[Siegmund]

Drunk hypotheses have a way of being somewhat ill-posed questions.

Given a series of N Bernoulli tries each with p=1/2,

E[(# succesful trials / # total trials) | 1st trial is a success] = (n+2)/(2n+2).

So, yes, if you pick a number of trials in advance, this is true. Similarly, if your poker room closes at a fixed time each day, winning your first hand does slightly increase the chance youll be ahead at closing time.

However, for any epsilon, there exists some K for which E < epsilon + 1/2 for all N>K.
Usually when a mathematician says something is true about the long-term behavior of a series, he means he can make the statement at the beginning, FIX epsilon, and show the statement to be *true* for all sufficiently large Ns.

Generally, results of the "I pick my constant and then let you pick yours" type are stronger results than results of the "pick any constant you like, and I can find..." type.

So, yes, your statement is true for finite sequences, but it's not exactly an earth-shattering result: in particular, it doesn't change the fact that lim (n+2)/(2n+2) as n->infinity is (exactly) 1/2, and it doesn't change your strategy playing a game based on Bernoulli trials, since knowing the outcome of the first trial doesn't help you decide whether calling heads or tails on the next trial is better.