A trader is using a system where, after each trade, he wins W units with a probability of p (0 < p < 1) or he loses L units with a probability of q = 1 - p. The trader has a starting bankroll of B0 units and he has in mind a surrender bankroll of X units (B0 > X). If the trader's bankroll drops to or below X, he admits defeat (i.e., he is RUINED). The trader will trade his system indefinitely until he is ruined. What is his risk of ruin? The bankroll after n trades is Bn = B0 + sum[t_i]i=1...n The expectation of each trade is E[t] = pW-q|L| The variance of each trade is V[t] = pq(W + |L|)Â² The expected value of the trader's bankroll is E[Bn] = E[ B0 + sum[t_i]i=1...n ] E[Bn] = B0 + nE[t] = B0 + n(pW-q|L|) The variance of the trader's bankroll is V[Bn] = nV[t] V[Bn] = npq(W + |L|)Â² Central Limit Theorem: Bn ~ N(E[Bn], V[Bn]) Bn ~ N(B0 + nE[t], nV[t]) The z-score of ruin is Zr = (X - E[Bn])/sqrt[V[Bn]] Zr = (X - B0 - nE[t])/sqrt[nV[t]] Zr = (-(B0-X)/sqrt[n] - sqrt[n]*E[t])/sqrt[V[t]] If E[t] <=0, then Zr will increase as n increases, making ruin more and more likely. BUT If E[t] > 0, then Zr becomes more negative as n increases, making ruin less and less likely. So the long-term risk of ruin is certain (100%) unless the trading system has positive expectation.