The "defense first" strategy in college football OT -- part 2

In a post a couple of days ago, I noted a study that found a strong advantage to choosing "defense first" in college football OT. The first couple of paragraphs of that previous post describe the overtime rule ... you can go back and read them if you're not familiar with it.

That study found that the team that goes on offense last in the first OT (which I'll call "the second team" or "team 2") beat the team that went on offense first ("the first team" or "team 1") 54.9% of the time.

With some of the numbers listed in the original study, and some additional assumptions, we can try to figure out a *theoretical* probability with which the second team will win, and compare that to the observed 54.9%. For this theoretical calculation, I'm assuming the two teams are equal in skill.

The original study gave the distribution of first team scoring. I've combined 6- and 7-point touchdowns to keep things simple (which won't affect the results much):

.235 – team 1 scores 0 points
.299 – team 1 scores FG
.466 – team 1 scores TD

What is the distribution of second team scoring? It depends what the first team does, and we have to guess a bit.

Suppose the first team scores a touchdown. Then, the second team never goes for a field goal. So it's in what otherwise would be a field goal situation .299 of the time, but will have to go for a touchdown anyway. Suppose from fourth-and-something, they will score a touchdown 50% of the time, and score 0 points 50% of the time. In those cases, that would change their distribution to:

.385 – team 2 scores 0 after team 1 TD
.000 – team 2 FG after team 1 TD
.615 – team 2 TD after team 1 TD

Now, suppose the first team scored a field goal. We'll assume the second team plays exactly the same way as the first team:

.235 – team 2 scores 0 after team 1 FG
.299 – team 2 FG after team 1 FG
.466 – team 2 TD after team 1 FG

Finally, suppose the first team scored zero. It had a 76.5% chance to score, but failed. The second team must have a greater than 76.5% chance to score, because it's going to go for a field goal in some cases where the first team might have chosen to go for a touchdown (and fumbled or something). Let's call it 80%.

.200 – team 2 scores 0 after team 1 scores 0
.800 – team 2 FG or TD after team 1 scores 0

The chances of the first team winning in the first OT is the sum of these probabilities:

.180 – team 1 TD, team 2 zero (.466 * .385)
.070 – team 1 FG, team 2 zero (.299 * .235)
.000 – team 1 TD, team 2 FG (never)
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.250 – Total chance team 1 wins in this OT

The chances of the second team winning in the first OT is this sum:

.188 – team 1 zero, team 2 TD/FG (.235 * .800)
.139 – team 1 FG, team 2 TD (.299 * .466)
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.327 – Total chance team 2 wins in this OT

And the chance of a tie is 1 minus the above two totals, which works out to

.423 – Total chance of this OT ending in a tie

Now, the chance of the (original) second team winning the game, is this sum:

Chance of winning in the first OT + (Chance the first OT is a tie * chance of winning the second OT) + (Chance the first two OTs are ties * chance of winning the third OT) + ...

Also, if a given OT ends in a tie, the first team has to go second this period, and the second team has to go first. So the probabilities are switched in the even-numbered OTs. Therefore, the above sum works out to:

.327 + (.423 * .250) + (.423^2 * .327) + (.423^3 * .250) + ...

The sum of that infinite series (which is actually two intertwined geometric series) works out to .526.

Under the assumptions listed above, the chance of the "defense first" team beating the other, equally matched team in OT is .526.

But as we saw in the other study, the actual chance was .549. Why is our estimate different?

One possibility is that we're assuming evenly-matched teams. In real life, college games are often mismatches. However, mismatches should minimize the effect. The more mismatched the teams, the less likely winning the coin toss should affect the result (if a team is good enough, it'll keep scoring TDs and win even if it has to go first). So the real life number for equally matched teams should be higher than .549, not lower.

So what's going on? Why do these estimates not match the observed results? One of three possibilities:

1. My calculations and logic are wrong;
2. The assumptions are wrong;
3. Favored teams won a lot of coin tosses just by luck.

I'm assuming it's not luck. Any suggestions?