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*}