Sure, but I think Hot Ones would prefer you buy all 10

You have been invited on as a guest and want to prepare for the show. However, you don’t feel like purchasing all 10 sauces in advance. Your plan is to purchase fewer sauces, and then to combine sauces together for any you are missing. For example, if you are missing sauce #7, then you can instead simultaneously consume sauces #3 and #4, since $3 + 4 = 7$. (I know the spiciness of the sauces isn’t linear, but for the purposes of this puzzle, let’s assume it is.)

After some pencil-and-paper scratch work, you realize you only need four spices. For example, here’s how you can generate all the values from 1 to 10 using the spices #1, #2, #3, and #4 at most once:

• $1=1$
• $2=2$
• $3=3$
• $4=4$
• $5=1+4$
• $6=2+4$
• $7=3+4$
• $8=1+3+4$
• $9=2+3+4$
• $10=1+2+3+4$

Including this particular set of four spices (i.e., #1 through #4), for how many sets of four spice numbers is it possible to generate all the numbers from 1 to 10 using each spice at most once?

The brute force way of doing this would be to check all $\binom{10}{4} = 210$ sets of four spice numbers with no repeats and check whether they are able to generate all the numbers from $1$ to $10$ while using each spice at most one. But that seems tiresome, so let's instead try to work on the general solution that we will use for the Extra Credit problem.

Let $[n] := \{ 1, 2, \dots, n \} \subset \mathbb{N}.$ For some subset $A \subseteq \mathbb{N},$ let's define the $$\Sigma(A) = \{ n \in \mathbb{N} \mid \exists ! i_1, \dots, i_k \in A, n = i_1 + \dots + i_k \},$$ whereas we will let $\sigma(A) = \sum_{a \in A} a$ and $M(A) = \max \{ a \in A \}.$ Let's define $$\mathcal{B} = \{ A \subseteq \mathbb{N} \mid [\sigma(A)] = \Sigma(A) \}.$$ For the Classic problem, we then want to find all $A \in \mathcal{B}$ with $|A| = 4$ and $\sigma(A) \geq 10.$

It will help to have one additional lemma of sorts: If $A \in \mathcal{B},$ then, for any $\ell \in \{ M(A)+1, \dots, \sigma(A) + 1 \},$ $A \cup \{\ell\} \in \mathcal{B}$ as well. To prove this point, since obviously $[\sigma(A)] = \Sigma(A) \subset \Sigma(A \cup \{\ell\}),$ we need to show that $$[\sigma(A) + \ell] \setminus [\sigma(A)] = \{\sigma(A)+1, \dots, \sigma(A)+\ell \} \subset \Sigma(A \cup \{\ell\}).$$ Let's take any $q \in [\sigma(A) + \ell] \setminus [\sigma(A)],$ then $q - \ell \lt \sigma(A),$ so $\exists ! i_1, i_2, \dots i_k \in A$ such that $i_1 + \dots + i_k = q - \ell,$ for some $k = 1, \dots, |A|.$ Furthermore, $\ell \gt M(A),$ so $i_j \ne q,$ for $j = 1, \dots, k,$ so $i_1, i_2, \dots, i_k,$ and $\ell$ are all unique elements of $A \cup \{\ell\}$ and sum to $q.$ Thus, $[\sigma(A) + \ell] \setminus [\sigma(A)] \subset \Sigma(A \cup \{\ell\}),$ therefore if $A \in \mathcal{B}$ then $A \cup \{\ell\} \in \mathcal{B}$ as well for $\ell \in \{M(A) + 1, \dots, \sigma(A) + 1 \}.$

Now let's start by indentifying all $A \in \mathcal{B}$ with $|A| = 4,$ well, er, ... I guess let's start a bit before that. Clearly, $\{1\} \in \mathcal{B}.$ From our lemma, we then see that $\{1,2\} \in \mathcal{B}.$ Using the lemma again gives the two members of $\mathcal{B}$ with cardinality of $3,$ that is $\{1,2,3\}$ and $\{1,2,4\}.$ Therefore, with one additional application of the lemma, we get \begin{align*}\mathcal{B}_4 = \{ A \in \mathcal{B} \mid |A| = 4 \} &= \Biggl\{ \{1,2,3,4\}, \{1,2,3,5\}, \{1,2,3,6\}, \{1,2,3,7\},\Biggr. \\ &\quad\quad\quad \Biggl. \{1,2,4,5\}, \{1,2,4,6\},\{1,2,4,7\}, \{1,2,4,8\}\Biggr\}.\end{align*} Since by inspection $\sigma(A) \geq 10$ for each $A \in \mathcal{B}_4,$ we conclude that there are 8 sets of four spice numbers where all the numbers from 1 to 10 can be generated using each spice at most once.

