How Many Trades ... to Win?
Suggested by Joe N.

Suppose we trade often and know (somehow) that there's a 65% probability of getting a 5% return (on a trade).
How many trades until we're pretty certain of winning?

>Huh?
Okay, here's the deal:

• Let N[x] be the standard Normal cumulative distribution. It has Mean = 0 and Standard Deviation (SD) = 1.
• Then, for any x, N[x] gives the probability that a random return (selected from a standard Normal distribution) is less than x.
• Example:
The probability of getting a return less than two SDs above the Mean, namely x = 2, is N[2.00] = 0.977 or 97.7%.
• Example:
The probability of getting a return less than two SDs below the Mean, namely x = -2, is N[-2.00] = 0.023 or 2.3%.
• Conclusion?
The probability of being within 2 SD of the Mean, namely having a return in -2 < x < 2, is 97.7% - 2.3% = 95.4%

>So our trades have a Mean return of 0 and a SD of 1?
No, that was just an example. Here's our scenario:

• We assume that, for our returns, the Mean Return (per trade) = M and the Standard Deviation = S.
• If we ask for the probability that our return is less than r then the distance from the Mean is r - M.
• How many SDs is that from the Mean? It's x = (r - M)/S.
• And what is the probability of that happening? It's N[(r-M)/S].

>So what about that 65% probability of getting a 5% return?
Okay, that means that we're looking for returns of greater than r = 0.05 (that's 5%).
There's a 65% probability of getting more than 5%, so the probability of getting less is 35%.
That's makes N[(0.05-M)/S] = 0.35 (that's 35%).
In Excel, that'd mean that:
N[(0.05 - M)/S] = NORMSDIST((0.05 - M)/S) = 0.35.
And that'd mean that:
(0.05 - M)/S = NORMSINV(0.35) = N-1[0.35]   where N-1 is the function which is inverse to N.
And that'd mean that:
M = 0.05 - S*NORMSINV(0.35)

>Huh?
You see, in order to get that 65% probability of making at least 5% (or, equivalently, a 35% probability of making less than 5%), we need an appropriate Mean and Standard Deviation (that's M and S). Not any will do. They should satisfy M = 0.05 - S*NORMSINV(0.35) so, in general:
 [A]     If there's a probability p of making a (Normally distributed) return of at least r,       then the returns must have a Mean M and Standard Deviation S which satisfy:       M = r - S*NORMSINV(1-p) = r - S*N-1[0.35]

>So, how long before you're guaranteed to win?
You're never guaranteed anything!
However, after n trades (assuming a Mean and SD as given by [A]), suppose your returns are r1, r2, ... rn so your winnings on a given \$1 investment are then r1 then r2 etc. ... each of which may be positive or negative!
After n trades (each investment equal to \$1), your winnings are r1 + r2 + ... + rn.

>And you need the distribution of that sum, eh?
Right on!
As it turns out, if the rs are normally distributed and they're independent (meaning zero correlation), then the sum is also Normally distributed with Mean M and Stand Deviation SQRT(n) S so the probability of having a total gain less than X is N[(X - M)/S].
Suppose we want this to be, say, 1%.

>So we're talking a 99% probability of making money?
Yes, "making money" means winnings greater than \$0, so a 1% probability of winnings less than X = 0. That's a 99% probability of making money, eh?
After n trades, the probability of having winnings less than 1% is:
N(- nM/SQRT(n)S) = N[-SQRT(n)M/S] = 0.01.
That means we must have (again using the Inverse Excel function):
-SQRT(n)M/S = NORMSINV(0.01) = N-1[0.01] = -2.326
That means that n, the number of trades, must be:
n = {S/M*NORMSINV(0.01)}2 = {S/M*N-1[0.01]}2 = {2.326S/M}2

In general:
 [B]     If there's a probability p of making a (Normally distributed) return of at least r,       and the returns have Mean M and Standard Deviation S,       then in order to have a 99% probability of making money after n trades, n must satisfy:       n = 1 + {(S / M)*NORMSINV(0.01)}2 = 1 + {(S / M)*N-1[0.01]}2 = 1 + {2.326 S / M}2