the most points in the league that week. What are the chances that I have played the number one scoring team in the league 4 times in 10 weeks. Another way to look at it could be what are the chances I roll a 1 on a 10 sided die (do those exist? who cares) 4 times in 10 rolls.

Note: I just want to get into the theoretical probabilities since going into more nuanced probabilities of losing these games would be far more difficult. In the four games I lost I was heavily favored to win all of them, and ended up being the 2nd or 3rd highest scoring team in the league each time in those weeks (twice 2nd, twice 3rd).

• "Do ten-sided dice exist?" Yes! The most common dice sizes include d4, d6, d8, d10 (and d%), d12, d20, d30, and d100 (though using a d10 and a d% are more commonly used than a d100. d100's are hard to read and don't stop rolling quickly enough). – JMoravitz Nov 14 '17 at 18:09
• As for the content of your question, The probability of rolling a $1$ exactly four times in ten rolls with a ten-sided die would be $\binom{10}{4}\left(\frac{1}{10}\right)^4\left(\frac{9}{10}\right)^{6}\approx 0.011$ as per the binomial distribution. Getting at least four would be $\approx 0.012$. This is of course looking at the simplification that all ten teams are equally likely to win. If your team really does have significant advantages over the others, then the odds shift more in your favor. – JMoravitz Nov 14 '17 at 18:13
• Minor typo: should be $(0.9)^6$ correct? – User8128 Nov 14 '17 at 18:16

## 1 Answer

Here is the probability that your opponent scores the most in the league exactly four times: there are ten weeks, so choose four of them where your opponent will score the most, this gives a factor of $\binom {10}4$. In those weeks, your opponent has a $0.1$ chance of scoring the most; since that had to happen all four times, you wind up with a factor of $(0.1)^4$. However, since we are assuming it happened exactly four times, you opponent also had to not score the most during the other weeks, this gives a factor of $(0.9)^6$. Thus the probability of it happening exactly four times is $$\binom{10}4 (0.1)^4(0.9)^6 \approx 0.01116$$ so about a $1\%$ chance. However, the probability that it happens at least four times is $$\sum_{k=4}^{10} \binom{10}k (0.1)^k(0.9)^{10-k} \approx 0.01279$$ so there is a roughly $1.3\%$ chance of something this bad or worse happening to you.

PS: fantasy owner myself and I have the opposite "problem". I'm $9$-$1$ and it's because no one can seem to score against me. This means I'll likely get stomped in the playoffs.