An N person class?! Now that's really deranged!

If there are $N$ students in the class, where $N$ is some large number, how likely is it that they form a single loop that includes the entire class, in terms of $N$?

In the classic problem, we actually went through the various types of assignments that could work, but even there we showed roughly enough to know that the number of single loop gift exchange assignments is $(N-1)!$. The somewhat trickier piece is for the total number of all acceptable gift exchange assignments. Here, if you were paying attention to the titles of these posts, you might have been alerted to the fact that an acceptable gift exchange assignment, that is a permutation where no number is assigned to itself, is called a derangement. The number of derangements of $N$ elements is sometimes called the sub-factorial, denoted $!N,$ which can be shown to have the formula $$!N = N!\sum_{i=0}^N \frac{(-1)^i}{i!}.$$ Thus, we see that the probability of having a single loop for some large number $N$ students is $$\frac{(N-1)!}{N! \sum_{i=0}^N \frac{(-1)^i}{i!} } \approx \frac{e}{N}$$ as $N \to \infty,$ since $\sum_{i=0}^\infty \frac{(-1)^i}{i!} = e^{-1}.$

You know, it honestly takes a smidgeon of the fun out of knowing the solution, rather than doing much work for it. So, I dunno, maybe let's derive the formula $$!N = N! \sum_{i=0}^N \frac{(-1)^i}{i!}.$$ Let $U$ be the set of all inappropriate gift exchange assignments, that is where someone receives themselves as the giftee. In particular, for each $k = 1, \dots, N,$ let $U_k$ be the set of assignments where $k$ receives herself as the giftee. Using the inclusion/exclusion principle, we have \begin{align*}|U| = |U_1 \cup U_2 \cup \dots \cup U_N| &= \sum_{i=1}^N |U_i| -\sum_{1 \leq i \lt j \leq N} |U_i \cap U_j| \\ &\quad\quad\quad+ \sum_{1 \leq i \lt j \lt k \leq N} |U_i \cap U_j \cap U_k | + \dots \\ &\quad\quad\quad\quad+ (-1)^{N+1} |U_1 \cap U_2 \cap \dots \cap U_N|.\end{align*} Now the $k$ summand in the above formula, e.g., $|U_{i_1} \cap U_{i_1} \cap \dots \cap U_{i_k}|$ counts all permutations where $i_1 \lt i_2 \lt \dots \lt i_k$ receive themselves as giftees leaving the other $N-k$ students free to explore all possible permutations. Therefore, we have $|U_{i_1} \cap \dots \cap U_{i_k}| = (N-k)!$, which is coupled with the fact that there are $\binom{N}{k}$ ways of choosing $i_1, \dots, i_k.$ So we have \begin{align*}|U| &= \binom{N}{1} (N-1)! - \binom{N}{2} (N-2)! + \dots + (-1)^{N+1} 0! \\ &= \sum_{i=1}^N (-1)^{i+1} \binom{N}{i} (N-i)! \\ &= N! \sum_{i=1}^N \frac{(-1)^{i+1}}{i!}.\end{align*} Since the number of total derangements would be the number of all permutations minus the non-derangements, we get $$!N = N! - |U| = N! \left(1 - \sum_{i=1}^N \frac{(-1)^{i+1}}{i!}\right) = N! \left( 1 + \sum_{i=1}^N \frac{(-1)^i}{i!} \right) = N! \sum_{i=0}^N \frac{(-1)^i}{i!}.$$ Pretty neat, if not anticlimactic, I guess.

A five person class! That's deranged!

You are participating in a holiday gift exchange with your classmates. You each write down your own name on a slip of paper and fold it up. Then, all the students place their names into a single hat. Next, students pull a random name from the hat, one at a time. If at any point someone pulls their own name from the hat, the whole class starts over, with everyone returning the names to the hat.

Once the whole process is complete, each student purchases a gift for the classmate whose name they pulled. Gifts are handed out at a big holiday party at the end of the year. At this party, you observe that there are “loops” of gift-giving within the class. For example, student A might have gotten a gift for B, who got a gift for C, who got a gift for D, who got a gift for A. In this case, A, B, C and D would form a loop of length four. Another way to have a loop of length four is if student A got a gift for C, who got a gift for B, who got a gift for D, who got a gift for A. And of course, there are other ways.

If there are a total of five students in the class, how likely is it that they form a single loop that includes the entire class?

For this particular case, we can just go through all of the possible outcomes. Let's first enumerate all of the possible single loops for all five students. Let's use the permutation notation $(ABCDE)$ stand for the loop where $A \to B \to C \to D \to E \to A.$ It doesn't necessarily matter where on this loop you start, so let's always assume that we start with $A$. Then we are free to choose any ordering of $B,$ $C$, $D,$ and $E$ to form a loops, so there are $4! = 24$ such loops.

