Trader with 60% chance of gaining 50% and 40% and 40% chance of dropping 50%. Average return per day calculation. I am reading Paul Wilmott's Frequently Asked Questions in Quantitative Finance, and there is a question that states the following:

Every day a trader either makes 50% with prob- ability 0.6 or loses 50% with probability 0.4. What is the probability the trader will be ahead at the end of a year, 260 trading days? Over what number of days does the trader have the maximum probability of making money?

He explains how to break even the trader's best chance is at ~164 days. This is simple, he solves for $1.5^n 0.5^{260-n}$. But then he says that the trader's average return per day is:

$1−e^{0.6 \ln1.5 + 0.4\ln0.5}$ = −3.34%

Where does this formula come from? Any idea how to get the formula on the left hand side?
 A: The $164$ days is the number of profitable days that are needed out of $260$ trading days for the trader to break even.  It comes from solving $1.5^n\cdot 0.5^{260-n}=1$.  You did not include the rest of the equation.  We can say the expected number of up days is $0.6 \cdot 260=156$.  This shows that his expectation is negative, as he needs more than the expected number of up days to break even.
To get the exponential, consider the log of the amount of the money the trader has.  It starts at $0$ because he has $1$ unit.  After one day, the expected value of the log is $0.6 \log 1.5 + 0.4 \log 0.5$.  The expected value of his account after one day is $e^{0.6 \log 1.5 + 0.4 \log 0.5}$ so the expected gain is this minus $1$, which comes out $-0.0334$
A: I do not follow where the confusion is. Let's say he had 1 rupee at the beginning of the day. He has probability of 0.6 to end with 1.5 rupee (50% gain) and 0.4 probability to end with 0.5 rupee (50% loss).
So at the end of the day, his money on an average will be $= (1.5)^{0.6}.(0.5)^{0.4} \approx 0.9666.$
He started the day with 1 rupee so his average daily return $= 0.9666-1 = -0.0334 = -3.34 \% $
A: $$e^{0.6\ln 1.5+0.4\ln 0.5}=e^{0.6\ln 1.5}\cdot e^{0.4\ln 0.5}=(1.5)^{0.6}\cdot (0.5)^{0.4}=0.966\ldots\quad(\because e^{n\ln x}=e^{\ln x^n}=x^n)$$ times the money at the beginning of the day becomes the money at the end of the day on average because the probability of getting $1.5$ times the money at the end of the day is $0.6$ and that of getting $0.5$ times the money is $0.4$. So, over a $10$(say)-day period, money gets $(1.5)^{6}\cdot (0.5)^{4}$ times (just a check).
So, the average loss per day is $[1-(1.5)^{0.6}\cdot (0.5)^{0.4}]\%$.

Also, it seems to me that $164$ are the days required in the best case to achieve a no-profit-no-loss situation in $260$ days. Hence, $1.5^{n}\cdot0.5^{260-n}=1\Rightarrow n=164$. Note that the best case is far from from the $10$-day logic as $(1.5)^{0.6\cdot260}(0.5)^{0.4\cdot260}\approx 0$.
