Binomial distribution

Repeat a yes-or-no trial $n$ times, independently and with the same chance $p$ of “yes” each time. The number of yeses $X$ then has the binomial distribution: $P(X = k) = \binom{n}{k}\,p^k(1 - p)^{n-k}$ for $k = 0, 1, \dots, n$. On the Galton board each row of pegs is one trial and a turn to the right is a yes, so the bin a ball lands in counts its right turns. The outlined bars are the exact probabilities; the balls you drop fill them in as frequencies.

Where do the two factors come from? A ball that ends in bin $k$ turned right $k$ times and left $n - k$ times, in some order. Because the turns are independent and $p$ never changes, every such path has the same probability, one factor per peg: $p^k(1 - p)^{n-k}$. The number of such paths is the number of ways to choose which $k$ of the $n$ rows are right turns, $\binom{n}{k}$; counted peg by peg, each peg is reached from the two above it, which is how Pascal’s triangle adds up. Without those conditions the product need not give a path’s probability. Dealing cards without replacement is the common case: the chance that the next card is a heart depends on the cards already dealt, so the number of hearts in a hand follows a hypergeometric distribution, not a binomial one.

Why is the bin with the most paths not always the most likely? Take $0 < p < 1$. The middle of the board has the most paths, but when $p \ne 1/2$ each of them is less likely than a path with more of the likelier turn. Split into factors shows the tug of war on a log scale, where the product becomes a sum: the path counts form an arch, one path’s chance is a straight line of slope $\ln\frac{p}{1-p}$, and the line tilts the arch. Neighbouring bins give the peak exactly: $\frac{P(k+1)}{P(k)} = \frac{n-k}{k+1}\cdot\frac{p}{1-p}$, which exceeds 1 while $k < (n+1)p - 1$. So the most likely count is $\lfloor (n+1)p \rfloor$; when $(n+1)p$ is a whole number, it ties with the bin just below.

What are the mean and the spread? Score each yes as 1 and each no as 0; then $X$ is the sum of $n$ scores, each with mean $p$ and variance $p(1 - p)$. Means always add, and for independent trials so do variances: $X$ has mean $np$ and standard deviation $\sqrt{np(1 - p)}$. For a fixed $n$ the spread is widest at $p = 1/2$ and shrinks toward zero as $p$ approaches 0 or 1; at the endpoints every trial gives the same answer. After a run the readout puts the balls’ mean and standard deviation beside these values; they vary from run to run, because the balls are a sample. Once $np(1 - p)$ is no longer small the bars look like a bell, and the central limit theorem says why.