# The Free Throws Problem

At a recent workshop on collaboration, the other participants and I were presented with a version of this problem:

1. Adam hits 60% of his free throws. He gets fouled just before the buzzer, and his team is down by one point. Based on the rules in play, he can shoot up to twice. If he misses the first shot, the game is over and his team loses. If he makes the first shot and misses the second, the game goes into overtime. If he makes both, his team wins. What is the probability of each outcome?
2. Bob is in the same situation, in a different game. The likelihood that his team will win is the same as that his team will lose. What is his free throw percentage?
3. Carl insists that his free throw percentage means that, in the same situation, the most likely event is that the game will go into overtime. Prove he’s wrong.

The first question is a straightforward probabilities question. The second involves some thought, as well as an opportunity to look at the problem from multiple ways. The third is the most interesting; we devised several proofs which are superficially different in nature, but equally sound.

Question 1 can most quickly be solved by taking $$P(E) = 0.6$$ and $$P(\overline{E}) = 0.4$$, where E represents making a free throw, W represents winning, T represents tying, and L represents losing:

• $$P(W) = P(E) \times P(E) = 0.6 \times 0.6 = 0.36$$
• $$P(T) = P(E) \times P(\overline{E}) = 0.6 \times 0.4 = 0.24$$
• $$P(L) = P(\overline{E}) = 0.4$$

Hence, the most likely situation is that Adam’s team will lose.

One of the people in my group used a visual approach to solving this. Since the moderator had discussed using an area model for multiplication, he asked how the area model could be applied to this approach. I adapted my groupmate’s approach as shown.

The percentages are the result of determining the area of the respective region, compared to the entire square. This provides an excellently visual way of showing what’s going on here: The Makes/Makes square is 60% by 60%, for instance, for a total area of 36%.

Question 2 is most definitively solved algebraically, but building a visual model in GeoGebra allows us to see dynamically what happens when we change the free throw percentage. Here’s what it looks like for percentages of 20%, 40%, 50%, and 80%:

At 62%, we have a decent approximation: The green square is 38.44% and the red rectangle is 38%.

To find the actual solution, let p be the probability of hitting a single shot, and w, t, and l be the respective probabilities of winning, tying, and losing. Then $w = p^2 \\ t = p(1-p) \\ l = 1 – p$

We want the case where winning and losing are equally possible, that is, where $$w = l$$. So $$p^2 = 1 – p$$ $$\Rightarrow p^2 + p – 1 = 0$$. Using the Quadratic Formula, we have solutions at $$p = \frac{-1 \pm \sqrt{1 + 4}}{2}$$; only one of these is positive, that is, $$p = \frac{\sqrt{5} – 1}{2} \approx 61.8\%$$.

So our visual method with a dynamic model yielded an answer that was decently close to the “actual” answer, and would serve well as an approximation.

Finally, let’s look at the third question. As a group, we generated four proofs to this, but I’ll skip the one involving calculus. The first proof is to simply graph the three functions. Looking at them on Desmos, for instance, we can quickly see that t (the blue line) will never be more than l (the green line). The only point at which they’re the same, they’re less than w (the red line). Here we can also quickly see that l crosses w (for good) at around 0.6, which is what we’d already determined.

The second proof algebraically demonstrates what the graph shows. First, when is $$l = t$$? Well, $$1 – p = p(1 – p)$$$$\Rightarrow 1 – p = p – p^2$$$$\Rightarrow p^2 – 2p + 1 = 0$$. This has solutions at $$p = \frac{2 \pm \sqrt{4 – 4}}{2} = 1$$. Since the coefficient on p is positive, the function is always greater than or equal to this vertex, that is, $$t < l$$ except at that solution. And at that solution, we can see that $$t = 0$$ and $$w = 1$$, so $$t < w$$. Thus, t is always less than at least one of the other two functions.

However, because I’m familiar with al-Khowarizmi’s visual method for solving quadratics, my mind fairly instantly connected the visual model with the question, and came up with a geometric proof.

Let’s label the points, and divide the red rectangle as shown. With one exception, it suffices to show that rectangles $$CEFI$$ and $$AGFH$$ are always congruent, because $$CEGB$$ (i.e., l) consists of $$CEFI$$ and $$IBGF$$. The exception is when $$IBGF$$ has no area at all, and that’s the trivial case that the entire square is green, i.e, that l and t are both 0 while w is 1.

So how do we prove that? Since $$DEFH$$ is a square, $$DE = EF = FH = HD$$. Since $$ABCD$$ is a square, $$AD = CD$$; we know $$DE = DH$$. Hence $$AD – DH = AH = EC = CD – DE$$. A fully rigorous proof would need to show that all the angles in the diagram are right, but that’s trivial. Since $$FH = EF$$ and $$CE = AH$$, the rectangles $$CEFI$$ and $$AGFH$$ are congruent.

I find this proof particularly elegant because it doesn’t require any algebra in the modern sense. It shows the meaning of algebra in al-Khowarizmi’s reality: We can find relationships between area by cutting shapes into smaller congruences and moving those congruent shapes around. And while this visual model, which is excellent at solving problems when numbers are given, struggles to find an exact answer for the second question, it gets us a decent enough estimate by which to check our answer.

The repeated message, of course, is that we shouldn’t feel compelled to find the “one true way” to solve problems. Exploring strategies and perspectives can give us tools far more powerful than simple rote memorization of algorithms.