On the other hand, since we cannot ever have a one person loop, where a person receives themselves as the giftee, the only other acceptable assingments for our gift exchange is that there is a combination of a 2-loop and a 3-loop. Here we see that there are $\binom{5}{2} = 10$ ways of choosing the two students to be in the 2-loop. By analogy, we see that there are $2 = (3-1)!$ ways of arranging the remaining members of our 3-loop. So there are a total of $\binom{5}{2} 2! = 20$ other acceptable gift exchange assignments.

Therefore, the probability that all of the giftees form a single loop when there are $5$ students is $$\frac{24}{24+20} = \frac{6}{11} = 54.5454545454\dots\%.$$

Millions of Peaches, Peaches for Free, .... or at least for Extra Credit

As before, your assistant intends to pick two random points along the circumference of the garden and run a hose straight between them.

This time, you’ve decided to contribute to the madness yourself by picking a random point inside the garden to plant a second peach tree. On average, how far can you expect this point to be from the nearest part of the hose?

Here let's again assume that the assistant's hose starts at $(1,0)$ and ends at $(\cos \theta, \sin \theta)$ for some uniform random $\theta \sim U(0,2\pi).$ Here however, we assume that there is a peach tree at the random point $(u,v)$ for some uniformly random point with $$f(u,v) = \frac{1}{\pi}\chi ( u^2+v^2 \leq 1 ).$$ We can again use the line $y = x\tan \frac{\theta}{2},$ which is the perpendicular bisector of the chord traces by the hose, or more precisely we will use parallel copies of it. Let's subdivide the circular area into three different spaces: \begin{align*}\Omega_1(\theta) &= \left\{ (u,v) \mid u^2 + v^2 \leq 1, \left| v - u \tan \frac{\theta}{2} \right| \leq \tan \frac{\theta}{2} \right\} \\ \Omega_2(\theta) &= \left\{ (u,v) \mid u^2 + v^2 \leq 1, v - u\tan \frac{\theta}{2} \geq \tan \frac{\theta}{2} \right\} \\ \Omega_3(\theta) &= \left\{ (u,v) \mid u^2 + v^2 \leq 1, v - u \tan \frac{\theta}{2} \leq - \tan \frac{\theta}{2} \right\}.\end{align*}

In general we get \begin{align*}\lambda^*(u,v,\theta) & = \arg\max \left\{ \| ( 1 - \lambda + \lambda \cos \theta - u, \lambda \sin \theta - v ) \|_2 \mid 0 \leq \lambda \leq 1 \right\} \\ &= \arg \max \left\{ (1-u)^2 + v^2 - 2 \lambda ( (1-u)(1-\cos \theta) + v\sin \theta ) \right.\\&\quad\quad\quad\quad\quad\quad\quad\quad \left.+ 2(1-\cos \theta) \lambda^2 \mid 0 \leq \lambda \leq 1 \right\} \\ &= \begin{cases} \frac{1}{2} \left( 1 - u + v \cot \frac{\theta}{2} \right), & \text{ if $(u,v) \in \Omega_1(\theta)$; } \\ 0, &\text{ if $(u,v) \in \Omega_2(\theta)$;} \\ 1, &\text{if $(u,v) \in \Omega_3(\theta).$}\end{cases}\end{align*} Plugging this back in we see that we have $$d(u,v,\theta) = \begin{cases} \frac{ \left| (1-u) \sin \theta + v(1-\cos \theta) \right| }{ 2 \sin \frac{\theta}{2} }, &\text{if $(u,v) \in \Omega_1(\theta)$;} \\ \sqrt{ (u-1)^2 + v^2 }, &\text{if $(u,v) \in \Omega_2(\theta)$;} \\ \sqrt{ (u- \cos \theta)^2 + (v - \sin \theta)^2 }, &\text{ if $(u,v) \in \Omega_3(\theta).$}\end{cases}$$

