This week the question is: Could You Have Won the Super Bowl?

[Football, football, football.] Every time your team is on offense, suppose there’s a 1-in-3 chance they score a touchdown (which we’ll say is worth a total of 7 points, as we won’t bother with 2-point conversions here), a 1-in-3 chance they score a field goal (worth 3 points), and a 1-in-3 chance they don’t score any points (i.e., they punt or turn the ball over on downs). After any of these three things happens, your team will then be on defense.

Now, here’s how overtime will work: Your team is on offense first. No matter how many points your team does or does not score, the other team then gets a chance at offense. If the game is still tied beyond this point, the teams will continue alternating between offense and defense. Whichever team scores next wins immediately.

Again, your team is on offense first. What is your team’s probability of winning?

And let’s include the extra credit puzzle too:

If your team happens to score a touchdown on its first possession, then it doesn’t make sense for your opponent to then attempt a field goal, since they’d be guaranteed to lose. Instead, they would attempt to score a tying touchdown.

So let’s add the following to our model: When either team is on offense, they now have a choice. They can still opt for a strategy that results in 7 points, 3 points, or 0 points, each with a 1-in-3 chance. Alternatively, they can opt for a more aggressive strategy that results in 7 points or 0 points, each with a 1-in-2 chance.

Your team remains on offense first. Assuming both teams play to maximize their own chances of Super Bowl victory, now what is your team’s probability of winning?

And here’s my suggested solution, highlight to reveal:

Read more: #ThisWeeksFiddler, 20240223

This is the basis of the situation: Every time a team is on offense, p(7 points) = 1/3, p(3 points) = 1/3 and p(0 points) = 1/3. I am on offense once, on defense once, and then it’s sudden death (we take turns, first to score wins, I begin on offense).

Let’s look at just the first 2 plays (I think it’s called?).

Team 1 \ Team 2	7 points	3 points	0 points
7 points	Tie	Win for team 1	Win for team 1
3 points	Loss for team 1	Tie	Win for team 1
0 points	Loss for team 1	Loss for team 1	Tie

Each column and each row has probability 1/3, hence each cell has probability 1/9. Summing the cells, we have probability 1/3 for a win, a loss and more plays for either team.

So what happens in sudden death? The points don’t matter anymore. Every time I’m on offense, I win with probability 2/3, otherwise there’s another play. Every time I’m on defense, I lose with probability 2/3, otherwise there’s another play.

What is the chance I win or lose?

	I win	I lose	Game continues
Play 1 + 2:	1/3	1/3	1/3
Play 3:	2/3	0	1/3
Play 4:	0	2/3	1/3
Play 5:	2/3	0	1/3
Play 6:	0	2/3	1/3

Actually it’s more interesting to look at the “real” probability, not just the “remember that play x only happens because all the preceding plays didn’t settle the score”.

	I win	I lose
Play 1 + 2:	1/3	1/3
Play 3:	1/3 * 2/3	0
Play 4:	0	1/32 * 2/3
Play 5:	1/33 * 2/3	0
Play 6:	0	1/34 * 2/3

Just a quick check, does all of these probabilities add up to 1? The first row is 2/3, and every row after that is 1/3ⁿ * 2/3, from n = 1. Actually, the whole sum can be written as beginning with 1/3ⁿ * 2/3, from n = 0. Sum of terms, first term 1, each next term multiplies previous term with r, infinitely many terms, the sum is 1 / (1 – r). In our case, r = 1/3. So the sum is 1 / 2/3 = 3/2. Sum of 1/3ⁿ * 2/3 (n from 0 to infinity) = 2/3 * sum of 1/3ⁿ = 2/3 * 3/2 = 1. Great.

However, I only want to sum the cases, where I win.

1/3 + 1/3 * 2/3 + 1/3³ * 2/3 … =

1/3 + 1/3 * (2/3 + 1/3² * 2/3 + 1/3⁴ * 2/3 …) =

1/3 + 1/3 * (1/9⁰ * 2/3 + 1/9 * 2/3 + 1/9² * 2/3 …) =

1/3 + 1/3 * 2/3 * (1/9⁰ + 1/9 + 1/9² …) =

1/3 + 1/3 * 2/3 * (1 / (1-1/9)) =

1/3 + 1/3 * 2/3 * (1 / 8/9) =

1/3 + 1/3 * 2/3 * 9/8 =

1/3 + 18/72 =

4/12 + 3/12 =

7/12 =

0.58333…

So the answer is 58.333%.

Just to make sure, I also wrote a program to simulate this thing. It also reports the probability as something like 58.3%.

And for extra credit:

Once we go into sudden death, the optimal strategy is to go for 7/3/0, as that means a 2/3 chance for points (compared to 7/0 giving a 1/2 chance for points). So, nothing changes. Once we reach this stage, it’s the same chance as above, 2/3 * 9/8 = 3/4 chance for me to win.

For my enemy team, if I originally scored 7 points, the optimal strategy is to go for 7/0, as that gives them 1/2 chance of equalizing (compared to 7/3/0 giving only 1/3 chance). If I scored 0 points, the optimal strategy is to go for 7/3/0, as that gives them 2/3 chance of winning (compared to 7/0 giving only 1/2 chance). But what about 3 points? It’s a little more complicated.

Going for 7/3/0 gives them 1/3 chance of win, 1/3 chance of tie and 1/3 chance of loss. Going for 7/0 gives them 1/2 chance of win, 1/2 chance of loss. Multiplying with the chances for winning the sudden death (1 – 3/4 = 1/4) gives them a chance to win in the first case of 1/3 + 1/3 * 1/4 = 4/12 + 1/12 = 5/12, compared to 1/2 in the second case. So the optimal strategy here is also 7/0.

Let’s look at the numbers so far, seeing how often I win.

If team 1 scores	Team 2 chooses	Team 2 scores	Team 2 scores	Team 2 scores	Team 1 wins
		7	3	0
7	7/0	Tie, p = 1/2		Win, p = 1/2	p = 1/2 * 3/4 + 1/2 = 7/8
3	7/0	Loss, p = 1/2		Win, p = 1/2	p = 1/2
0	7/3/0	Loss, p = 1/3	Loss, p = 1/3	Tie, p = 1/3	p = 1/3 * 3/4 = 1/4

And now we’ve worked our way back to my initial choice. If I go for 7/3/0, my chance of winning is 1/3 * 7/8 + 1/3 * 1/2 + 1/3 * 1/4 = 13/24 = 26/48. If I go for 7/0, my chance of winning is 1/2 * 7/8 + 1/2 * 1/4 = 9/16 = 27/48. So my optimal strategy is 7/0. And then my chance of winning is 9/16 = 0.5625 = 56.25%. Ouch. Neat. (I had a math mistake, so at first got less than 50%.)