just putting this here wrote:
What you've done isn't correct.
Even keeping your assumption that overtime means a 50% probability of winning, the expression for the probability of winning when down 3 would be T3 + F/2. And when down 4, it's T4.
1-T4 is probability of LOSING.
The Cowboys were the ones making the supposed decision (go up 3 or 4) so logically I was talking about their chances of winning. You are talking about the Lions' chances of winning.
1-(F+T3)/2 is not meaningful. For one thing, you've assumed that a touchdown could lead to overtime
That's true, but it was late at night, whereas your above error was made during the day.
But even beyond that, you've made an assumption that this is the last possession of the game. You can't assume that.
I explicitly stated that assumption on the implied grounds that it's close enough to zero to be negligible with 2 minutes left, which it is. Unless they scored a TD, already an unlikely event since they don't have to, the Lions would surely run down the clock before their FG attempt.
So, the revised probability of Cowboy victory at 3 points ahead is (just using T for the sake of clarity)
1 - (F + T) {don't tell me the Lions might score twice!}
+ F/2 {home field advantage won't be enough to matter}
= 1 - T - F/2
The magic website says that equals 0.82. If so,
T + F/2 = 1 - .82 = .18
F + 2T = .36
Is this realistic in light of the points-per-drive averages I linked before? Twelve teams had a scoring % better than that even during regulation, with punting an option:
http://www.footballdb.com/stats/drives.htmlThe expected long term result of the offense's drive can be expressed as
3F + 7T
If rojo's source is correct, this equals
3(.36 - 2T) + 7T = 1.08 + T
Since surely F >= T, we have
.36 = F + 2T >= T + 2T = 3T
.12 >= T
for a maximum expected score of 1.20, among the worst points-per-drive for any team in regulation during 2014. Basically your correction helped my argument. Good looking out.
About that homefield advantage, even if the Cowboys had a .60 chance of winning OT, you'd have
T + .40F = .18
2.5T + F = .45, an expectation of
3(.45 - 2.5T) + 7T = 1.35 - .5T, and since T can't be less than zero, thus a drive expectation of at best 1.35 - still very low.
More simply, since few people will read all that, suppose fate set the chance of a Lion TD at 0, which should maximize the Cowboys' chances. To have an 82% chance of winning, they'd need the Lions' chance of a FG to be only 0.36 - barely one in three - or a significant overtime advantage. I can't find stats on this, but I doubt many NFL defenses are 2 for 3 in that situation or that many teams are much better than .50 in OT.
If there are any more errors, I don't care, it's still better to be up by 4.