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, P[X=+1] = q, P[X=-1] = 1-q.

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  [S - E(S)] / stdev(X)*\sqrt(N) =: Z is normally can be approximated by a normal distribution N(0,1).

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

So
P[S >= 600] ~= P[Z >= [600 -(-600)] / (1*\sqrt(N)) ] = P[Z>= 1200/244] = 1-P[Z<=4.89922] = 4.8x10^-7.

That probability is almost 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.

Edit1...
Edit2...
Woot my mistakes! See my post here.