# Is bold play optimal in football?

It has been 18 months since my last blog post. At that point I was very angry about Trump’s mishandling of the covid pandemic and the fact that people wouldn’t wear masks, while on other days I was saying goodbye to two former colleagues who were mentors, colleagues, and good friends. Not much has changed: now I am angry about people who won’t get vaccinated and I spend my time sticking pins into my Aaron Rodgers voodoo doll hoping that a covid outbreak on his team will keep him from winning the Super Bowl.

To calm myself I have decided to do some math and relax. It is a well-known result (but not easy for an impatient person to find on the internet) that if you are playing a game that is biased against, bold play is optimal. Specifically, if  you want to reach a fixed target amount of money when playing such a game then the optimal strategy is to bet all the money you have until you reach the point where winning will take you beyond your goal and then bet only enough to reach your goal.

For a concrete example, suppose (i) you have \$1 and your wife wants you to take her from brunch at the Cheesecake Factory which will cost you \$64, and (ii) you want to win the necessary amount of money by betting on black at roulette where you win \$1 with probability 18/38 and lose \$1 with probability 20/38. A standard calculation, which I’ll omit since it is not very easy to type in Mircosoft Word, (see Example 1.45 in my book Essentials of Stochastic Processes) shows that the probability I will succeed is 1.3116 x 10 -4. In contrast the strategy of starting with \$1 and “letting it ride” with the hope that you can win six times in a row has probability (18/38)6 = 0.01130. This 86 times as large as the previous answer, but still a very small probability.

Consider now the NFL football game between the Green Bay Packers and the Baltimore Ravens held on Sunday December 19, 2021. After trailing 31-17 the Ravens scored two touchdowns to bring the score to 31-30. To try to win the game without going to overtime they went for a two-point conversion, failed and lost the game. Consulting google I find that surprisingly 49.4% of two point conversions are successful versus 94.1% of kicks for one point. In this game under consideration the two point conversion would not necessarily win the game, since there were about 45 seconds left on the clock with Green Bay having one time out, so there is some chance (say 30%) that Green Bay could use passes completed near the sideline to get within range to make a field goal and win the game 34-32. Rounding 49.4 to 50, going for two results in a win probability for the Ravens of 35%. With a one-point conversion their win probability is 0.94 x 0.7 x p, where p is the probability of winning in overtime. If p = ½ this is 33%. However, if the 8-6 Ravens feel like the 11-3 Packers have a probability significantly bigger than ½ to win in overtime then the right decision was to go for two points.

A second example is provided by the Music City Bowl game between Tennessee and Purdue, held December 30, 2021. After a fourth quarter that saw each team score two touchdowns in a short amount of time (including a two-point conversion by Purdue to tie the score), each had 45 points. The pre-overtime coin flip determined that Tennessee would try to score first (starting as usual from the opponent’s 25 yard line). Skipping over the nail biting excitement that makes football fun to watch, we fast-forward to Tennessee with fourth down and goal on the 1 yard line. The timid thing to do would be to kick the field goal which has a probability that is essentially 1. In this case if Purdue

(i) scores a touchdown (with probability p7) Tennessee (or T for short) loses

(ii) kicks a field goal (with probability p3), they go to a second overtime period

(iii) does not score (with probability p0) T wins

Using symmetry the probability T wins is p0 + p3/2 = 1 – p7 – p3/2

Case 1. If T fails to score (which is what happened in the game) then the Purdue will win   with high probability, since they only need a field goal. In the actual game, three running plays brought the ball 8 yards closer and then the kicker made a fairly routine field goal.

Case 2. If T scores (with probability q) then Purdue must score a touchdown, an event of probability P7 > p7 so the probability T wins when they try to score a touchdown is q[(1-P7) +P7/2]

There are a few too many unknowns here, but if we equate p7 with scoring a touchdown when the team is in the red zone (inside the 20) then the top 10 ranked teams all have probabilities of > 0.8. If we take q=0.5, set p7=0.8 then the probability T wins in Case 2 is 0.3 versus 0.2 – p3/2 in Case 1 which is 0.15 if p3=0.1 (half the time they don’t score a touchdown they get a field goal.

Admittedly like a student on exam with a question they don’t know the answer to, I have laid down a blizzard of equations in hopes of a better score on the problem. But in simple terms since p7 is close to 1, the Tennessee coach could just skip all the math and assume Purdue would go on to score a touchdown on their possession and realize he needs his team to score a touchdown, which regrettably they did not.

Like many posts I have written, the story ended differently than I initially thought but you have to follow the math where it takes you.