Extra Credit LearnedLeague Defensive Efficiency

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?

In the classic problem we costructed a table for $N_2.$ Here we need to then construct similar tables for $N_1$, $N_3,$ $N_4$ and $N_5.$ For $i=1,$ there is exactly one way apiece to score zero or $3$ and two ways apiece to score $1$ or $2$, so we have the following table:

$N_1$	$DE_1$	$\mathbb{P} \{ N_1 \}$
$0$	$1$	$\frac{1}{6}$
$1$	$\frac{2}{3}$	$\frac{1}{3}$
$2$	$\frac{1}{3}$	$\frac{1}{3}$
$3$	$0$	$\frac{1}{6}$

Similarly, we can construct the other tables as follows:

$N_3$	$DE_3$	$\mathbb{P} \{ N_3 \}$
$2$	$1$	$\frac{1}{20}$
$3$	$\frac{4}{5}$	$\frac{1}{5}$
$4$	$\frac{3}{5}$	$\frac{1}{4}$
$5$	$\frac{2}{5}$	$\frac{1}{4}$
$6$	$\frac{1}{5}$	$\frac{1}{5}$
$7$	$0$	$\frac{1}{20}$

$N_4$	$DE_4$	$\mathbb{P} \{ N_4 \}$
$4$	$1$	$\frac{2}{15}$
$5$	$\frac{3}{4}$	$\frac{1}{5}$
$6$	$\frac{1}{2}$	$\frac{1}{3}$
$7$	$\frac{1}{4}$	$\frac{1}{5}$
$8$	$0$	$\frac{2}{15}$

$N_5$	$DE_5$	$\mathbb{P} \{ N_5 \}$
$6$	$1$	$\frac{1}{6}$
$7$	$\frac{2}{3}$	$\frac{1}{3}$
$8$	$\frac{1}{3}$	$\frac{1}{3}$
$9$	$0$	$\frac{1}{6}$

In particular we see that

$$\mathbb{P} \{ DE_i \gt \frac{1}{2} \} = \begin{cases} \frac{1}{2}, &\text{if $i \in \{1,3,5\}$;}\\ \frac{1}{3}, &\text{if $i \in \{2,4\}$.}\end{cases}$$ So if your opponent answers $I$ questions where $I$ is uniformly distributed on the set $\{1, 2,3,4,5\},$ then the probability of having a defensive efficiency score greater than $50\%$ is equal to $$\begin{align*} \mathbb{P} \{ DE_I \gt \frac{1}{2} \} &= \sum_{i=1}^5 \mathbb{P} \{ DE_i \gt \frac{1}{2} \} \mathbb{P} \{ I = i \} \\ &= \frac{1}{5} \sum_{i=1}^5 \mathbb{P} \{ DE_i \gt \frac{1}{2} \} \\ &= \frac{1}{5} \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{2} + \frac{1}{3} + \frac{1}{2} \right) \\ &= \frac{13}{30} = 43.\bar{3}\%.\end{align*}$$