Pascal Triangle Ball Conundrum

Question

Imagine we have a Pascal Triangle Pin Board:

A ball is dropped, and at every pin it has an equal chance of falling left or right.

If we drop $32$ balls, what is the probability that the final column count is $1\;5\;10\;10\;5\;1$?

Answer 1

We compute the probability of any particular output $(a_0,\ldots,a_5)$, given that we drop $a_0+\cdots+a_5=m$ balls. Each ball lands in slot $i$ with probability $p_i=\binom5i /2^5$, and the balls are independent, so the probability of seeing counts $(a_0,\ldots,a_5)$ is $$ \binom{m}{a_0,\ldots,a_5}\prod_{i=0}^5 p_i^{a_i}$$ where $\binom{m}{a_0,\ldots,a_5}$ is the multinomial coefficient $m!/(a_0!\cdots a_5!)$. For the output $(1,5,10,10,5,1)$, we have $m=32$ and the answer is $$\frac{2524139530371646396815776824951171875}{272225893536750770770699685945414569 1648}\approx 0.000927222.$$

Answer 2

For each ball this is essentially doing 5 coin flips and counting number of heads/tails: first bucket is 5 lefts, second is 1 right 4 lefts...

So for any single ball, prob of it landing in bucket $n=0,...,5$ is $$\frac{{5 \choose n}}{2^5}$$

As the events are independent just multiply up the probabilities raising them to the power of the values in that bucket (easier to say than write). But there are $32!$ possible orderings of the balls, divided by $1! \times 5! \times 10! \times 10! \times 5! \times 1!$ for the duplicates gives the final probability.