Let's first attack $A(\theta) = \iint_{\Omega_1(\theta)} d(u,v,\theta) f(u,v) \,du\,dv.$ We see that, so long as $0 \leq \theta \leq \pi$, by rotating the plane counterclockwise about the origin by $\frac{\pi-\theta}{2}$ that the chord is now parallel to the $x$-axis and stretches from $(-\sin \frac{\theta}{2}, \cos \frac{\theta}{2})$ to $(\sin \frac{\theta}{2}, \cos \frac{\theta}{2}).$ In this case, it is clear to see that the distance $d(u^\prime, v^\prime, \theta) = |v^\prime - \cos \frac{\theta}{2}|.$ So the transformed integral is \begin{align*}A(\theta) &= \frac{2}{\pi} \int_0^{\sin \theta/2} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left|y - \cos \frac{\theta}{2}\right| \,dy \,dx \\ &= \frac{2}{\pi} \int_0^{\sin \theta/2} \left( \int_{-\sqrt{1-x^2}}^{\cos \theta/2} \left(\cos \frac{\theta}{2} - y \right)\,dy + \int_{\cos \theta/2}^{\sqrt{1-x^2}} \left( y - \cos \frac{\theta}{2} \right) \,dy \right) \,dx \\ &= \frac{2}{\pi} \int_0^{\sin \theta/2} \left(1 + \cos^2 \frac{\theta}{2} - x^2\right) \,dx \\ &= \frac{2}{\pi} \sin \frac{\theta}{2} \left( 1 + \cos^2 \frac{\theta}{2} \right) - \frac{2}{3\pi} \sin^3 \frac{\theta}{2} \\ &= \frac{4}{3\pi} \sin \frac{\theta}{2} \left( 1 + 2 \cos^2 \frac{\theta}{2} \right).\end{align*}

We see from symmetry that $\iint_{\Omega_2(\theta)} d(u,v,\theta) f(u,v)\,du\,dv = \iint_{\Omega_3(\theta)} d(u,v,\theta) f(u,v) \,du\,dv,$ so let's let \begin{align*}B(\theta) = \frac{2}{\pi} \iint_{\Omega_2(\theta)} d(u,v,\theta)\,du\,dv &= \frac{2}{\pi} \int_{-\cos \theta}^1 \int_{-\sqrt{1-x^2}}^{\tan \frac{\theta}{2} (x - 1)} \sqrt{ (x-1)^2 + y^2 }\,dy\,dx \\& = \frac{2}{\pi} \int_{\pi + \theta/2}^{3\pi/2} \int_0^{-2\cos \phi} \rho^2 \,d\rho \,d\phi,\end{align*} where the final change of variables represents the transformation $(u,v)$ into $(1 + \rho \cos \phi, \rho \sin \phi)$ for $\phi \in ( \pi + \theta/2, 3\pi/2)$ and $0 \leq \rho \leq -2\cos \phi.$ Therefore, we see that $$B(\theta) = \frac{2}{\pi} \int_{\pi + \theta/2}^{3\pi/2} \left(-\frac{8}{3} \cos^3 \phi\right) \,d\phi = \frac{2}{\pi} \left(\frac{16}{9} - \frac{8}{3} \sin \frac{\theta}{2} + \frac{8}{9} \sin^3 \frac{\theta}{2}\right).$$

Putting these two together we see that the average distance from the random point $(u,v)$ to the chord from $(1,0)$ to $(\cos \theta, \sin \theta)$ is given by $d(\theta) = A(\theta) + B(\theta).$ Therefore, through symmetry, we see that the average distance is given by $$\hat{d} = \frac{1}{\pi} \int_0^\pi d(\theta) \,d\theta = \frac{1}{\pi} \int_0^\pi A(\theta) \,d\theta + \frac{1}{\pi} \int_0^\pi B(\theta)\, d\theta.$$ We have \begin{align*}\frac{1}{\pi} \int_0^\pi A(\theta) \,d\theta &= \frac{4}{3\pi^2} \int_0^\pi \sin \frac{\theta}{2} \left( 1 + 2 \cos^2 \frac{\theta}{2} \right)\,d\theta \\ &= \frac{8}{3\pi^2} \left.\left( -\cos \frac{\theta}{2} - \frac{2}{3} \cos^3 \frac{\theta}{2} \right)\right|_0^\pi = \frac{40}{9\pi^2}.\end{align*} Similarly, we have \begin{align*}\frac{1}{\pi} \int_0^\pi B(\theta) \,d\theta &= \frac{2}{\pi^2} \int_0^\pi \left(\frac{16}{9} - \frac{8}{3} \sin \frac{\theta}{2} + \frac{8}{9}\sin^3 \frac{\theta}{2} \right) \,d\theta \\ &= \frac{2}{\pi^2} \left.\left( \frac{16}{9} \theta + \frac{32}{9} \cos \frac{\theta}{2} + \frac{16}{27} \cos^3 \frac{\theta}{2} \right)\right|_0^\pi = \frac{96\pi - 224}{27\pi^2}.\end{align*} Combining these together gives the average distance between the randomly place hose and your randomly placed peach tree at a slightly larger $$\hat{d} = \frac{40}{9\pi^2} + \frac{96\pi - 224}{27\pi^2} = \frac{96\pi - 104}{27\pi^2} \approx 0.741494295364\dots$$ furlongs.

Peaches, peaches, peaches, peaches, peaches … I likely won’t water you

You and your assistant are planning to irrigate a vast circular garden, which has a radius of 1 furlong. However, your assistant is somewhat lackadaisical when it comes to gardening. Their plan is to pick two random points on the circumference of the garden and run a hose straight between them.

