How much would you bet on this trade

http://traders-library.com/download...k. How to Supercharge Your Option Trading.pdf
See table at top of page 13

 

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.

 

How did you arrive at the win rate and payoff ratio?

 

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.

 

I see what you mean, thanks.

 

Combination of educated guesswork, backtesting, and historical performance.

 

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?

 

Yes, fair point.

 

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.