Aren't there are two cases required here, requiring two separate lemmas, one for when n is even and the other for when n is odd? I don't quite follow this proof. Jim On Monday, March 21, 2016, Erich Friedman <erichfriedman68@gmail.com> wrote:
this follows by induction from the easy lemma: if p>1/2 and q>1/2, then pq+(1-p)(1-q)>1/2 proof: pq+(1-p)(1-q) is larger than its complement p(1-q)+(1-p)q since this reduces to (2p-1)(2q-1) > 0.
erich
On Mar 21, 2016, at 11:06 PM, James Propp <jamespropp@gmail.com <javascript:;>> wrote:
Show that, for any bias p < 1/2, and any positive integer n, if you take a biased coin that shows heads with probability p and toss it n times, the number of times the coin shows heads is likelier to be even than odd.
I know one proof (by way of an exact formula for the probability in question) but I'll bet some of you will find other ways to prove this.
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