You’re concerned that different parts of your garden—especially your prized peach tree at the very center—will be too far from the hose to be properly irrigated.

On average, how far can you expect the center of the garden to be from the nearest part of the hose?

Without loss of generality, let's assume that we choose a coordinate system such that one end of the hose is located at the point $(1,0)$ and the other end is located at $(\cos \theta, \sin\theta)$ for some uniformly random $\theta \sim U(0,2\pi).$ The minimal distance from the origin, where the prized peach tree is, and the chord traced by the hose occurs along the ray that perpendicularly bisects the chord. We can arrive here either by calculus and finding $$\lambda^* = \arg\max \{ \| (1-\lambda + \lambda \cos \theta, \lambda \sin \theta ) \|_2 \mid 0\leq \lambda \leq 1 \} = \frac{1}{2}$$ or by spatial reasoning about Lagrange multipliers and optimizers occurring when contours and contraints are normal to one another or by any other means, I suppose. Whichever Feynman-esque path we take to arrive at the perpendicular bisector, we then see through some trigonometry that this minimal distance is this $$d(\theta) = \left| \cos \frac{\theta}{2}\right|,$$ as a function of the random $\theta.$

So, the average distance between the randomly placed hose and your precious peach tree is \begin{align*}\bar{d} &= \frac{1}{2\pi} \int_0^{2\pi} d(\theta) d\theta \\ &= \frac{1}{\pi} \int_0^\pi \cos \frac{\theta}{2} \,d\theta \\ &= \frac{2 \sin \frac{\pi}{2} - 2 \sin 0}{\pi} = \frac{2}{\pi}\approx 0.636519772\dots\end{align*} furlongs, which is about 420.3 feet. This doesn't seem too terribly bad, but given that the spread of an average peach tree is only about 20 feet (according to a quick Googling), your assistant's method is not expected to provide a large amount of water to your peaches.

Even more so, seems like an actual uniform generator would be simpler…

Randy has an updated suggestion for how the button should behave at door 2. What hasn’t changed is that if a contestant at door 2 and moves to an adjacent door, that new door will be 1 or 3 with equal probability.

But this time, on the first, third, fifth, and other odd button presses that happen to be at door 2, there’s a 20 percent the contestant remains at door 2. On the second, fourth, sixth, and other even button presses that happen to be at door 2, there’s a 50 percent chance the contest remains at door 2.

Meanwhile, the button’s behavior at doors 1 and 3 should in no way depend on the number of times the button has been pressed.

As the producer, you want the chances of winding up at each of the three doors—after a large even number of button presses— to be nearly equal. If a contestant presses the button while at door 1 (or door 3), what should the probability be that they remain at that door?

In this case, let $q$ be the probability of remaining at door 1 (or at door 3), then we can treat the two different behaviors at door 2 sequentially in order to come with the two-step transition matrix \begin{align*}Q & = \begin{pmatrix} q & 1-q & 0 \\ 0.4 & 0.2 & 0.4 \\ 0 & 1-q & q \end{pmatrix} \begin{pmatrix} q & 1-q & 0 \\ 0.25 & 0.5 & 0.25 \\ 0 & 1-q & q \end{pmatrix} \\ & = \begin{pmatrix} q^2 -\frac{1}{4}q +\frac{1}{4} & -q^2 + \frac{1}{2} q +\frac{1}{2} & -\frac{1}{4}q + \frac{1}{4}\\ \frac{2}{5}q + \frac{1}{20} & -\frac{4}{5}q + \frac{9}{10} & \frac{2}{5}q + \frac{1}{20}\\ -\frac{1}{4}q + \frac{1}{4} & -q^2 + \frac{1}{2} q + \frac{1}{2} & q^2 -\frac{1}{4}q + \frac{1}{4}\end{pmatrix}.\end{align*}

We will lean upon our own great (?) shoulders from the Classic problem to show that we need to solve for $q$ that makes the transition matrix symmetric. In this case, that requirement yields $$\frac{2}{5}q + \frac{1}{20} = -q^2 + \frac{1}{2} q + \frac{1}{2},$$ or equivalently, $$q^2 -\frac{1}{10} q - \frac{9}{20} = 0.$$ Solving this quadratic for the positive root (since after all we need $q\in[0,1]$ as a probability), gives that the appropriate probability to remain at door 1 in this even more complicated Markov scheme is $$q=\frac{ \frac{1}{10} + \sqrt{ \frac{1}{100} + 4 \frac{9}{20} } }{2} = \frac{1+\sqrt{181}}{20} \approx 0.722681202354\dots$$

Seems like an actual uniform generator would be simpler…

You are a producer on a game show hosted by Randy “Random” Hall (no relation to Monty Hall). The show has three doors labeled 1 through 3 from left to right, and behind them are various prizes.

