The Binomial Distribution

You used the product rule to count lunches and distributions to describe payouts. Now let X count wins in a book of 10 coconut-shy goes. Assume the results are independent and every go has win probability 0.2. A binomial model describes this count. It needs fixed trials, success or not on each trial, constant success probability and independence. Knowing the success count beforehand is not a condition.

Check the four conditions separately. Trials may have several physical outcomes grouped as success or not, such as six or not-six on a die. Resetting a stall does not prove constant probability or independence: learning, fatigue or different players may change its chances. A calculation can return a number when conditions fail. Check whether it is an exact model or a useful approximation.

For n independent trials, let p be the success probability and r the success count. One ordered result has probability p^r × (1-p)^(n-r). In our book, WWWMMMMMMM has probability 0.2³ × 0.8⁷ = 0.0016777216. Every order with three wins uses the same factors, so it has the same probability. Independence allows multiplication; constant p makes the factors match.

A drum holds 20 tickets, including 5 winners. Draw three uniformly without replacement. The second win chance is 4/19 after a win but 5/19 after a miss, showing dependence. Before seeing earlier results, each draw still has marginal win probability 5/20. Returning each ticket and mixing thoroughly before a fresh uniform draw gives independent draws with the same win probability.

Write X ~ B(n,p) for this binomial count. C(n,r), read 'n choose r', counts sets of r positions among n. Then P(X = r) = C(n,r) × p^r × (1-p)^(n-r). C(n,0) = C(n,n) = 1. Multiply the count by one arrangement's probability; adding the count to a probability does not total the alternatives. Define success first: when counting misses, p must be the miss probability.

Equal marginal probabilities do not prove independence. Toss one fair coin and copy its result into six positions. Each entry has head probability 0.5, but learning one entry reveals every other entry. The head count is only 0 or 6, unlike B(6,0.5). Independence requires earlier results to leave the probabilities of later results unchanged.

There is one all-success arrangement and one no-success arrangement. Their probabilities are p^n and (1-p)^n. At least one success is the complement of no successes, so P(X ≥ 1) = 1 - (1-p)^n. With n = 1, the complete distribution is P(X = 0) = 1-p and P(X = 1) = p.

Arrangement counts are symmetric because selecting successes also selects the remaining failures. Consecutive counts have ratio C(n,r+1)/C(n,r) = (n-r)/(r+1), so they rise while this ratio exceeds 1. For even n, the middle count r = n/2 is largest; for odd n, the two middle counts tie. This is about arrangement counts, not necessarily the most likely count when p differs from 0.5.

For six trials, the remaining counts mirror these: 15, 6, 1. The middle count is largest.

Multiple uniform draws without replacement from a mixed finite population are dependent. A binomial can approximate the count when the sample is small relative to the population; check accuracy for the event needed. Exceptions include a single draw, which gives B(1,p), and all-success or all-nonsuccess populations, whose counts match B(n,1) or B(n,0).