Riddler Nation for the Win!

Riddler Nation is competing against Conundrum Country at an Olympic archery event. Each team fires three arrows toward a circular target 70 meters away. Hitting the bull’s-eye earns a team $10$ points, while regions successively farther away from the bull’s-eye are worth fewer and fewer points.

Whichever team has more points after three rounds wins. However, if the teams are tied after each team has taken three shots, both sides will fire another three arrows. (If they remain tied, they will continue firing three arrows each until the tie is broken.)

For every shot, each archer of Riddler Nation has a one-third chance of hitting the bull’s-eye (i.e., earning $10$ points), a one-third chance of earning 9 points and a one-third chance of earning $5$ points. Meanwhile, each archer of Conundrum Country earns $8$ points with every arrow.

Which team is favored to win? What is the probability that the team you identified as the favorite will win?

Despite its name the score of Conundrum Country's archery team is anything but. In any $3$-shot round, its perfectly imprecise archers will score $24$ points. While in expectation, the more volatile Riddler archers also average $24$, we need to figure out the probability that Ridder Nation will score more than $24$ points in any round.

Let $R_i$ be the score of the $i$th Riddler nation arrow, which is i.i.d. uniformly distributed over the discrete set $\Omega = \{5,9,10\}$ for $i = 1,2,3.$ There are $27 (=3^3)$ total outcomes for $(R_1, R_2, R_3) \in \Omega^3,$ which result in $10$ distinct possible outcomes for Riddler nation in any round, \begin{align*} 15 &= 5 + 5 + 5\\ 19 &= 5 + 5 + 9 = 5 + 9 + 5 = 9 + 5 + 5 \\ 20 &= 5 + 5 + 10 = 5 + 10 + 5 = 10 + 5 + 5\\ 23 &= 5 + 9 + 9 = 9 + 5 + 9 = 9 + 9 + 5\\ 24 &= 5 + 9 + 10 = 5 + 10 + 9 = 9 + 5 + 10 \\ & \,\,\,= 9 + 10 + 5 = 10 + 5 + 9 = 10 + 9 + 5 \\ 25 &= 5 + 10 + 10 = 10 + 5 + 10 = 10 + 10 + 5 \\ 27 & = 9 + 9 + 9 \\ 28 &= 9 + 9 + 10 = 9 + 10 + 9 = 10 + 9 + 9 \\ 29 &= 9 + 10 + 10 = 10 + 9 + 10 = 10 + 10 + 9 \\ 30 &= 10 + 10 + 10\end{align*}

So we see that \begin{align*}\mathbb{P} \{\text{loss}\} &= \mathbb{P} \{ R_1 + R_2 + R_3 \lt 24 \} = \frac{10}{27}\\ \mathbb{P} \{ \text{tie} \} &= \mathbb{P} \{ R_1 + R_2 + R_3 = 24 \} = \frac{6}{27}; \,\text{and}\\ \mathbb{P}\{\text{win}\} &= \mathbb{P} \{R_1 + R_2 + R_3 \gt 24\} = \frac{11}{27}.\end{align*} Since we can safely ignore the tie scenario, as this would (outside of a measure zero event) result in a finite number of rounds before eventually ending up in either a win or a loss for Riddler Nation. Therefore, we have $$\mathbb{P} \{\text{win} \mid \text{no ties}\} = \frac{\frac{11}{27}}{1 - \frac{6}{27}} = \frac{11}{21},$$ which implies that Riddler Nation should be the favorite to win the archery competition.