The math behind Warren Buffett's famous rule – never lose money
There are two things I should point out. One is that the arithmetic mean doesn't properly measure annual growth rate. The second is why.
The correct calculation for average annual growth is geometric mean.
Let $r_1,r_2,r_3,\ldots,r_n$ be the yearly growth of a particular investment/portfolio/whatever. Then if you invest $P$ into this investment, after $n$ years your final amount of money is $Pr_1r_2\cdots r_n$. The (yearly) average growth rate of this investment is the number $r$ such that if the investment grew at a constant rate of $r$ every year then after $n$ years we'd have the same amount as we actually ended up with. In other words it is $r$ such that $Pr_1r_2\cdots r_n=Pr^n$. Thus we have $$r=\sqrt[n]{r_1r_2r_3\cdots r_n},$$ which is the geometric mean, not the arithmetic mean.
If we use the geometric mean, we see that Turtle's average yearly growth is $\sqrt[5]{1.39}\approx 1.07$, and Hare's average yearly growth rate is $\sqrt[5]{1.36}\approx 1.06$, which is more in line with our expectations.
Why doesn't the arithmetic mean behave as expected?
Well, let's look at something over two years. Say its arithmetic mean growth is 1. Then the growth rate for one year will be $1+x$ and the other year will be $1-x$. Multiplying these together, we see that total growth is $1-x^2$. In other words, actual growth is always less than or equal to that predicted by the arithmetic mean (this is true for $n$ years as well, see the AM-GM inequality). Note further that the actual growth is closer to that predicted by the arithmetic mean when the individual annual growth rates are closer together. Thus if you are more consistent (your annual growth rates are closer together) then your arithmetic mean growth rate will closely approximate your true average annual growth rate as in Turtle's case. On the other hand, if your annual growth rates are more spread out, then your true average annual growth rate will be much lower than the arithmetic average growth rate (as in Hare's case).
After a 50% loss you need a 100% gain to break even. In that scenario the arithmetic average return is 25% and the geometric average return is 0%.
It is more important to maximize geometric rather than arithmetic average return -- and this is intimately connected with the concept of risk-adjusted return and mean-variance optimization.
Given a set of returns $R_1,R_2, \ldots, R_n$ we have the arithmetic average and variance
$$A = \frac{1}{n} \sum_{k=1}^nR_k, \quad\quad V = \frac{1}{n}\sum_{k=1}^n(R_k - A)^2$$
A useful approximation that relates the geometric and arithmetic average return is
$$G = \left[\prod_{k=1}^n(1+R_k) \right]^{1/n}- 1 \approx A - \frac{V}{2}$$
This is in part a motivating factor for constructing a portfolio that maximizes expected return subject to a an upper constraint on variance or minimizes variance subject to a lower constraint on expected return.
This looks like the difference between an arithmetic mean and a geometric mean.
Each year, if you invest $\$1000$ at the start and cash out at the end of the year, you want to use the arithmetic mean pointing at Mr Hare
But if instead you invest $\$1000$ at the start of year $1$ and keep all the money invested until the end of year $5$ then you want to use the geometric mean pointing at Mr Turtle
The geometric mean is particularly sensitive to low values: in the worst case of you losing all your money in a particular year ($0$ in your table), you can never make it back. But you can if you restart each year with the same amount
If you want to do even better, each year give half your money to Mr Turtle and half to Mr Hare to invest, rebalancing every year. With these results, you will be $43\%$ better off after five years, a compound $7.4\%$ a year, better than both Mr Turtle's $39\%$ and $6.8\%$ and Mr Hare's $36\%$ and $6.4\%$. Half-and-half is not quite optimal but it is simple and is close to optimal with these particular numbers