[Goose]

I think the formula is the value for going it if successful times the probability of success minus the value of going for it but failing times (1 - p) has to be greater than the value of punting.

So it becomes 3.83p - 2.72(1 - p) > -1.8 (the value to the other team of having the ball on the 20, if you punt).  Solving for p, you get

p > 0.92 / 1.11; p > 83%

It's a bit late, so I could be wrong on the math; I would have thought the breakeven point would be lower, but maybe that's because I would have expected the value of being on the 40 compared to the 20 to be smaller.

BC

UPDATE: Your equation is correct. My short-lived initial response was nonsense :-). I was evaluating play as if there was an option to do nothing. There isn't. If you don't go for it, you have to punt, which has a presumed -1.8 fixed value. 3.83p - 2.72 + 2.72p > -1.8 becomes 6.55p > 0.92 or p > 0.140. That yields a pretty low 14% success probability as the point above which going for it makes sense. My guesstimate of our success probability being zero still says don't do it, but it makes it much more arguable. I have evaluated the formula at p=90% and p= 20% with reasonable results.  Of course if one believes the ball can be downed inside the 20 that moves the required success probability up.

A more "conservative" result that may have some value as a "worst case" scenario is what does p need to be completely discounting the benefits of not giving the ball back to your opponent. IOW -2.72 (1 -p) > -1.8 or 2.72p > .92. That yields a p of .338.

[Goose]

I think the formula is the value for going it if successful times the probability of success minus the value of going for it but failing times (1 - p) has to be greater than the value of punting.

So it becomes 3.83p - 2.72(1 - p) > -1.8 (the value to the other team of having the ball on the 20, if you punt).  Solving for p, you get

p > 0.92 / 1.11; p > 83%

It's a bit late, so I could be wrong on the math; I would have thought the breakeven point would be lower, but maybe that's because I would have expected the value of being on the 40 compared to the 20 to be smaller.

BC

UPDATE: Your equation is correct. My short-lived initial response was nonsense :-). I was evaluating play as if there was an option to do nothing. There isn't. If you don't go for it, you have to punt, which has a presumed -1.8 fixed value. 3.83p - 2.72 + 2.72p > -1.8 becomes 6.55p > 0.92 or p > 0.140. That yields a pretty low 14% success probability as the point above which going for it makes sense. My guesstimate of our success probability being zero still says don't do it, but it makes it much more arguable. I have evaluated the formula at p=90% and p= 20% with reasonable results.  Of course if one believes the ball can be downed inside the 20 that moves the required success probability up.

A more "conservative" result that may have some value as a "worst case" scenario is what does p need to be completely discounting the benefits of not giving the ball back to your opponent. IOW -2.72 (1 -p) > -1.8 or 2.72p > .92. That yields a p of .338.