How much would you bet on this trade

Quote from acrary:

However, Std. dev. of trades is not a major determinant of drawdowns.
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Would you be able to elaborate a bit on this as I'm not quite sure what you're getting at? Seems like max allowable loss and probability of losing would be, I'm just not sure how that relates/applies.:)
 
Quote from Ghost of Cutten:

It's a spread play between two related markets. I was thinking of risking 5% and wondering if it was too big. My drawdown tolerance is in the 20-25% range.
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How did you arrive at the win rate and payoff ratio?
 
Quote from Soon2Bgreat:

Would you be able to elaborate a bit on this as I'm not quite sure what you're getting at? Seems like max allowable loss and probability of losing would be, I'm just not sure how that relates/applies.:)
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In the example, 1500 is used for the wins and 1000 for the losses. In the real world we'll see a distribution of wins and losses such that the mean of the wins would be approx. 1500 and the mean of the losses would be 1000. I don't know that to be the case here, but in general it has been for my systems. Absolute loss and absolute gain targets are usually sub-optimal in trading. I was merely pointing out that if there is a distribution with say a 50% level of std. dev. on both win amount and loss amount, then the drawdown levels wouldn't be drastically different. In this case with the 5% risked the drawdown would increase to 26.6% at the 95% confidence level versus the 23% with no variability in the trades. For the 3% risk level the drawdown would increase to 16.7% at the 95% confidence level.
 
Quote from acrary:

I was merely pointing out that if there is a distribution with say a 50% level of std. dev. on both win amount and loss amount, then the drawdown levels wouldn't be drastically different.
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I see what you mean, thanks.
 
Quote from abattia:

http://traders-library.com/download...k. How to Supercharge Your Option Trading.pdf
See table at top of page 13
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Ok, thanks for the link - but this seems to give the probability of seeing X consecutive losers, not the probability of a given drawdown. You could lose 10 in a row, see 1 winner, then another 6 in a row etc.

Is there any method other than Monte Carlo simulations that can calculate confidence intervals for max drawdown?
 
Quote from Ghost of Cutten:

... this seems to give the probability of seeing X consecutive losers, not the probability of a given drawdown. You could lose 10 in a row, see 1 winner, then another 6 in a row etc...
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Yes, fair point.
 
Quote from abattia:

At 75% win rate, you have something like 2% - 9% chance of seeing a run of 5 - 6 consecutive losers in a 100-trade period... risking 5% per trade puts you 25% - 30% down after 5 - 6 consecutive losers.
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It's a binomial distribution with P=0.75. The probability of 5 trials and no successes is:

b(x=0; 5, 0.75) = 5C0 * (0.75)^0 * (0.25)^5 = 0.0009765625.

Next you can use the above probability of losing 5 times in a row as a new binomial game. The probability of not seeing a 5 event losing streak in 100 trials is the same as the above event not occurring in 20 trials:

b(x=0; 20; 0.0009765625) = 20C0 * (0.0009765625)^0 * (1 - 0.0009765625)^20 = 0.9806488908.

The cumulative probability of b(x>=0; 20; 0.0009765625) is simply 1 - 0.9806488908 = 0.0193511092.

So there's about a 1.935% chance of seeing a 5 event losing streak in 100 trials.

If you drop your original edge down to P=0.55, things get much worse:

b(x=0; 5, 0.55) = 0.0184528125.
b(x=0; 20, 0.0184528125) = 0.6890070402.
b(x>=0; 20, 0.0184528125) = 0.3109929598.

so, with P=0.55, you have fully a 31% chance of seeing a 5 event losing streak in 100 trials.
 
Note my error above. 20 independent trials of 5 can miss a losing streak that crosses the 5 event trial boundaries: [11100] followed by [00011] does indeed have a [00000] 5 streak loser crossing the boundary. Some additional recursive math is needed. Probabilities are higher as a result.
 
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