Kaufmann Risk of Ruin

Discussion in 'Risk Management' started by virtualmoney, Nov 8, 2011.

  1. kut2k2


    That formula is wrong on two counts:

    (1) it's oversimplified;
    (2) Edge is not the probability of winning.

    Here's a much better formula proposed by Peter A. Griffin in his book The Theory of Blackjack:

    risk_of_ruin = RoR = ((1 - z)/z)^(C/U),

    C = current trading account
    U = (p*W² + q*L²)^½
    p = win rate of the trading system
    q = loss rate = 1 - p
    W = average winning trade
    L = average losing trade
    z = 0.5*(1 + (p*W - q*|L|)/U)
  2. Ok, this formula looks 'cooler' but how does it look like for a fair coin toss?
    i.e. p=q & L=W?
  3. kut2k2


    For a "fair coin toss", presumably you mean p = q = 0.5 and W = |L|. In that case, z = 1-z = 0.5 and so RoR = 1, as expected.

    As I said above, setting the edge equal to the win probability in the Kaufmann case leads to an error.

    Derivation to be posted later.
  4. Ok, so ruin is certain for coin toss. But this formula still does not take into account bet size for each bet? Looking for formula to work backwards for optimal bet size each time I bet , taking into account p,q,W,L,C,N where N = max consecutive/string of losses.
    N is an important variable to roughly estimate where your account still can implode even when p>0.5
  5. N54_Fan


    Interesting concept. But wouldn't you have to define what RoR is for the trader in question? For me 25% may be ruin and for others it may be 50%. How does this equation take that into account?
  6. kut2k2


    This equation is the risk of going broke (100% drawdown) if there is no target account limit or a time limit. There is supposedly another formula that accounts for what you're asking but I don't know what or where it is.
  7. Or more importantly, some kind of Nstop% you can preset ahead to hint to yourself when results no longer seem random i.e. losses too clustered together, and time to leave the table or stop trading algo.
  8. kut2k2


    The bet size is U. If you still want that derivation, let me know.
    N = C/U
    #10     Nov 9, 2011