Contestants pick one of the three doors at which to start, and then they press an electronic button many, many times in rapid succession. Each time they press the button, they either stay at their current door or move to an adjacent door. If they’re at door 2 and move to an adjacent door, that new door will be 1 or 3 with equal probability.

Randy has decided that when a contestant presses the button while at door 2, there should be a 20 percent chance they remain at door 2.

As the producer, you want the chances of a contestant ultimately winding up at each of the three doors to be nearly equal after many button presses. Otherwise, mathematicians will no doubt write you nasty letters complaining about how your show is rigged.

If a contestant presses the button while at door 1 (or door 3), what should the probability be that they remain at that door?

Firstly, so true ... if there were a meaningful bias in the supposedly uniformly random process of selecting which door, I would take notice. Though rather than calling to complain I would be more likely to try to get on the show to exploit this bias, but I guess potato-potato.

Moving on, per Randy's specifications and your desire for the appearance of uniform randomness, we have a three state Markov chain setup where the transition matrix $$P= \begin{pmatrix} p & 1-p & 0 \\ 0.4 & 0.2 & 0.4 \\ 0 & 1-p & p \end{pmatrix}$$ and we are wondering for which values of $p$ will $P^n \to U,$ as $n\to \infty$, where the limiting matrix $U$ is the $3\times 3$ matrix where each entry is $\frac{1}{3}.$

There are many ways to solve for $p$ here. For instance, though we would obviously start with some specific position, e.g., $\pi_0 = (1,0,0)$, if $P^n \to U,$ as $n\to \infty$, then we would necessarily need to have $u=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ satisfy $uP=u$. We could obviously just solve here and get three linear equations (that hopefully collapse well since there is only one unknown), but instead let's math out!

Since $P$ is a transition matrix its rows sum to one, so we see that $Pu^T=u^T,$ which if we combine with $uP=u$ implies the $$Pu^T= u^T = (uP)^T = P^Tu^T.$$ So $P$ must be symmetric $(P=P^T)$, which leaves the much easier linear equation $1-p=0.4,$ that is, in order to provide the illusion of a uniform distibution of the door's landing spot through this elaborate Markov scheme you must have the probability of remaining on door 1 when the button is pressed be $p=60\%.$

Of course, armed with the knowledge that we have a symmetric transition matrix, we can then justify that this works by using the Spectral Theorem. We have already seen that $\lambda=1$ and $u^T$ is an eigenpair. We could certainly additionally calculate the other two eigenpairs as well, or simply argue that the eigenvalues must have absolute value less than $1$ and that the eigenvectors can be chosen to be an orthonormal basis of $\mathbb{R}^3$, such that $P=VDV^T,$ where $D = diag(1, \lambda_2, \lambda_3)$ and $$V= \begin{pmatrix} \sqrt{3}/3 & v_{12} & v_{13} \\ \sqrt{3}/3 & v_{22} & v_{23} \\ \sqrt{3}/3 & v_{32} & v_{33} \end{pmatrix}$$ satisfies $V^TV=VV^T=I.$ Since this $V$ represents an orthonormal basis we can change bases and represent any initial $\pi_0$ in $V$-coordinates, where the first coordinate is also $\sqrt{3}/3$ whenever $\pi_0$ sums to $1.$ Let's generically assume that $\pi_0 v_2 = c_2 \in \mathbb{R}$ and $\pi_0 v_3 = c_3 \in \mathbb{R}.$ Then we see that \begin{align*}\pi_n = \pi_{n-1} P &= \pi_{n-1} VDV^T \\ &= \cdots = \pi_0 V D^n V^T \\ &= u + c_2 \lambda_2^n v_2^T + c_3 \lambda_3^n v_3^T\end{align*} Since $|\lambda_2|, |\lambda_2| \lt 1,$ we see that for any value of $\pi_0$ we get $\pi_n \to u,$ as desired.

Extra Credit swinging the probabilities, or ... Hey, how'd the Catalan numbers show up here???

Instead of a best-of-seven series, now suppose the series is much, much longer. In particular, the first team to win $N$ games wins the series, so technically this is a best-of-($2N−1$) series, where $N$ is some very, very large number.

In the limit of large $N$, what is the probability swing for Game 1 in terms of $N$?

