Big Trapezoidal Bean Machine

Suppose you have the board below, which starts with a row with six pins (and five slots), followed by a row with five pins (and six slots), and then back to six pins in an alternating pattern. Balls are only allowed to enter through the middle three slots on top.

Your goal is to create a trapezoidal distribution along one of the rows with six slots, which have been highlighted with dotted lines in the diagram above. More precisely, the frequency of balls passing through the six slots from left to right should be $x−y, x, x+y, x+y, x, x−y,$ for some values of $x$ and $y$ with $x \geq y.$

Suppose the ratio by which you drop balls into the top three middle slots, from left to right, is $a : b : c,$ with $a + b + c = 1.$ Find all ordered triples $(a, b, c)$ that result in a trapezoidal distribution in at least one row with six slots.

As opposed to the classic bean machine, let's go back to assuming that beans have an equal probability of going either to the left or the right. In this case, given that we want to end up with a symmetric probability distribution, we can reasonably deduce that $a = c.$ In this case, we then can reduce the dimensionality by also recognizing that $a + b + c = 2a + b = 1,$ that is, $b = 1 -2a.$

Let's define $\pi_i^{(k)}$ as the probability of a bean passing through $i$th slot from the left on the $k$th dotted line row from the top. We can set up a recurrence formula by going through the various routes from one row to the next. In particular, we get

$$\begin{align*} \pi^{(k+1)}_1 &= \frac{1}{2} \pi_1^{(k)} +\frac{1}{4} \pi_2^{(k)}\\ \pi^{(k+1)}_2 &= \frac{1}{2} \pi_1^{(k)} +\frac{1}{2} \pi_2^{(k)} +\frac{1}{4} \pi_3^{(k)}\\ \pi^{(k+1)}_3 &= \frac{1}{4} \pi_2^{(k)} +\frac{1}{2} \pi_3^{(k)} +\frac{1}{4} \pi_4^{(k)}\\ \pi^{(k+1)}_4 &= \frac{1}{4} \pi_3^{(k)} +\frac{1}{2} \pi_4^{(k)} +\frac{1}{4} \pi_5^{(k)}\\ \pi^{(k+1)}_5 &= \frac{1}{4} \pi_4^{(k)} +\frac{1}{2} \pi_5^{(k)} +\frac{1}{2} \pi_6^{(k)}\\ \pi^{(k+1)}_6 &= \frac{1}{4} \pi_5^{(k)} +\frac{1}{2} \pi_6^{(k)} \end{align*}$$ If we have a trapezoidal distribution with $\pi^{(k)}_2 = x,$ then since $\sum_{i=1}^6 \pi^{(k)}_i = 6x = 1,$ it must mean that we have $\pi^{(k)}_2 = \frac{1}{6}.$

For the first dotted line, given ordered triples of $(a, 1-2a, a),$ we have

$$\pi^{(1)} = \left(0, \frac{a}{2}, \frac{1-a}{2}, \frac{1-a}{2}, \frac{a}{2}, 0\right).$$ Solving $\pi_2^{(1)} = \frac{a}{2} = \frac{1}{6},$ we get the ordered triple $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}).$ For the second dotted line, we have $$\pi^{(2)} = \left( \frac{a}{8}, \frac{a+1}{8}, \frac{3-2a}{8}, \frac{3-2a}{8}, \frac{a+1}{8}, \frac{a}{8} \right).$$ Here again, solving $\pi^{(2)}_2 = \frac{a+1}{8} = \frac{1}{6},$ we get the ordered triple $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}).$ For the third dotted line, we have $$\pi^{(3)} = \left( \frac{3a+1}{32}, \frac{2a+5}{32}, \frac{10-5a}{32}, \frac{10-5a}{32}, \frac{2a+5}{32}, \frac{3a+1}{32} \right).$$ Solving $\pi^{(3)}_2 = \frac{2a+5}{32} = \frac{1}{6},$ we get the ordered triple $(\frac{1}{6}, \frac{2}{3}, \frac{1}{6}).$

Going further, we get

$$\pi^{(4)} = \left(\frac{8a+7}{128}, \frac{5a+22}{128}, \frac{35-13a}{128}, \frac{35-13a}{128}, \frac{5a+22}{128}, \frac{8a+7}{128}\right).$$ Here though, we come to the equation $\frac{5a+22}{128} = \frac{1}{6},$ which cannot be solved by an $a \geq 0,$ since $\frac{22}{128} \gt \frac{1}{6}.$ So there are no ordered triples that can provide a trapezoidal distribution on the fourth dotted line. Similarly, we get $$\pi^{(5)} = \left( \frac{21a+36}{512}, \frac{13a+93}{512}, \frac{127-34a}{512}, \frac{127-34a}{512}, \frac{13a+93}{512}, \frac{21a+36}{512}\right),$$ where $\frac{93}{512} \gt \frac{1}{6}$ so again no ordered triple can provide a trapezoidal distribution on the fifth dotted line. Finally, we have $$\pi^{(6)} = \left( \frac{55a+165}{2048}, \frac{34a+385}{2048}, \frac{474-89a}{2048}, \frac{474-89a}{2048}, \frac{34a+385}{2048}, \frac{55a+165}{2048}\right),$$ where $\frac{385}{2048} \gt \frac{1}{6}$ so again no ordered triple can provide a trapezoidal distribution on the sixth dotted line.

Therefore, the only possible ordered triples that result in a trapezoidal distribution are $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$ and $(\frac{1}{6}, \frac{2}{3}, \frac{1}{6}).$