A new season of LearnedLeague recently kicked off! Many folks may not be familiar with this daily trivia platform. I learned about it a few years ago and joined after my turn as a game show contestant.
Anyway, here’s how it works: Every day, you and your opponent for the day are presented with the same six trivia questions. You each do your best to answer these, and you assign point values for your opponent, without knowing (until the following day) which questions your opponent answered correctly. You must assign point values of 0, 1, 1, 2, 2, and 3 to the six questions.
For example, suppose I assigned values of 1, 2, 0, 3, 2, and 1 to the six questions, in order. Unbeknownst to me, my opponent answers the first, third, and fourth questions correctly. That means they get 1 + 0 + 3, or 4 points for the match. My own score depends on which questions I got right, and how these were scored by my opponent. If I get more than 4 points, I win the match.
Now, when someone answers three questions correctly, like my opponent just hypothetically did, the fewest points they can earn is 0 + 1 + 1 = 2, while the most points they can earn is 2 + 2 + 3 = 7. The fact that they got 4 points wasn’t great (from my perspective), but wasn’t terrible. In LearnedLeague, my defensive efficiency is defined as the maximum possible points allowed minus actual points allowed, divided by the maximum possible points allowed minus the minimum possible points allowed. Here, that was (7−4)/(7−2), which simplified to 3/5, or 60 percent.
By this definition, defensive efficiency ranges somewhere between 0 and 100 percent. (That is, assuming it’s even defined—which it’s not when your opponent gets zero questions right or all six questions right.)
Suppose you know for a fact that your opponent will get two questions right. However, you have absolutely no idea which two questions these are, and so you randomly apply the six point values to the six questions.
What is the probability that your defensive efficiency for the day will be greater than 50 percent?
WLOG, assume you always choose assigned values of $0,1,1,2,2,3$. There are ${6 \choose 2} = 15$ ways your opponent can choose two questions. The distribution of achieving each score is detailed below:
A score of 1 or 2 are necessary to have defensive efficiency $\gt 50\%$.
$$\boxed{\frac{5}{15} = 33.\overline{3}\%}$$Now suppose your opponent is equally likely to get one, two, three, four, or five questions correct.
As before, you randomly apply the six point values (0, 1, 1, 2, 2, 3) to the six questions.
What is the probability that your defensive efficiency will be greater than 50 percent?
Lets go through all!
There are ${6 \choose 1} = 6$ ways your opponent can choose one question. The distribution of achieving each score is detailed below:
A score of 0 or 1 are necessary to have defensive efficiency $\gt 50\%$.
$$\boxed{\frac{3}{6} = 50\%}$$From Fiddler, a score of 1 or 2 are necessary to have defensive efficiency $\gt 50\%$.
$$\boxed{\frac{5}{15} = 33.\overline{3}\%}$$There are ${6 \choose 3} = 20$ ways your opponent can choose three questions. The distribution of achieving each score is detailed below:
A score of 2, 3, or 4 are necessary to have defensive efficiency $\gt 50\%$.
$$\boxed{\frac{10}{20} = 50\%}$$There are ${6 \choose 4} = 15$ ways your opponent can choose four questions. The distribution of achieving each score is detailed below:
A score of 4 or 5 are necessary to have defensive efficiency $\gt 50\%$.
$$\boxed{\frac{5}{15} = 33.\overline{3}\%}$$There are ${6 \choose 5} = 6$ ways your opponent can choose five questions. The distribution of achieving each score is detailed below:
A score of 6 or 7 are necessary to have defensive efficiency $\gt 50\%$.
$$\boxed{\frac{3}{6} = 50\%}$$Since all are equally likely,
\begin{align*} P &= \frac{3}{5}\cdot 50\% + \frac{2}{5}\cdot 33.\overline{3}\% \\ &= \frac{3}{5}\cdot \frac{1}{2} + \frac{2}{5}\cdot \frac{1}{3} \\ &= \boxed{\frac{13}{30} = 43.\overline{3}\%} \\ \end{align*}