Riskier Risk, Now with Wilds!

The full deck of Risk cards also contains two wildcards, which can be used as any of the three types of cards (infantry, cavalry, and artillery) upon trading them in. Thus, the full deck consists of $44$ cards.

You must have at least three cards to have any shot at trading them in. Meanwhile, having five cards guarantees that you have three you can trade in.

If you are randomly dealt cards from a complete deck of $44$ one at a time, how many cards would you need, on average, until you can trade in three? (Your answer should be somewhere between three and five. And no, it’s not four.)

Let $S$ be the random number of cards you have when you can first turn in a set of three cards. Obviously we have $3\leq S\leq 5,$ so lets go about calculating $p_s = \mathbb{P} \{ S= s\},$ for $s \in \{3, 4, 5\},$ which we can then compute $$\mathbb{E} [S] = \sum_{s=3}^5 sp_s.$$

From the Classic wildless Risk problem, we see that there remain $3836$ possible combinations of three card sets without using any wilds. To this we must at $\binom{2}{1} \binom{42}{2} = 1722$ possible combinations of three card sets with ome wild card and $\binom{2}{2}\binom{42}{1}=42$ possible combinations of three card sets with two wild cards. In total, there are $5600$ possible sets of three cards, out of a total $\binom{44}{3}=13244$ possible hands of three Risk cards, so we have $$p_3 = \frac{5600}{13244}=\frac{200}{473}\approx 42.283\dots\%$$

Rather than the messiness of figuring out the exact combinations for $S=4$, let's instead tackle $S=5.$ We see that if you do not get a set on your fourth card then you will automatically get a set by your fifth card, so we only need to figure out what situations lead you to not have a set after four cards. This can only happen whenever you have two of one type and two of a different type (with no wilds, obviously). There are $\binom{14}{2}\binom{14}{2}\binom{3}{2}=24843$ possible hands that have exactly two cards in two different types. Since there are $\binom{44}{4} = 135751$ possible combinations of four distinct Risk cards, we thus have $$p_5=\frac{24843}{135751}=\frac{3549}{19393}\approx 18.300\dots\%$$

Since we can reason that $p_4=1-p_3-p_5 = \frac{7644}{19393}\approx 39.416\dots\%,$ we have the expected number of Risk cards needed to turn in a set is $$\mathbb{E}[S]= 3 \frac{200}{473} + 4 \frac{7644}{19393} + 5 \frac{3549}{19394} = \frac{72921}{19393} \approx 3.760171\dots$$

Risk without Wilds

There are $42$ territory cards in the deck—$14$ that depict an infantry unit, $14$ that depict a cavalry unit, and $14$ that depict an artillery unit. Once you have three cards that either (1) all depict the same kind of unit, or (2) all depict different kinds of units, you can trade them in at the beginning of your next turn in exchange for some bonus units to be placed on the board.

If you are randomly dealt three cards from the $42$, what is the probability that you can trade them in?

In this case, we see that there are $\binom{42}{3}=11480$ possible hands combined of three Risk cards if you remove the wild cards. There are $\binom{14}{3}=364$ possible combinations of three cards within any kind and three different kinds (that is, infantry, cavalry and artillery), so a total of $1092$ different three of a kind sets that can be formed. There are also $14^3 = 2744$ different one of each sets. So there a total of $3836$ sets of three Risk cards without wild cards, which implies that the probability of being able to trade in a randomly dealt set of three cards when the wild cards have been removed is $$p = \frac{3836}{11480} = \frac{137}{410} \approx 33.41463\dots \%$$

Letters Boxed, Like Surgeons General

Consider the following diagram, which consists of eight points (labeled $A$ through $H$), two on each side of the square. A valid “letter boxed” sequence starts at any of the eight points, and proceeds through all of the other points exactly once. However, adjacent points in the sequence can never be on the same side of the square. The first and last points in the sequence can be on the same side, but do not have to be.

As an illustration, $AFBCHEDG$ is a valid sequence of points. However, $AFBCHGED$ is not a valid sequence, since $H$ and $G$ are adjacent in the sequence and on the same side of the square. How many distinct valid “letter boxed” sequences are there that include all eight points on the square?

