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August 26, 2005

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Analysis of the NCAA Overtime Format

In this article we calculate the theoretical advantage of winning the coin toss, and the optimal strategy for two-point conversions, in the NCAA overtime format. Since the model on which the analysis is based is calibrated to NFL data, the results would be directly applicable only if the NFL adopted the NCAA format. However, we would not be surprised if similar conclusions apply to college football.

In the NCAA format, if the score is tied at the end of regulation, there is a coin toss followed by a series of overtime periods. In an overtime period, each team gets one untimed possession starting at the opponent's 25-yard line (plus a try if the possession culminates in a touchdown). In the first overtime period and in all subsequent odd-numbered overtime periods, the winner of the coin toss can choose whether to have the first or second possession. The coin-toss loser gets the choice in the even-numbered overtime periods. Play continues until, at the end of some overtime period, the score is no longer tied. Beginning with the third overtime, a team that scores a touchdown must attempt a two-point try. Details are contained in the rulebook.

The team with the second possession in an overtime period has a strategic advantage, because they know what they have to do to win. For example, if your opponent fails to score during the first possession of an overtime period, you know you need only a field goal to win. If instead your opponent scores a touchdown, you know a field goal is useless. Intuition suggests that the advantage to starting on defense in the first overtime should not be completely dissipated by having to start on offense in the even-numbered overtimes. Consequently, in practice, the coin-toss winner almost always elects to begin on defense. The primary goal of our analysis is to determine, as a theoretical matter, the coin-toss winner's probability of winning the game.

The calculation is a straightforward application of our model for the two-minute drill. Since that model was originally intended for late-game drives under time pressure, it might seem strange that it's useful for studying a format in which possessions are untimed. However, time pressure isn't inherent to that model. One can remove the time pressure altogether by letting the available time be very large, or (to minimize the computations) assuming that a play consumes only one second. The key property of that model is that if we input the offense's win probabilities conditional on the drive resulting in a touchdown, field goal, or no score, the model will calculate the offense's win probability at the start of the drive, assuming that both the offense and defense use optimal strategies.

Results

Overtime
number
Win probability for team
that starts on defense
1 0.5212
2 0.5203
≥ 3 0.5227

We will begin with the results, and defer the details of the calculations until the next section. For each overtime period, the Table on the right shows the win probability at the beginning of the period for the team that starts that period on defense. The win probability is almost the same in each period: a little over 0.52. To put this in perspective, it's a smaller advantage than a team would have if they could start a game with a 1-point lead (0.53 according to the footballcommentary.com Dynamic Programming Model), and is a much smaller advantage than the coin-toss winner has in the NFL's current sudden-death format, which we estimate to be 0.57. Still, a coach who mistakenly selects the first possession after winning the coin toss lowers his team's win probability by more than 0.04, which qualifies as a major coaching blunder. For those whose distaste for sudden death derives mainly from the importance of the coin toss, the NCAA format would not appear to be a good substitute, particularly when there are alternatives that truly render the coin toss irrelevant.

Two-point conversions are not required during the first two overtimes. Let p2 denote the success probability on a two-point conversion. Suppose your opponent scores 7 points in the first possession of the 2nd overtime, and you then score a touchdown of your own. If you kick the extra point, you move to the 3rd overtime. However, in that overtime you will have to start on offense, and your win probability will be 0.4773. Since kicked extra points have about a 0.985 success probability, p2 has to exceed 0.47 before you can justify going for two. In the realistic case in which p2 ≤ 0.47, the teams should never go for two prior to the 3rd overtime. (The Table entries, generated using p2= 0.4, correspond to that case.) If p2 is between 0.47 and 0.57, the team with the second possession during an overtime period should go for two if their opponents score 7 points, but the team with the first possession should kick if they score a touchdown. Consequently, as p2 rises from 0.47 to 0.57, the format becomes steadily more favorable to the team that wins the coin toss. Their win probability peaks at 0.54 when p2 = 0.57. When p2 > 0.57, even the team with the first possession should go for two following a touchdown.

Details of the Calculations

Let U(p0, pFG, pTD) denote the offense's win probability at the beginning of a possession, if pTD, pFG, and p0 are the offense's win probabilities conditional on the possession resulting in a touchdown, field goal, or no score respectively. The function U can be derived from our model for the two-minute drill.

For n < 4 let F(n) be the win probability, at the start of overtime n, for the team that has the first possession in that overtime. If that team scores i points, let S(n,i) be the win probability for the team that has the second possession, at the start of its possession. We will show how to compute F(n) and each S(n,i) given F(n+1), which we assume has already been determined. Since every overtime after the second is identical, the boundary condition is F(3)=F(4). The unique value for F(4) that satisfies this condition and all the equations listed below can be found by a simple iteration, as shown in the MATLAB® source code.

The first step is to compute S(n,i) for each i. As an example, we will show how to compute S(n,3), the win probability for the team that goes second given that the team that went first scored a field goal. If the team with the second possession fails to score, their win probability is 0. If they score a field goal, the game continues to the next overtime, where they have the first possession. Their win probability is therefore F(n+1). If they score a touchdown, their win probability is 1. Hence

(1)
S(n,3) = U(0, F(n+1), 1).

As another example, we will show how to compute S(n,7). (This is meaningful only when n equals 1 or 2. Beginning with the third overtime, teams have to attempt two-point conversions.) If the team with the second possession fails to score or scores a field goal, their win probability is 0. If they score a touchdown, they can kick the extra point or go for two. If they kick, their win probability is p1F(n+1), where p1 is the success probability on a kicked extra point. If they go for two, their win probability is simply p2, the success probability on a two-point conversion. Since they will choose the larger of these, their win probability is max{p1F(n+1), p2} if they score a touchdown. Therefore,

(2)
S(n,7) = U(0, 0, max{p1F(n+1), p2}).

In a similar manner, we find

(3)
S(n,0) = U( F(n+1), 1, 1),
(4)
S(n,8) = U(0, 0, p2F(n+1)).

If n < 3 we have

(5)
S(n,6) = U(0, 0, p1+(1−p1)F(n+1)),

whereas if n ≥ 3,

(6)
S(n,6) = U(0, 0, p2+(1−p2)F(n+1)).

All that remains is to compute F(n). For notational simplicity, define f(n,i) = 1−S(n,i). This is the win probability for the team that has the first possession, if the possession results in i points.

If the team with first possession scores no points, their win probability is f(n,0), whereas if they score a field goal, their win probability is f(n,3). If they score a touchdown, they can either kick, in which case their win probability is

(7)
p1f(n,7) + (1−p1)f(n,6),

or go for two, in which case their win probability is

(8)
p2f(n,8) + (1−p2)f(n,6).

(The first option is available only when n < 3.) Therefore

(9)
F(n) = U( f(n,0), f(n,3), max{p1f(n,7)+(1−p1)f(n,6), p2f(n,8)+(1−p2)f(n,6)}).

Copyright © 2005 by William S. Krasker