There are $6$ peas in a glass, $4$ floating on the top and $2$ sitting on the bottom. At each five second interval, a random number of pas from $0$ to $2$ sink from the top to the bottom and a random number from $0$ to $2$ rise from the bottom to the top. (If there is only $1$ pea left, it moves or stays put with equal probability.) What is the probability that all six peas are on the top before all six are on the bottom?

Here's what I did. We set up the following equations and want to solve for $p_4$: $$\begin{align*} p_5 &= {1\over6} + {1\over3}p_5 + {1\over3}p_4 + {1\over6}p_3, \\ p_4 &= {1\over9} + {2\over9}p_5 + {1\over3}p_4 + {2\over9}p_3 + {1\over9}p_2\\ p_3 &= {1\over9}p_5 + {2\over9}p_4 + {1\over3}p_3 + {2\over9}p_2 + {1\over9}p_1,\\ p_2 &= {1\over9}p_4 + {2\over9}p_3 + {1\over3}p_2 + {2\over9}p_1, \\ p_1 &= {1\over6}p_3 + {1\over3}p_2 + {1\over3}p_1. \end{align*}$$ But this requires setting up a system of $5$ linear equations and a lot of algebra assuming we brute force, so I'm wondering if there are any shortcuts. In particular, is there a way we might be able to exploit the (almost) symmetry here to speed up the calculation?

  • $\begingroup$ By symmetry you get that $p_3 = 1/2$ (if you start with $3$ up, $3$ down, the probability that you get $6$ top before $6$ bottom is equal to the probability of $6$ bottom before $6$ top, as the problem is symmetric to interchange of top and bottom). By the same token, $p_{(6-n)} = 1 - p_{n}$ for all $n$. (you're as likely to end up with $6$ bottom first starting from $4$ bottom as you are to end up with $6$ top first starting from $4$ top). This doesn't completely solve the problem, but it does reduce the number of unknowns (from 5 to 2). $\endgroup$ Commented Aug 11, 2021 at 17:59

1 Answer 1


It is a simple observation that $P_{k, 6-k} = 1 - P_{6-k, k}$ where $P_{a,b}$ represents the probability of all peas reaching the top before the bottom when $a$ is the number of peas at the top and $b$ is the number of peas at the bottom.

Simple calculations reveal that $P_{4,2} = \frac{1}{9}P_{6,0} + \frac{2}{9}P_{5,1} + \frac{1}{3} P_{4,2} + \frac{2}{9} P_{3,3} + \frac{1}{9}P_{2,4}$

We know $P_{3,3}$ to be $\frac{1}{2}$, and after a bit of simplification (substituting $P_{3,3}$ for $\frac{1}{2}$, substituting $P_{2,4}$ for $1 - P_{4,2}$, making $P_{4,2}$ the subject) we reach

$\frac{7}{9}P_{4,2} = \frac{1}{3} + \frac{2}{9}P_{5,1}$

But $P_{5,1} = \frac{1}{6}P_{6,0} + \frac{1}{3}P_{5,1} + \frac{1}{3}P_{4,2} + \frac{1}{6}P_{3,3}$

Therefore $\frac{2}{3}P_{5,1} = \frac{1}{4} + \frac{1}{3}P_{4,2}$ and $\frac{2}{9}P_{5,1} = \frac{1}{12} + \frac{1}{9}P_{4,2}$

Finally then, $\frac{2}{3}P_{4,2} = \frac{5}{12}$ and you find $P_{4,2} = \frac{5}{8}$, which agrees with numerical tests.


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