Well in total, there are $8! = 40,320$ different words that can be made up of the letters $A$ through $H$ with no repeated letters. One possible way is to get this answer is to just enumerate all $40,320$ possible words and throw away any words with forbidden doublets, like $AB$, $EF$ or $HG$. This is not too terribly time consuming in a short Python script, which yields that there are a total of $13,824$ different letters boxed that include all eight points on the square.

Instead of two points on each side of the square (and eight points in total), now there are three points on each side (and twelve points in total), labeled A through L in the diagram below. How many distinct valid “letter boxed” sequences are there that include all $12$ points on the square?

While $8!$ might be well within a modern programming languages wheelhouse to try to iterate through, $12!$ is more than $500$ million possible words that are composed of the letters $A$ through $L$ with no repeated letters. So unless we want to run a script all weekend, let's try a different approach. We can encode the 12 points on the square and the adjacency of the letter boxed game into a graph $\mathcal{G},$ where for instance there is an edge from $A$ to each letter $D, E, F, \dots, L,$ but no edge between $A$ and $B$, $A$ and $C$ or $B$ and $C,$ since they all share the same side of the square. A letter boxed that contains all twelve letters would then be a Hamiltonian path on $\mathcal{G},$ so let's liberally borrow the Python3 code provided by user Gaslight Deceive Subvert from this StackOverflow post related to enumerating all Hamiltonian paths on a graph and get to countings. Since I don't particularly care about the paths themselves and just the number of them, instead of printing the entire list of all Hamiltonian paths, I modified it to just print the number of distinct Hamiltonian paths.

To confirm that everything seemed above board, I started out with the graph from the Classic problem given by \begin{align*}\textsf{G} &\textsf{= \{'A':'CDEFGH', 'B':'CDEFGH', 'C':'ABEFGH', 'D':'ABEFGH',}\\ &\quad\quad \textsf{'E':'ABCDGH', 'F':'ABCDGH','G':'ABCDEF', 'H' : 'ABCDEF'\}}.\end{align*} THankfully, the code retrieved the answer of $13,824$ letters boxed, so I then felt confident enough to encode the Extra Credit graph as \begin{align*}\tilde{G} &\textsf{= \{'A':'DEFGHIJKL', 'B':'DEFGHIJKL', 'C':'DEFGHIJKL',}\\ &\quad\quad\quad \textsf{'D':'ABCGHIJKL', 'E':'ABCGHIJKL', 'F':'ABCGHIJKL',}\\ &\quad\quad\quad\quad \textsf{'G':'ABCDEFJKL', 'H' : 'ABCDEFJKL', 'I':'ABCDEFJKL',}\\ &\quad\quad\quad\quad\quad \textsf{ 'J' : 'ABCDEFGHI', 'K' : 'ABCDEFGHI', 'L' : 'ABCDEFGHI'\}}.\end{align*}.$$ Here, after running the code for roughly 15 minutes, we get a grand total of $53,529,984$ letters boxed with all $12$ points on the square.

High-Low Hijinks Extra Credit

Your friend is playing an epic game of “high-low” and has made it incredibly far, having racked up a huge number of points. Given this information, and only this information, what is the probability that your friend wins the next round of the game?

Let's define this problem as the following probabilistic statement, what is the conditional probability that $N \geq n+1$ given that $N \geq n,$ when $n$ is large? That is, find $$\lambda = \lim_{n \to \infty} \mathbb{P} \left \{ N \geq n+1 \mid N \geq n \right\} = \lim_{n\to \infty} \frac{\mathbb{P} \{ N \geq n+1 \}}{\mathbb{P} \{ N \geq n \} } = \lim_{n\to\infty} \frac{p_{n+1}}{p_n}.$$

One method that we can do is to just compute successive values of $p_n,$ as per the following table:

$n$	$p_n$
1	$\frac{3}{4}$
2	$\frac{13}{24}$
3	$\frac{25}{64}$
4	$\frac{541}{1920}$
5	$\frac{1561}{7680}$
6	$\frac{47293}{322560}$
7	$\frac{36389}{344064}$
8	$\frac{7087261}{92897280}$
9	$\frac{34082521}{619315200}$
10	$\frac{1622632573}{40874803200}$

Even by this point, it seems that the ratio of $p_{n+1} / p_n$ approaches roughly $0.72134752\dots,$ but let's see if we can make sense of this otherwise mysterious limit.

