An hypothetical trading system with 60% win rate and 1:1 payoff has if you bet $100, $20 expectation/trade.

A system with 40% win rate but with 2:1 payoff also has $20 expectation if you bet $100.

The standard deviation for the first system for one trade is the square root of 0.6 * 0.4, which means the SD is 0,4898, and if the bet is $100, the SD is $48,98.

A risk of ruin calculator tells me you need a bigger bankroll for the second system. I thought the SD is the same at the second system but the calculator tells me it isn’t.

What is the formula for the SD from the second system?

And if I want to see how many trades I need to be sure I win I should use this formula from “Gambling theory and other topics”, right?

0 = N * 20 – sqrt (N) * 3 * 48,98

The standard deviations are not the same, and you didn't compute the standard deviation correctly for the first case. sqrt(.6*.4) is the sd of a Bernoulli trial, which means you win 1 bet or lose 0 bets, not win 1 bet or lose 1 bet. Your actual standard deviation is twice what you computed. You can also get it from

var = .6*100^2 + .4*(-100)^2 - 20^2

sd = sqrt(var) ≈ 97.98

In the second case it would be

var = .4*200^2 + .6*(-100)^2 - 20^2

sd = sqrt(var) ≈ 146.97

So you need a bigger bankroll for the second system by a factor of (146.97/97.98)^2 = 2.25 for any ror. To compute the bankroll you need for a given ror, you should use

B = -sd^2/(2*20) * ln(ror)

rather than the one in GTAOT. Where did you get the formula you used for sd? The equation you gave for N, if you substitute the correct value for the sd, will give you the number of bets to have a 99.87% chance of being ahead since being behind would mean being more than 3 sd below the mean at that point which is 0.13%.

I've found the formula in a statistics book. I suspected it was wrong since it didn't take bets into account.

I've read that the stock market doesn't have a normal distribution:
http://www.investopedia.com/articles/04/021804.asp

"We can observe two differences between the normal distribution and actual returns. First, the actual returns have taller peaks - meaning a greater preponderance of returns near the average. Second, actual returns have fatter tails. (Our findings align somewhat with more extensive academic studies, which also tend to find tall peaks and fat tails; the technical term for this is kurtosis)."

What does this mean? Is it good for the standard deviation of a winning trading system (supposing such system exists and supposing we know it has a $20 expectation at a $100 bet)? Does this mean the SD is smaller (because the returns are more likely to be near average)? Or is worse (because it's more likely that the results may be bigger than 3 SD)?

They're not talking about betting a fixed amount like $100 every time with 2 possible outcomes. They're talking about the return on an investment that you hold over time with all profits reinvested. When you just bet $100 with 2 outcomes, your return will approximate a normal distribution after enough bets by the central limit theorem because your return will be the sum of the results of each bet. You can use the standard deviation to compute your ror and time to reach the long run as above. But with the investment, at each time interval you are multiplying the value of the investment by some amount that varies randomly, either greater than 1 for a gain, or less than 1 for a loss. Over time your bankroll will be multiplied by a bunch of random factors. So instead of adding a random amount to your investment each time, you are adding a log to the log of your investment since the log of a product is a sum of the logs. By the central limit theorem, the sum of the logs will be normally distributed over a long enough time, which means the log of the investment is normally distributed, which means the value of the investment is lognormally distributed.

The standard deviation is still the standard deviation, but you can't use that standard deviation with a normal bell curve to find a probability that your investment will be a certain value because it won't follow the normal bell curve. Instead it will follow the lognormal curve, and the article shows the difference in the graph.

You can still use the standard deviation to measure volatility for the investment. If the fraction of the bankroll you have invested is EV/sd^2, then you are betting full Kelly. This maximizes the exponential growth rate of your bankroll; however, you would expect your bankroll to be reduced to x% of it's original value with probability x%. You will lose half your bankroll before it is doubled 1/3 of the time. This is too volatile for most people, so they will bet some fraction of Kelly, say half Kelly (k=1/2) or 1/3 Kelly (k=1/3), etc. Full kelly would be k=1. In general for starting bankroll B

P(bankroll reaches yB before xB) = [x^((2/k)-1) - 1]/[(x/y)^((2/k)-1) - 1].

For half Kelly, you only lose half your bankroll 1/8 of the time (set k = 1/2, y = 1/2, x = something really big like 1000). You would lose half your bankroll before doubling it only about 11.1% of the time (set k=1/2, y=1/2, x=2).

Note that you would increase or decrease the amount you have invested to keep the desired Kelly fraction.

If you are making fixed bets with 2 possible outcomes, then you can still use the Kelly criterion to compute your bets. The fraction of your bankroll to bet for full Kelly is

f = p - b/q

Where p is the probability you win, q is the probability you lose, and b is the payoff odds (1 for 1:1, 2 for 2:1, etc.). This is the same as saying that the Kelly fraction is edge over odds.