Help with understanding an example from the book 'Fooled by Randomness' This is an example from the book "Fooled by Randomness":
(...)We know a priori that he is an excellent investor, and that he will be expected to earn a return of 15% in excess of Treasury bills, with a 10% error rate per annum (what we call volatility). It means that out of 100 sample paths, we expect close to 68 of them to fall within a band of plus and minus 10% around the 15% excess return, i.e., between 5% and 25% (to be technical; the bell-shaped normal distribution has 68% of all observations falling between -1 and 1 standard deviations). It also means that 95 sample paths would fall between -5% and 35%.
A 15% return with a 10% volatility (or uncertainty) per annum translates into a 93% probability of success in any given year. But seen at a narrow time scale, this translates into a mere 50.02% probability of success over any given second.
Table 3.1 Probability of success at different scales
Scale      Probability
1 year     93%
1 quarter  77%
1 month    67%
1 day      54%
1 hour     51.3%
1 minute   50.17%
1 second   50.02%

How do I calculate the probability of success at different scales (Table 3.1)?
E.g. where does 77% for a quarter come from?
 A: Here's a thought that seems to give results in agreement with some table values and is close to others (see note at end).  
For example in the case of quarterly return: It appears that we should take $\mu_{\text{quarter}}=\frac{\mu_{\text{year}}}{4}=.0375$ (measured in return above the baseline of T-bills);
and $\sigma_{\text{quarter}}=\frac{\sigma_{\text{year}}}{\sqrt{4}}=.05$
This comes from looking at a year as a sample of size $n=4$ quarters.
The probability of the quarterly $X$ return being above $0$ (i.e., better than T-bills) is computed based on $X\ge 0$ from the quarterly distribution above.  This would be a normal distribution, assuming that the distribution for the annual return is normal.
I don't have $100\%$ confidence in this answer, but it works for quarters and months; however, I get a bit of disagreement with the table values as we take finer time subdivisions.  (Perhaps this has to do with the look-up for the normal probabilities, and/or rounding of some of the decimals involved.)
A: I was also reading this book and spent some time to understand it, using the @paw88789 answer, a more formal approach:
$$
\mu_{month}=\frac{15}{12}=1.25\\
\sigma_{month}=\frac{10}{\sqrt{12}}=2.89\\
$$
To X be greater than zero, you have to find the z-score for this point in the normal distribution:
$$
k=\frac{\mu_{month}}{\sigma_{month}}=0.43\\
k=\text{how many standard deviations for zero return}
$$
Once you are looking for the left side of the curve, consider this as a negative value -0.43 $(-0.43*\sigma_{month}\approx-1.25=z_{score})$
And from z-table (z = -1.25), you retrieve the probability of 0.33360, or (1-0.33360) that gives you 0.6664 very close to the book 67%
For the other time scale, please consider @José David Sánchez recommendation for days (260 trading days instead of 365) and 8-hours/per-day as stated also in the book.
A: I had the same doubts, but if you use 260 instead of 365 for calculating the probability for 1 day you get the exact number (there are about 260 trading days in a year). Also, you have to use 8 hour per day for the same reason.
A: Essentially what he's saying is that the returns per period required to achieve a total 15% average would be the sum of the incremental returns therefore the mean is divided in equal parts. Therefore as you segment it into smaller and smaller intervals the mean return per interval is approaching zero - therefore in any given interval at the limit half of the observations would be showing a loss and half show a profit. The distribution is gradually shifted from centred at 15% to 0%. A pointed out above the sigma value would also be adjusted here by the square root of the number of intervals. The reduction in the mean though has the bigger effect on the probabilities as it is divided by n not square root n.
