Here’s an extension to the problem in my previous post. Time has run out, and a player is at the free throw line. If he makes the first shot, he gets a second try. If he makes both shots, his team wins; if he misses the first, his team loses. Otherwise, it’s a tie game, and it goes into overtime.
- Dave tracks his averages for his first and second free throws separately. He says that, using those numbers, the probability of a tie game in the scenario is greater than either of the other two possibilities. Does he do better or worse on his second free throw?
If we used the visual area model, we could see that this scenario is now possible because the green area can be any rectangle, so the congruence discussed in the previous post no longer necessarily holds. But it doesn’t tell us anything immediate about the relationship between the first and second free throw percentages.
Let’s use a completely algebraic method. Let p and q be the probability of hitting the first and second free throws, respectively, and w, t, and l against be the respective probabilities of winning, tying, and losing. Then \[w = pq \\ t = p(1-q) = p – pq \\ l = 1 – p\]
We want t to be greater than both l and w. Let’s solve those individually.
In the first case, \(t > l\) so \(p – pq > 1 – p \Rightarrow -pq > 1 – 2p \Rightarrow q < (2p – 1)/p\).
In the second case, \(t > w\) so \(p – pq > pq \Rightarrow p > 2pq \Rightarrow 1/2 > q \Rightarrow q < 1/2\).
Since \(q > 0\), \((2p – 1)/p > 0 \Rightarrow 2p – 1 > 0 \Rightarrow p > 1/2\).
So t is only greater than both l and w when \(p > 1/2\) and \(q < 1/2\), meaning that \(q < p\). That is, Dave has a worse average on his second free throws than on his first.
We can see the exact domain for p and q by graphing the two inequalities for q in Desmos. The overlapped area are the cases where p (x-axis) and q (y-axis) make the probability of tying the highest. It’s not sufficient that q merely be below 0.5 and p be above 0.5. We can calculate the intersection point for the red and blue dotted lines: \((2p – 1)/p = 1/2 \Rightarrow 2(2p – 1) = p \Rightarrow 4p-2 = p \Rightarrow 3p = 2\). So when \(p = 2/3\) and \(q = 1/2\), the three values (w, l, and t) ought to be the same.
Let’s check that that’s true. \(w = pq = (2/3)(1/2) = 1/3\); \(t = p – pq = 2/3 – 1/3 = 1/3\); and \(l = 1 – p = 1 – 2/3 = 1/3\). This is the only case where all three probabilities are the same.
One more question: What is the minimum difference between p and q? Clearly, for values \(p > 2/3\), \(q < 1/2\), so any value of p above 2/3 has a greater minimal difference with q than \(p = 2/3\). At \(p = 2/3\), the difference is \(2/3 – 1/2 = 1/6\). At \(p = 1/2\), the difference is \(1/2 – 0 = 1/2\).
Adding the line \(q = p – 1/6\), we can easily see that this intersection point represents the minimal difference.
So if Dave’s free throw averages are such that the chance of a tie in the scenario given is greater than the other two possibilities, his second throw average has to be at least 1/6 less than his first throw average.