Applying the same logic used in the Classic Fiddler problem, we want to first find $p_{1,N} = \mathbb{P} \{ \text{win best of (2N-1) series} \mid \text{win game 1} \},$ from which we get the probability swing of game 1 in a best of $(2N-1)$ series as $\Delta_N = 2p_{1,N} - 1.$ Again following in the Classic Fiddler's solution's footsteps, if you win games $1$ and $k$, then there are $\binom{k-2}{N-2}$ ways of arranging another $N-2$ wins in the other $k-2$ games, so $$p_{1,k,N} = \mathbb{P} \{ \text{winning a best of $(2N-1)$ series in $k$ games} \mid \text{win game 1} \} = \binom{k-2}{N-2} \frac{1}{2^{k-1}}.$$ Summing over all possible values of $k = N, N+1, \dots, 2N-1,$ we get $$p_{1,N} = \sum_{k=N}^{2N-1} \binom{k-2}{N-2} \frac{1}{2^{k-1}}.$$ We could try to go further and define some generating function f_N, but this would lead to some escalating number of derivatives, that gets messy fast.

Instead let's set up a recursive formula. We note that $p_{1,1} = 1,$ which makes sense since it is a winner-takes-all one game playoff. For some $N \geq 1,$ let's take a look at $$p_{N+1} = \sum_{k=N+1}^{2N+1} \binom{k-2}{N-1} \frac{1}{2^{k-1}}.$$ The standard binomial coefficient recursion formula (which comes from the Pascal triangle) gives $$\binom{k-2}{N-1} = \binom{k-3}{N-1} + \binom{k-3}{N-2},$$ so we have \begin{align*} p_{N+1} & = \sum_{k=N+1}^{2N+1} \left(\binom{k-3}{N-1} + \binom{k-3}{N-2} \right) \frac{1}{2^{k-1}} \\ &= \left( \sum_{k=N}^{2N} \binom{k-2}{N-1} \frac{1}{2^k} \right) + \left( \sum_{k=N}^{2N} \binom{k-2}{N-2} \frac{1}{2^k} \right) \\ &= \frac{1}{2} \left( \sum_{k=N+1}^{2N} \binom{k-2}{N-1} \frac{1}{2^{k-1}} \right) + \frac{1}{2} \left( \sum_{k=N}^{2N-1} \binom{k-2}{N-2} \frac{1}{2^{k-1}} \right) + \binom{2N-2}{N-2} \frac{1}{2^{2N}} \\ &= \frac{1}{2} p_{1,N+1} - \binom{2N-1}{N-1} \frac{1}{2^{2N+1}} + \frac{1}{2} p_{1,N} + \binom{2N-2}{N-2} \frac{1}{2^{2N}}.\end{align*} Pulling the copy of $\frac{1}{2}p_{1,N+1}$ back onto the lefthand side and then multiplying by 2, we get the recursion formula \begin{align*} p_{1,N+1} &= p_{1,N} + \binom{2N-2}{N-2} \frac{1}{2^{2N-1}} - \binom{2N-1}{N-1} \frac{1}{2^{2N}} \\ &= p_{1,N} + \binom{2N-2}{N-2} \frac{1}{2^{2N-1}} - \left( \binom{2N-2}{N-1} + \binom{2N-2}{N-2} \right) \frac{1}{2^{2N}} \\ &= p_{1,N} - \frac{1}{4^N} \left( \binom{2N-2}{N-1} - \binom{2N-2}{N-2} \right) \\ &= p_{1,N} - \frac{1}{4^N} C_{N-1}, \end{align*} where $C_n = \frac{1}{n+1} \binom{2n}{n} = \binom{2n}{n} - \binom{2n}{n+1},$ for $n \in \mathbb{N}$ is the standard Catalan number.

Since we start with $p_{1,1} = 1,$ we then see that $$p_{1,N} = 1 - \sum_{k=1}^{N-1} \frac{C_{k-1}}{4^k} = 1 - \frac{1}{4} \sum_{k=0}^{N-2} \frac{C_k}{4^k}, \,\, \forall N \in \mathbb{N}.$$ We can rely on the fact that the generation function of the Catalan numbers is $$c(x) = \sum_{n=0}^\infty C_n x^n = \frac{1 - \sqrt{1-4x}}{2x},$$ so that $$\frac{1}{4} \sum_{k=0}^\infty \frac{C_k}{4^k} = \frac{1}{4} c(\frac{1}{4}) = \frac{1}{4} \frac{1 - \sqrt{1 - 4 \cdot \frac{1}{4}}}{2 \cdot \frac{1}{4}} = \frac{1}{2}.$$ Therefore, we see that $$p_{1,N} = 1 - \frac{1}{4} \sum_{k=0}^{N-2} \frac{C_k}{4^k} = 1 - \frac{1}{4} \sum_{k=0}^\infty \frac{C_k}{4^k} + \frac{1}{4}\sum_{k=N-1}^\infty \frac{C_k}{4^k} = \frac{1}{2} + \frac{1}{4} \sum_{k=N-1}^\infty \frac{C_k}{4^k},$$ for all $N \in \mathbb{N}.$ Now when $k$ is large we have $$C_k \sim \frac{4^k}{k^{3/2} \sqrt{\pi}},$$ from repeated applications of Stirling's approximation, so when $N$ is sufficiently large we have $$\frac{1}{4} \sum_{k=N-1}^\infty \frac{C_k}{4^k} \approx \frac{1}{4\sqrt{\pi}} \sum_{k=N-1}^\infty k^{-3/2} \approx \frac{1}{2\sqrt{\pi (N-1)}},$$ where the last approximation is due to the fact that $\sum_{k=N}^\infty k^{-p} \sim \int_N^\infty x^{-p} \,dx.$ Therefore, in a fairly concise way, we have $$p_{1,N} \approx \frac{1}{2} + \frac{1}{2\sqrt{\pi(N-1)}},$$ when $N$ is large, so the probability swing of winning the first game is $$\Delta_N = 2p_{1,N} -1 \approx \frac{1}{\sqrt{\pi(N-1)}}$$ when $N$ is large.

