# The Wednesday Morning Math Challenge: Week 11 Answers

In week 11, we looked at a scenario that football coaches face all the time: when to go for two. If you haven’t had a chance to tackle the problems yet, **click here. **

**1.** Most football coaches might try to answer this question by mumbling about momentum or trusting your players. But you are too smart to do that. You don’t have to trust your gut. You can use mathematics to figure out which gives you a higher probability of winning.

You know that Gruff has a .97 probability of successfully kicking the extra point, and that your team has a .47 probability of success if you go for two. You also know that you’ll need to score twice in order to give yourself a chance to win the game. (If you tie the game and go into overtime, you have a 50% chance of winning.)

The way to think through this is to look at the different scenarios. You’ll end up with what mathematicians call a decision tree. Here are your options:

If you go for one and do not make it, then you’ll have to go for two after the second score in order to tie the game and send it into overtime. To determine the probability that this will play out, we have to look at the probability that all those things will happen — which we can find by multiplying the probability of each. So if .1-.97 is the probability that Gruff misses the kick after the first score, and .47 is the probability that the Falco offense converts two after the second score to tie the game, and .5 is the chance of winning in overtime, then we have .5 x (1-.97) x.47=.00705.

If you go for one and make it, then you’ll have the choice of going for two points, which would win the game, or one point, which would send the game into overtime, where you’d have a 50% chance of winning. Let’s look at the option of going for two first. The chance you’ll get the extra point and then convert two after the second score is .97x.47=.4559.

The chance that you’ll get the extra point kick, get a second extra point after, and then win the game in overtime is .5x.97x.97=.47045.

Now, in order to figure out our winning probability if you kick the extra point after the first score (since that’s the decision facing you), we have to put this all together. Since you have a choice of going for two or one after the second score, you’ll obviously want to go with the one that gives us the best chance of winning a game, which, in this case, is kicking the second extra point. So the probability that you’ll win if you try to kick the extra point after the first score is equal to the probability of success if you miss the first point, plus the probability that you make it, kick another extra point, and win in overtime: .00705+.47045.=.4775.

Then we need to look at what happens if you go for two after the first score. If you don’t make it, then you’ll have to go for two after the second score in order to tie the game and send it to overtime. The chance of this happening is (1-.47)x.47x.5=.12455.

If you go for two and do make it, then after you’ve scored again, you only have to kick a point after in order to win. The chance that this will happen is .47x.97=.4559.

Now, we can put this together to determine our win probability if we go for two after the first score: .12455+.4559.=.58045.

Go for two!

**2.** To figure out the lowest two point conversion rate for which you would want to go for two, we need to think about this more generally. We can reword the question: what is the lowest possible conversion rate that would give us a better probability of winning than if we kick the extra point?

Return to your decision tree. Call your two point conversion rate x, and your one point conversion rate y. If you kick the extra point, you have a win probability of 1/2(1-y)x+1/2y^2.

If you go for two, your win probability is 1/2x(1-x)+xy.

So we can turn our question into an inequality: 1/2(1-y)x+1/2y^2<1/2x(1-x)+xy.

We already know that y=.97. So (1/2)(1-.97)x+(1/2)(.97^2)<1/2x(1-x)+.97x

.015x+.47045<.5x-.5x^2+.97x

Then we can set it to zero and use the quadratic equation to solve for x.

0=-.5x^2+1.455x -.47045

As long as you’re converting two points at a higher rate than 37%, you should go for two.

As it happens, the two point conversion rate in the NFL is MUCH higher than this — closer to the Falco offense’s 47 percent. And yet coaches are still often hesitant to go for two.

**3.** Now we want to generalize the problem even more.

We can restrict ourself to the region where y>x, because if y

We have 1/2(1-y)x+ymax{1/2y,x}<1/2x(1-x)+xy, where the lefthand side is your winning probability if you go for one and your righthand side is the probability if you go for two.

You might be wondering, what’s that “max” business? Remember that if you make the extra point after the first score, you have a choice of whether to go for one or two, and you want to choose the one that gives you the higher probability for winning (or maximizes your chances).

So let’s take a look at this max function. You can quickly tell that when x>1/2 y, the right hand side is greater than the left, so you should still go for two.

Now we only need to look at x<1/2y

1/2(1-y)x+1/2y^2<1/2(1-x)x+xy

0 < -1/2 y^2+3/2 x y-1/2x^2

or

0 < y^2-3xy+x^2.

You can find the demarcation between the two regions — the region for which you should go for two versus one — when you set it equal to zero and take the portion above the line y = 2 x

Recognize anything? Problem 2 is a specific case of the generalization.