Let X_i be the random variable modeling the profit of each bet.

Assumptions:
1) profits went from 260 to -330, that is manlteo's profit = (+/-) 600btc
2) N=60k bets of 1btc each @ 2x payout
3) math bullshit (iid random variables), q=0.495

Let S= X_1+X_2+ ... + X_60000. We should expect E(S)= 60000 x ( 1x0.495 + (-1)x0.505 ) = -600 (a loss).

We want to know P[S >= 600]. Central limit theorem states that  \sqrt(N) x [ S - E(S) ] / stdev(X) =: Z is normally distrubuted N(0,1).

var(X) = E(X^2) - E(X)^2 = 1 - (2q-1)^2 = 4q(1-q) = 0.9999, so stdev ~ 1

So
P[S >= 600] = P[Z >= \sqrt(N) [600 -(-600)] / 1 ] = P[Z>= 244x1200] = 1-P[Z<=294000] = 1-1 = 0.

That probability is zero. I'd say you are few orders of magnitude more likely to die, along with 1 million people at the same time, before you finish reading this, than to have maeonlgerry's luck.