Big Skewed Bean Machine

Suppose you have a board like the one shown below. The board’s topmost row has three pins (and two slots for a ball to pass through), while the bottommost row has two pins (and three slots for a ball to pass through). The remaining rows alternate between having three pins and two pins. But instead of the $12$ rows of pins in the illustrative diagram, suppose the board has many, many rows. And at the very bottom of the board, just below the two bottommost pins, are three buckets, labeled $A,$ $B,$ and $C$ from left to right.

Whenever a ball encounters one of the leftmost pins, it travels down the right side of it to the next row. And whenever a ball encounters one of the rightmost pins, it travels down the left side of it to the next row. But this isn’t your garden variety bean machine. Whenever a ball encounters any of the other pins, it has a $75$ percent chance of traveling down the right side of that pin, and a $25$ percent chance of traveling down the left side of that pin.

A single ball is about to be dropped into the left slot at the top of the board. What is the probability that the ball ultimately lands in bucket $A,$ the leftmost slot at the bottom?

Let's break down the many, many, many rows of this bean machine into repetitions of a two pin row followed by a three pin row. Let's zoom in on one of these blocks. Assume that the probability that bean will hit the left pin with probability $p$ (and hence the probability it hits the right pin is $1-p$). Let's further say that the probability that after it passes through the three pin row that it is leaving down the left slot is $q = q(p).$ Let's work out the relationship between $p$ and $q.$

As shown in the figure, there are three possible ways that it leaves down the left slot: (a) it hits the left pin in the upper row, goes to the left hitting the leftmost pin in the lower row, then leaves through down the left slot; (b) it hits the left pin in the upper row, goes to the right hitting the middle pin in the lower row, then goes to the left leaving through the left slot; or (c) it hits the right pin in the upper row, goes to the left hitting the middle pin in the lower row, then goes to the left, leaving through the left slot. The total probability of option (a) is $p \cdot \frac{1}{4} \cdot 1 = \frac{p}{4}$. The total probability of option (b) is $p \cdot \frac{3}{4} \cdot \frac{1}{4} = \frac{3p}{16}$. The total probability of option (c) is $(1-p) \cdot \frac{1}{4} \cdot \frac{1}{4} = \frac{1-p}{16}.$ Therefore, we see that $$q = q(p) = \frac{p}{4} + \frac{3p}{16} + \frac{1 - p}{16} = \frac{6p + 1}{16}.$$

Let's say that the probability of hitting the left pin in the $n$th repitition of these blocks of consecutive two and three pin rows is $p_n.$ Then, since the probability of leaving the $n$th block through the left slot is the same as the probability of hitting the left pin in the $(n+1)$th block, we have $$p_{n+1} = \frac{6p_n + 1}{16}.$$ If we have a total of $N$ of these blocks, then the probability of ending up in $A$ is given by $p_A = \frac{p_{N+1}}{4}.$ While it isn't entirely super impactful when $N$ is large, we should note that since we are dropping a ball in the left slot that we start with $p_0 = 1,$ from which we can derive the exact probability of getting ending up in $A$ for any $N.$

For this particular problem, though, where we assume that $N \gg 1$ we can just take $p_\infty = \lim_{n \to \infty} p_n$ which we can get by plugging $p_\infty$ into both sides of the recursion formula, that is, $$p_\infty = \frac{6 p_\infty + 1}{16},$$ or equivalently $p_\infty = \frac{1}{10}.$ Therefore, when the number of such blocks grows large, the probability of ending up in $A$ is $$p_A = \frac{p_\infty}{4} = \frac{1}{40} = 2.5\%.$$