Here is how I have been trying to figure this out. Suppose we wanted to figure out what a daily winning percentage had to be in order to observe a 50 percent probability of a 63 day winning streak. It would be (.5)^(1/63), because the probability of 63 straight wins would be Pr(one win)^(63). It turns out that (.5)^(1/63)=.989, which I will round to .99.
Now lets say a firm has a proprietary trading model that is correct 51 percent of the time. This means that on the average day, it will come out ahead (suppose all trades are $1 trades). But if a trader makes one trade a day, he will close the day ahead only 51 percent of the time. If he makes 100 trades a day, however, while his winning percentage per trade remains the same, put his winning percentage per day goes up a lot. Specifically, the standard error for a daily outcome goes down by 1/10, from sqrt(.51*.49) to sqrt(.51*.49/100), or from about .25 to .025. The chance of finishing the day losing on average is based on how many standard deviations away .5 is from .51. In this case, it does from .01/.25 (or not far at all) to .01/.025, or .4 standard deviations away. In a normally distributed world, this means there is a 65 percent chance of finishing the day ahead, assuming each trade has a .51 batting average and 100 trades per day.
To get to winning 99 percent of days, we need to get the standard error for the day to be sufficiently low that .5 is more 2.4 standard deviations away from .51, so the standard error needs to be .01/2.4 or about .004. So we need to find X such that sqrt((.49*.51)/X)=.004. or X=.25/(.004^2)=15,625 trades per day.
Three big assumptions go into this calculations. First, it assumes a stable model. Over the course of one quarter, this may be reasonable. Second, it assumes a model with a 51 percent winning percentage. This is a huge assumption (I do not know what a reasonable number might be). Third, it assumes normality. This is probably not too bad; we do know that Chebyshev’s Inequality says that (1-1/k^2) share of any distribution must be within k standard deviations of the mean. This means that 99 percent of any distribution is within 10 standards deviations, but that is an extreme outcome.