Swinging the probabilities

You and your opponent are beginning a best-of-seven series, meaning the first team to win four games wins the series. Both teams are evenly matched, meaning each team has a 50 percent chance of winning each game, independent of the outcomes of previous games.

As the team manager, you are trying to motivate your team as to the criticality of the first game in the series (i.e., “Game 1”). You’d specifically like to educate them regarding the “probability swing” coming out of Game 1—that is, the probability of winning the series if they win Game 1 minus the probability of winning the series if they lose Game 1. (For example, the probability swing for a winner-take-all Game 7 is 100 percent.)

What is the probability swing for Game 1?

Let's break this down as follows. Let $p_1 = \mathbb{P} \{ \text{win series} \mid \text{win game 1} \}.$ In order to win the series, you must win it in $k$ games for some $k=4, 5, 6, 7,$ so let's further let $$p_{1,k} = \mathbb{P} \{ \text{win series in $k$ games} \mid \text{win game 1} \},$$ where here we see that $p_1 = \sum_{k=4}^7 p_{1,k}.$ Now, the total number of ways to win the first and $k$th games and two others somewhere in games $2$ through $k-1$ is given by $\binom{k-2}{2}$ and the overall probability of any particular combination of $k$ games is $\frac{1}{2^k},$ so $$p_{1,k} = \frac{\mathbb{P} \{ \text{win series in $k$ games and win game $1$} \}}{\mathbb{P} \{ \text{win game 1 } \}} = \binom{k-2}{2} \frac{1}{2^{k-1}}.$$ Therfore, $$p_1 = \sum_{k=4}^7 \binom{k-2}{2} \frac{1}{2^{k-1}} = \frac{1}{2} \sum_{k=4}^7 \binom{k-2}{2} \frac{1}{2^{k-2}} = \frac{1}{2} \sum_{k=2}^5 \binom{k}{2} \frac{1}{2^k}.$$

Now one way of computing $p_1$ would be using some generating function wizardry. Define the function $f(x) = \sum_{k=2}^5 \binom{k}{2} x^k,$ in which case, $p_1 = \frac{1}{2} f(\frac{1}{2}).$ Now we also see that \begin{align*} f(x) &= \frac{1}{2} x^2 \sum_{k=2}^5 k(k-1) x^{k-2} \\ &= \frac{1}{2} x^2 \frac{d^2}{dx^2} \left( \frac{1-x^6}{1-x} \right) \\ &= \frac{1}{2} x^2 \frac{d}{dx} \left( \frac{1-6x^5+5x^6}{(1-x)^2} \right) \\ &= \frac{1}{2} x^2 \frac{2(1-15x^4+24x^5-10x^6}{(1-x)^3} \\ &= \frac{ x^2 ( 1 - 15 x^4 + 24x^5 -10x^6) }{(1-x)^3}.\end{align*} So we have $$p_1 = \frac{1}{2} f(\frac{1}{2}) = \frac{1}{2} \frac{ \frac{1}{4} \left( 1 - \frac{15}{16} + \frac{24}{32} - \frac{10}{64} \right) }{ \frac{1}{8} } = \frac{ 42}{64} = \frac{21}{32}.$$

Now from symmetry, we see that the probability of you winning having lost the first game, let's say $q_1 = \mathbb{P} \{ \text{win series} \mid \text{lose game 1} \}$ is the same as the probability of you winning the series having lost the series having won the first game. That is $q_1 = 1 - p_1.$ So the proabbility swing of the first game is $$\Delta = p_1 - q_1 = p_1 - (1- p_1) = 2p_1 - 1 = 2 \frac{21}{32} - 1 = \frac{5}{16} = 32.125\%.$$