# Could You Have Won the Super Bowl?

## This Week's Fiddler

For the initial offense-defense round, there is a $\frac{1}{3}$ chance of a tie, forcing the game to proceed into the "sudden death" phase. Due to symmetry, that means the remaining $\frac{2}{3}$ chance is split evenly amongst our team and theirs, giving us a $\frac{1}{3}$ chance to win in the first offense round and $\frac{1}{3}$ chance to lose in the first defense round.

In the sudden death phase, whoever scores at all wins immediately, meaning there is a $\frac{2}{3}$ chance the team on offense wins a given round. As we start, there is a $\frac{2}{3}$ chance we win outright. If we lose ($\frac{1}{3}$) our offense round, we must defend one round ($\frac{1}{3}$) to get back to the same situation. So this becomes a geometric series:

$\frac{2}{3}+\biggl(\frac{1}{9}\biggl)\frac{2}{3}+\biggl(\frac{1}{9}\biggl)^2\frac{2}{3}+\cdots$Which resolves to:

$\frac{2}{3}\cdot\frac{1}{1-\frac{1}{9}}=\frac{2}{3}\cdot\frac{9}{8}=\frac{3}{4}$We can confirm this because our opponent's chance of victory in the sudden death phase involves surviving an initial defense round ($\frac{1}{3}$) and getting into the same position we were. In the first offense-defense round, this gives them a $\frac{1}{3}\cdot\frac{2}{3}$ chance of outright victory. This makes their odds of winning:

$\frac{2}{9}+\biggl(\frac{1}{9}\biggl)\frac{2}{9}+\biggl(\frac{1}{9}\biggl)^2\frac{2}{9}+\cdots \\ \frac{2}{9}\cdot\frac{1}{1-\frac{1}{9}} \\ \frac{2}{9}\cdot\frac{9}{8} \\ \frac{1}{4}$Which is the complement of our sudden death chances, which helps verify our answer's validity.

Our overall chance of winning then is $\frac{1}{3}+\frac{1}{3}\cdot\frac{3}{4}=\frac{7}{12}$, or about 58.33%. (Their overall chance of winning is $\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}=\frac{5}{12}$.)

**Answer:** 7/12

## Extra Credit

First thing to note is that once we go to the sudden death rounds, each team simply wishes to maximize their chance of scoring at all, so each team would *always* go for the 7-3-0 odds, giving them the same $\frac{2}{3}$ chance of winning each offense round. So our team's chance of winning sudden death is still $\frac{3}{4}$ as per the previous part.

The interesting thing would be what each team opts for in the first round. We can first consider how the opposing team will respond to each of our possible outcomes. From there, we can use mini-max to determine what our team's probability of victory is given ideal play from both teams.

If we score 0 points, the other team only needs to score at all, so they would opt for the 7-3-0 odds, giving them a $\frac{2}{3}$ chance of outright victory and a $\frac{1}{3}$ chance of proceeding to the sudden death phase.

If we score 7 points, the other team *needs* to tie with us, so they would opt for the 7-0 odds to maximize their chance of getting 7. This gives them a $\frac{1}{2}$ chance of tying and proceeding to sudden death, and it gives them a $\frac{1}{2}$ chance to outright lose.

If we score 3 points, let's consider the opposing team choosing between 7-0 odds and 7-3-0 odds. For 7-0 odds, there is a $\frac{1}{2}$ chance for them to outright win and a $\frac{1}{2}$ chance to outright lose. For 7-3-0 odds, there is a $\frac{1}{3}$ chance for them to outright win, a $\frac{1}{3}$ chance to outright lose, and a $\frac{1}{3}$ to continue to sudden death. Because sudden death is in our favor, 7-3-0 odds would be in our favor, meaning the opposing team should opt for 7-0 odds.

This means that if we choose 7-3-0 odds, our chance to outright win is $\frac{1}{3}\bigl(\frac{1}{2}+\frac{1}{2}+0\bigl)=\frac{1}{3}$, our chance to outright lose is $\frac{1}{3}\bigl(0+\frac{1}{2}+\frac{2}{3}\bigl)=\frac{7}{18}$, and our chance to proceed to sudden death is $\frac{1}{3}\bigl(\frac{1}{2}+0+\frac{1}{3}\bigl)=\frac{5}{18}$. This brings our overall chance of victory to $\frac{1}{3}+\frac{3}{4}\cdot\frac{5}{18}=\frac{13}{24}$. (Conversely, the opposing team's chance of victory is $\frac{7}{18}+\frac{1}{4}\cdot\frac{5}{18}=\frac{11}{24}$, in line with expectations.)

On the other hand, if we instead opt for 7-0 odds, our chance to outright win is $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$, our chance to outright lose is $\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}$, and the chance to proceed to sudden death is $\frac{1}{2}\bigl(\frac{1}{2}+\frac{1}{3}\bigl)=\frac{5}{12}$. This brings our overall chance of victory to $\frac{1}{4}+\frac{3}{4}\cdot\frac{5}{12}=\frac{9}{16}$. (Conversely, the opposing team's chance of victory is $\frac{1}{3}+\frac{1}{4}\cdot\frac{5}{12}=\frac{7}{16}$, in line with expectations.)

Because $\frac{9}{16}>\frac{13}{24}$, we thus opt for 7-0 odds, making for $\frac{9}{16}$ (56.25%) overall odds of victory.

**Answer:** 9/16