Kaufmann Risk of Ruin

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

  1. Looks logical. For coin toss or 50-50 system,
    If I set d=1/4, and 100% prob. of drawdown,
    => C/U=4.
    i.e. If I do not want drawdown above 25%, for equal sized bet U, Stop betting when there are 4 consecutive? losses.
    Is C/U just no. of losses or nonstop losses? Why is U fixed, not user changeable ?
     
    #21     Nov 10, 2011
  2. kut2k2

    kut2k2

    C/U is the maximum number of consecutive losses leading to ruin. U is fixed because it is an artificial bet. As shown in that other thread that shall remain nameless (but you know which one I'm talking about), the Gambler's Ruin formula is based on the assumption of equal-payoff bets (W = |L|). What Griffin did was to take the general situation where W and L can be any values and to create an artificial "equal-payoff equivalent" system to model it. The key was to set the expectation and the variance of the artificial system equal to their counterparts in the real system. That enables you to solve for U and z, or to solve for U and edge.

    I think the main thing to get here (which they clearly never got in that other thread) is that the risk of ruin for a positive-expectation trading system is not 100%. If it was, there'd be no real point to trading for any of us.
     
    #22     Nov 10, 2011
  3. kut2k2

    kut2k2

    Edited for accuracy :cool:
     
    #23     Nov 10, 2011
  4. How positive must expectation be for risk of ruin to be 50% or less?
     
    #24     Nov 11, 2011
  5. No, Being the King of copy n paste in ET doens't make you an expert in anything. In that other thread you went and you made a mess because of the tacticts you follow and for that you should be permanently banned from these thread. In that thread they proved

    --- If a system with positive edge has no time stop or bankroll stop then the probability of ruin as time gets large goes to 1.

    --- Actually, intrabill asked the conditions for the general question but you are a dumbass. He asked under what conditions, if any, the risk of ruin for a positive expectancy system with quit time and/or bankroll stop is 1. Even in this case, DontMissTheBus gave a general answer that the risk of ruin is path dependent.
     
    #25     Nov 11, 2011
  6. ronblack

    ronblack

    This formula makes no sense.
     
    #26     Nov 14, 2011
  7. What I find most hilarious is that the same ET folks who would otherwise smirk with derision that "those egghead academics still use their stupid normal distribution" then falls on even more simplified binomial casino math as the cutting edge.

     
    #27     Nov 14, 2011
  8. kut2k2

    kut2k2

    What I find most hilarious is that you can't find the obvious flaw in your "proof" of universal ruination yet you sneer at an empirical formula that is much closer to reality than anything you've posted here or in that other thread.

    As already stated, there is no closed-form solution to the general risk of ruin problem. For those of us with common sense, this means either an empirical method must be found (e.g., Griffin's formula) or the heavy usage of Monte Carlo sims must be performed. But just throwing up one's hands and saying it can't done because it isn't exact is what losers do.
     
    #28     Nov 14, 2011
  9. No kidding. I made the same point all over that thread and in the earlier thread by intradaybill that led to this one.

    This entire thread's discussion is based on intradaybill's somewhat outlandish, but clearly stated assumptions.

    By the way, you've done nothing to show my proof was wrong (other than pointing out a trival typo) - you actually copy and pasted something from google that actually showed I was right.

     
    #29     Nov 14, 2011
  10. He still fails to understand that given infinite time, if a system can generate losses, it can generate enough losses to cause ruin. If he understood simple coin tossing he should know that but he doesn't.

    My question was under what conditions, when a quit time or an equity stop is enforced the risk of ruin is 100%.

    This lunatic never dealt with the problem I posed. He just copied and pasted an equation for risk of ruin for blackjack which is based on numerous assumptions, the most prominent is that of normality. The other is of a probability of win > 0.5.

    He was told these facts over and over again and he still refuses to deal with reality.

    Edit: The reason he does not understand this and many others like thim is because they confuse the probability of a sequence of losses, which is very small in some cases, with the probability of ruin.

    If someone tosses a fair coin, wins $1 for heads and pays $1 for tails, starts with $10, the probability of 10 tails in a row is tiny, yet the probability of ruin is 100% if the game goes without stop because the probability of such sequence coming up in infinite time is 100%.

    He does not understand that. Many do not.
     
    #30     Nov 14, 2011