Sorry to be dense, but I'm still confused. (Actually, in the course of composing this email I think I un-confused myself; see below.) On Tuesday, March 22, 2016, Erich Friedman <erichfriedman68@gmail.com> wrote:
if you believe the purely algebraic lemma, then your result about an even number of heads being more likely than an odd number of heads follows easily.
let p be the probability of heads on n flips,
p is already defined to be the bias of the coin. But I can be flexible about notation. Anyway, I'm unsure what you mean by "the probability of heads on n flips"; I think you mean the probability of an even number of heads on n flips. and let q be the probability of heads on 1 more flip. Actually, I think you want to let q be the probability of a single toss (the n+1st) coming up tails (see below).
by the inductive hypothesis, p>1/2 and q>1/2.
q<1/2 if q denotes the probability of heads on a single flip. So I'll proceed on the assumption that you meant that q is the probability of getting an even number of heads (which is to say none at all) on the last toss, i.e., the probability of getting tails.
and therefore the probability of an even number of heads on all n+1 flips, pq+(1-p)(1-q) > 1/2 as well.
Yes, I think I get it now. Nice! Jim