That was an interesting essay on binomial coefficients by Lunnon. https://www.dropbox.com/s/anykne0pd55ehjg/binomial.pdf It had never occurred to me before that anybody could consider anything other than Lunnon's preferred EQ 12=assertion 7, but he indicates that mathematica and MAPLE both apparently are brain dead about this. Incredible. Consulting Feller vol.1 as Lunnon says re "ballot numbers"... we find THEOREM by W.A.Whitworth 1878 then J.L.F.Bertrand 1887: If candidate P gets p votes and rival candidate Q gets q votes, p>=q, then the probability that throughout the count P is always ahead, is (p-q)/(p+q). "THE BALLOT THEOREM": Let a>0, b>0 be integers. The number of paths -- on the usual integer square lattice where each step adds either (+1, +1) or (+1, -1) to current location -- from (0,0) to (a,b) which always lie above the x-axis, is (b/a) * Binomial(a, (a+b)/2). Equivalently, the number which never go below the x-axis is ((b-1)/(a-1)) * Binomial(a-1, (a+b)/2-1). This is proved with the brilliantly simple "reflection lemma" that (if a,b,c,d positive integers) the number of paths from (c,d) to (a,b) which cross the x-axis at least once, is the same as the number of unrestricted paths from (c,-d) to (a,b). --------------------- QUESTION: If candidate P gets p votes and rival candidate Q gets q votes, p>=q, what is the probability distribution of the number of "lead changes" as we count ballots? Or (almost equivalently) of the number of times P & Q tied for the lead? I do not know the answer, but one can write down certain expressions for prob(0), prob(1), prob(2),... and one can also get an approximate answer by using THEOREM OF L.TAKACS (1967): If you sum i.i.d. random integer variables each with mean=m, then the probability every partial sum is positive (forever), equals m if m>0, otherwise 0. Applying Takacs' theorem with the random variable which is +1 with prob=(p-q)/(p+q) and -1 with prob=2q/(p+q) we see that the chance that P stays ahead forever equals m=(p-3q)/(p+q) which it makes sense is a lower bound on the chance (p-q)/(p+q) that P stays ahead during the first p+q votes... although this is not a proof. The chance that P stays enters the lead exactly n>=1 times (over infinite time), is from Takacs theorem equal to (1-m)^(n-1) * m so this perhaps is a lower bound on the chance there are <=n "P enters lead" moments during the count of the p+q votes. QUESTION: If candidate P gets p votes and rival candidate Q gets q votes, p>=q, what is the probability distribution of the number of moments (among p+q total moments) at which P is not in the lead? It is possible these questions have some nice partial answers, albeit so far my attempts to answer them have only yielded horribly complicated expressions.