LearnedLeague Defensive Efficiency

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.

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?

For each $i = 1, 2, 3, 4, 5,$ let's $N_i$ be the random score of your opponent, conditioned on getting exactly $i$ answers correct. Given the set of weights we see that \begin{align*}0 \leq & N_1 \leq 3 \\ 1 \leq & N_2 \leq 5 \\ 2 \leq & N_3 \leq 7 \\ 4 \leq & N_4 \leq 8 \\ 6 \leq & N_5 \leq 9,\end{align*} though the distributions are not uniformly distributed. In this case, $$DE_i(N_i) = \begin{cases} \frac{3 - N_1}{3}, &\text{if $i=1;$}\\ \frac{5 - N_2}{4}, &\text{if $i=2;$}\\ \frac{7-N_3}{5}, &\text{if $i=3;$}\\ \frac{8-N_4}{4}, &\text{if $i=4;$}\\ \frac{9 - N_5}{3}, &\text{if $i=5.$}\end{cases}$$

The only thing that we need to do is determine the probabilities for each value of $i=2$.

Since there are two instances of the weight $1,$ there are two possible outcomes that give a total score of $N_2 = 1.$ Let's call them $(0,1_A)$ and $(0, 1_B).$ Similarly, there are three possible outcomes that give a total score of $N_2 = 2$, that is $(0,2_A),$ $(0,2_B)$ and $(1_A, 1_B).$ There are five different ways to get an outcome of $N_2=3$: $(0,3)$, $(1_A, 2_A)$, $(1_B, 2_A)$, $(1_A, 2_B),$ and $(1_B, 2_B).$ There are three was to get $N_2=4$: $(1_A, 3)$, $(1_B, 3)$ and $(2_A, 2_B).$ Finally, we have two was to get $N_2=5$: $(2_A, 3)$ and $(2_B, 3).$ There are $15$ possible outcomes, so we can construct the following table:

$N_2$	$DE_2$	$\mathbb{P} \{ N_2 \}$
$1$	$1$	$\frac{2}{15}$
$2$	$\frac{3}{4}$	$\frac{1}{5}$
$3$	$\frac{1}{2}$	$\frac{1}{3}$
$4$	$\frac{1}{4}$	$\frac{1}{5}$
$5$	$0$	$\frac{2}{15}$

Based on the table above, we see that if your opponent will answer two questions correctly, then the probability of having a defensive efficiency greater than $50\%$ is $$\mathbb{P} \{ DE_2 \gt \frac{1}{2} \} = \mathbb{P} \{ N_2 \leq 2 \} = \frac{2}{15} + \frac{1}{5} = \frac{1}{3}.$$