Let's focus on the recursion $h_{n+1}(t) = \int_0^1 I(t,x) h_n(x) \,dx$ that we saw in the Classic problem. Let's assume that we have $h_{n+1} (x) \approx \lambda h_n (x),$ for large values of $n \gg 1.$ Then the recursion formula would settle down to something like $$\lambda h^*(t) = \int_0^1 I(t,x) h^*(x) \,dx = \begin{cases} \int_t^1 h^*(x)\,dx, &\text{if $t \lt \frac{1}{2};$}\\ \int_0^t h^*(x) \,dx, &\text{if $t \gt \frac{1}{2};$}\end{cases}.$$ Differentiating both sides of the equation, we get $$\lambda \frac{d}{dt} h^*(t) = \begin{cases} -h^*(t), &\text{if $t \lt \frac{1}{2};$}\\ h^*(t), &\text{if $t \gt \frac{1}{2}.$}\end{cases}.$$ Let's focus on when $t \gt \frac{1}{2},$ in which case we have $\lambda \frac{d}{dt} h^*(t) = h^*(t),$ so we would have $h^*(t) = C \exp \left( \frac{t}{\lambda} \right)$ for $t \gt \frac{1}{2},$ with $h^*(t) = h^*(1-t) = C \exp \left( \frac{ 1-t}{\lambda} \right)$ for $t \lt \frac{1}{2}.$ In this case, plugging back in, if our recursion formula still holds for $t \gt \frac{1}{2},$ we would need to have \begin{align*}\lambda C e^{t/\lambda} &= \int_0^\frac{1}{2} C \exp \left( \frac{1-t}{\lambda} \right)\,dt + \int_\frac{1}{2}^t C\exp \left(\frac{t}{\lambda}\right) \,dt \\ &= \lambda C \left( e^{1/\lambda} - e^{1/(2\lambda)} \right) + \lambda C \left( e^{t/\lambda} - e^{1/(2\lambda)} \right) \\ &= \lambda C e^{t/\lambda} + \lambda C \left( e^{1/\lambda} - 2 e^{1/(2\lambda)} \right).\end{align*} Therefore, we must have either $\lambda = 0$ or $e^{1/\lambda} - 2 e^{1/(2\lambda)} = 0$ in order tof the recursion formula to hold, which implies that the conditional probability of winning the next round in a long game of high-low is $$\lambda = \lim_{n \to \infty} \frac{p_{n+1}}{p_n} = \frac{1}{2\ln 2} \approx 0.721347520444\dots.$$

High-Low Hijinks

You’re playing a game of “high-low,” which proceeds as follows:

• First, you are presented with a random number, $x_1$, which is between $0$ and $1.$
• A new number, $x_2$, is about to be randomly selected between $0$ and $1$, independent of the first number. But before it’s selected, you must guess how $x_2$ will compare to $x_1$. If you think $x_2$ will be greater than $x_1$ you guess “high.” If you think $x_2$ will be less than $x_1$, you guess “low.” If you guess correctly, you earn a point and advance to the next round. Otherwise, the game is over.
• If you correctly guessed how $x_2$ compared to $x_1$ then another random number, $x_3$, will be selected between $0$ and $1$. This time, you must compare $x_3$ to $x_2$, guessing whether it will be “high” or “low.” If you guess correctly, you earn a point and advance to the next round. Otherwise, the game is over.

You continue playing as many rounds as you can, as long as you keep guessing correctly. You quickly realize that the best strategy is to guess “high” whenever the previous number is less than $0.5,$ and “low” whenever the previous number is greater than $0.5.$ With this strategy, what is the probability you will earn at least two points? That is, what are your chances of correctly comparing $x_2$ to $x_1$ and then also correctly comparing $x_3$ to $x_2$?.

