Riddler Classic: March 26, 2021

David Ding

March 26, 2021

March Mathness Classic

Answer: Two profiles.

Explanation:

Let $(p,q,r)$ denote a shooting profile as defined by an ordered triplet of $0\le p,q,r,\le 1$, each denoting the free throw probabily of the player's first, second, and third free throws in the RCAA, respectively. As seen from the rule of the game, the expected score of a trip to the free throw line is:

$$\begin{array}{rl}E(p,q,r)& =p(1-q)+2pq(1-r)+3pqr\\ & =p-pq+2pq-2pqr+3pqr\\ & =p+pq+pqr\end{array}$$

Now, there are exactly $3!=6$ ways to permute the shooting profile composed of the ordered triplet, and so our answer lies between 1 and 6, inclusive. The following lemma will pare this down to three:

Lemma: two profiles cannot produce the same expected score if both of them contain at least one probability in the same position.

Proof: Clearly if two of the three probabilities are in the same position, the third one must be as well. Therefore, without loss of generality, let $p$ for two profiles be in the same position. Therefore, we are looking at the profiles $(p,q,r)$ and $(p,r,q)$. Their respective expected scores are $p+pq+pqr$ and $p+pr+prq$. Clearly, $pqr=prq$, and so in order for those two expected scores to be equal, $q=r$, which is not possible as it is given in the problem that $p,q,r$ are all distinct. Similar arguments can be made for the other two positions. Q.E.D.

Corollary: this means that at most $(p,q,r)$, $(q,r,p)$, and $(r,p,q)$ can yield equal expected scores, which means the answer is at most three.

Now let's formally prove that two profiles is the greatest number of distinct shooting profiles.

Achievability

Let's consider two profiles $(p,q,r)$ and $(q,r,p)$. Their respective expected scores are $p+pq+pqr$ and $q+qr+qrp$. Now, as $pqr=qrp$, we essentially only need to look at the equation $p+pq=q+qr$. This is one equation with three unknowns, and so we can have up to two free variables. For simplicity, let $r$ be the fixed variable as it only shows up once in the equation. Re-arranging, we get:

$$\begin{array}{rl}p+pq& =q+qr\\ qr& =p+pq-q\\ r& =\frac{p+pq-q}{q}\end{array}$$

So we can pick $0\le p\le 1$, $0\le q\le 1,q\ne p$, such that $0\le r\le 1,r\ne p,r\ne q$. There could be strategies to pick out those values systematically, but for achievability we only need one example, so through trial and error we can have:

$$\begin{array}{rl}p& =0.5\\ q& =0.4\\ r& =\frac{p+pq-q}{q}\\ & =\frac{0.5+0.2-0.4}{0.4}\\ & =\frac{0.3}{0.4}\\ \\ & =0.75\end{array}$$

Let's verify those profiles yield the same expected score:

$$\begin{array}{rl}E(p,q,r)& =p+pq+pqr\\ & =0.5+(0.5)(0.4)+(0.5)(0.4)(0.75)\\ & =0.5+0.2+0.15\\ & =0.85\\ E(q,r,p)& =q+qr+qrp\\ & =0.4+(0.4)(0.75)+(0.4)(0.75)(0.5)\\ & =0.4+0.3+0.15\\ & =0.85\\ \\ E(p,q,r)& =E(q,r,p)\end{array}$$

Hence achievability part is proven!

Converse

Now we have to show that the two profiles answer is the tightest upper bound. To do this, we show that it is not possible to have three distinct profiles, as by our lemma we showed it is not possible to have more than three. Assuming, by contradiction, that three distinct profiles is possible, then by using our lemma from earliear this can only result from:

$$\begin{array}{rl}E(p,q,r)& =E(q,r,p)=E(r,p,q)\\ p+pq& =q+qr=r+pr\end{array}$$

Re-arranging for $r$, we get $$r=\frac{p+pq}{1+p}$$

We now substitute the above equation into $p+pq=q+qr$, and with a few algebraic re-arrangements:

$$\begin{array}{rl}p+pq& =q+qr\\ (1+p)(p+pq)& =q(1+2p+pq)\\ p+p^2+pq+p^{2}q& =q+2pq+p{q}^2\\ p+p^2-pq+p^{2}q-p{q}^2-q& =0\\ (p-q)(1+p+pq)& =0\end{array}$$

Since it is given that $p\ne q$, we must have $1+p+pq=0$. This means $q=-\frac{1+p}{p}$. As $0\le p\le 1$, this means $q<0$, a clear impossibility as it is given that $0\le q\le 1$. Hence the converse is proven.

Therefore, two distinct profiles is the tightest upper bound. Q.E.D.