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?

$$e^{0.6\ln 1.5+0.4\ln 0.5}=e^{0.6\ln 1.5}\cdot e^{0.4\ln 0.5}=(1.5)^{0.6}\cdot (0.5)^{0.4}=0.966\ldots\quad(\because e^{n\ln x}=e^{\ln x^n}=x^n)$$ 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 $(1.5)^{0.6\cdot260}(0.5)^{0.4\cdot260}\approx 0$.