Let's define the indicator function $$I(s,t) = \begin{cases} 1, &\text{if $(t-s)(s-\frac{1}{2}) \leq 0;$} \\ 0, &\text{otherwise,}\end{cases}$$ such that $I(x_1,x_2) = 1$ whenever you successfully earned a point in the first round and $I(x_2,x_3) = 1$ whenever you successfully earned a second point. Let $N = N(x_1, x_2, \dots, x_n, \dots)$ be the random variable that corresponds to your score from playing this game of high-low. We see that based on this setup we have $$\mathbb{P} \{ N \geq n \} = \int_0^1 \int_0^1 \cdots \int_0^1 \prod_{i=1}^n I(x_i, x_{i+1}) \,dx_{n+1} \,dx_n \,dx_{n-1}\, \cdots \,dx_2 \,dx_1.$$ Further, let's define $h_0(t) = 1$ and $$h_1(t) = \int_0^1 I(t, x) \,dx = \begin{cases} \int_t^1 \,dx = 1-t, &\text{if $t \lt \frac{1}{2};$}\\ \int_0^t \,dx = t, &\text{if $t \geq \frac{1}{2};$} \end{cases} = \max \{ t, 1-t \}.$$ Finally, for any $n \in \mathbb{N},$ define $$h_{n+1}(t) = \int_0^1 I(t,x) h_n(x) \,dx.$$

In particular, if we want to see the probability of haveing a score of at least $2$ is \begin{align*}p_2 = \mathbb{P} \{ N \geq 2 \} &= \int_0^1 \int_0^1 \int_0^1 I(x_1,x_2) I(x_2, x_3) \,dx_3\,dx_2\,dx_1 \\ &= \int_0^1 \int_0^1 I(x_1,x_2) \left(\int_0^1 I(x_2, x_3) \,dx_3 \right) \,dx_2\,dx_1 \\ &= \int_0^1 \int_0^1 I(x_1,x_2) h_1(x_2) \,dx_2 \,dx_1\\ &= \int_0^1 h_2(x_1) \,dx_1.\end{align*} If we extrapolate further, which we will need primarily for the Extra Credit problem, we see that $$p_n = \mathbb{P} \{ N \geq n \} = \int_0^1 h_n(t) \,dt, \,\, \forall n \in \mathbb{N}.$$

Returning to the Classic problem here, we have $h_1(t) = \max \{ t, 1-t \}$ and \begin{align*}h_2(t) = \int_0^1 I(t,x) \max \{x, 1-x\} \,dx &= \begin{cases} \int_t^1 \max \{ x, 1 - x\} \,dx, &\text{if $t \lt \frac{1}{2};$} \\ \int_0^t \max \{x, 1-x\}\,dx, &\text{if $t \geq \frac{1}{2};$}\end{cases} \\ &= \begin{cases} \int_t^\frac{1}{2} (1-x) \,dx + \int_\frac{1}{2}^1 x\,dx, &\text{if $t \lt \frac{1}{2};$} \\ \int_0^\frac{1}{2} (1-x) \,dx + \int_\frac{1}{2}^t x \,dx, &\text{if $t \geq \frac{1}{2};$}\end{cases} \\ &= \begin{cases} \frac{3}{4} - t + \frac{t^2}{2}, &\text{if $t \lt \frac{1}{2};$} \\ \frac{1}{4} + \frac{t^2}{2}, &\text{if $t \geq \frac{1}{2};$}\end{cases} \\ &= \max \left\{ \frac{3}{4} -t + \frac{t^2}{2}, \frac{1}{4} + \frac{t^2}{2} \right\}.\end{align*} So we have the probability of earning at least two points is \begin{align*}p_2 = \mathbb{P} \{ N \geq 2 \} &= \int_0^1 h_2(t) \,dt\\ &= \int_0^\frac{1}{2} \left( \frac{3}{4} - t + \frac{t^2}{2} \right) \,dt + \int_\frac{1}{2}^1 \left( \frac{1}{4} + \frac{t^2}{2} \right) \,dt\\ &= \left[ \frac{3}{4}t -\frac{1}{2}t^2 + \frac{1}{6} t^3 \right]^{t=\frac{1}{2}}_{t=0} + \left[ \frac{1}{4} t + \frac{1}{6} t^3 \right]^{t=1}_{t=\frac{1}{2}} \\ &= \left(\frac{3}{8} - \frac{1}{8} + \frac{1}{48}\right) + \left(\frac{1}{4} + \frac{1}{6}\right) - \left( \frac{1}{8} + \frac{1}{48} \right) \\ &= \frac{13}{24}.\end{align*}