What you're looking for is called Zipf's law. This law says that many distribution curves in which the data values are placed in rank order on the horizontal axis by frequency (or, equivalently, percent) follow a power law. The most famous use of Zipf's law is to describe the frequency of word usage in any given language, although the Wikipedia article specifically mentions income rankings as you ask for.
It can be thought of as a discrete version of the Pareto distribution, so you're right about that. Added: This is because Zipf's law is the discrete power law distribution, and Pareto is the continuous power law distribution.
(Update, in response to the OP's request for more on the relationship between Zipf and Pareto.)
I'm going to do this in the general case. The argument will also be for numbers and amounts, rather than probabilities, with the understanding that we can convert the functions involved to pdfs or pmfs by scaling by the appropriate constants.
Suppose we have the density function $p(x)$ for dollars (although it could be any good) allocated among people in a group, so that $\int_a^b p(x) dx$ gives the number of people in the group who have between $a$ and $b$ dollars. Now, rank the people in the group by wealth, and let $z(y)$ denote the wealth that the person ranked $y$ has. The question then is, "What is the relationship between $p(x)$ and $z(y)$?"
Consider the number of people who have more than $M$ dollars. Using $p(x)$, that is given by $\int_M^{\infty} p(x) dx$. But this is also $R$, where $R$ is the largest rank of a person who has at least $M$ dollars. (In other words, if the 34th person has at least $M$ but the 35th does not, then $R = 34$.) So $z(R+1) < M \leq z(R)$. If the population is large enough, we can say $z(R) = M$ without losing much accuracy. Thus $R = z^{-1}(M)$. So there's our relationship (approximately): $$\int_M^{\infty} p(x) dx = z^{-1}(M).$$
Thus the ranking function $z(y)$ is the inverse of the wealth tail cumulative distribution function $\int_M^{\infty} p(x) dx$.
How does this relate to power laws? Well, in this special case, if $p(x) = \frac{C}{x^{\alpha+1}}$ (i.e., a Pareto distribution) for some $\alpha > 0$ and constant $C$, then we have $$z^{-1}(M) = \int_M^{\infty} \frac{C}{x^{\alpha}} dx = \frac{C}{\alpha M^{\alpha}},$$
which means, for some constant $K$, $$z(y) = \frac{K}{y^{\frac{1}{\alpha}}}.$$
Thus $z(y)$ is also a power law. Thus a Pareto (power law) distribution function for some good produces a power law ranking function for people with that good (i.e., Zipf).
Suppose you pay a trillion dollars to enter the game. The following table contains some of the values of the net payoff you can possibly end up with and the corresponding probabilities:
\begin{align*}
\begin{array}{rr}
1/2&-\,1\mathord{,}000\mathord{,}000\mathord{,}000\mathord{,}000\\
1/4&-\,999\mathord{,}999\mathord{,}999\mathord{,}998\\
1/8&-\,999\mathord{,}999\mathord{,}999\mathord{,}996\\
1/16&-\,999\mathord{,}999\mathord{,}999\mathord{,}992\\
\vdots\\
1/2^{10}&-\,999\mathord{,}999\mathord{,}999\mathord{,}488\\
\vdots\\
1/2^{20}&-\,999\mathord{,}999\mathord{,}475\mathord{,}712\\
\vdots\\
1/2^{30}&-\,999\mathord{,}463\mathord{,}129\mathord{,}088\\
\vdots\\
1/2^{40}&-\,450\mathord{,}244\mathord{,}186\mathord{,}112\\
1/2^{41}&99\mathord{,}511\mathord{,}627\mathord{,}776\\
1/2^{42}&1\mathord{,}199\mathord{,}023\mathord{,}255\mathord{,}552\\
\vdots\\
1/2^{50}&561\mathord{,}949\mathord{,}953\mathord{,}421\mathord{,}312\\
\vdots\\
1/2^{100}&633\mathord{,}825\mathord{,}300\mathord{,}114\mathord{,}114\mathord{,}699\mathord{,}748\mathord{,}351\mathord{,}602\mathord{,}688\\
\vdots\\
1/2^{200}&803\mathord{,}469\mathord{,}022\mathord{,}129\mathord{,}495\mathord{,}137\mathord{,}770\mathord{,}981\mathord{,}046\mathord{,}170\mathord{,}581\mathord{,}301\mathord{,}261\mathord{,}101\mathord{,}496\mathord{,}890\mathord{,}396\mathord{,}417\mathord{,}650\mathord{,}688\\
\vdots
\end{array}
\end{align*}
The paradox lies in the following observation. If you take out a loan of one trillion dollars to play this game, you will go bankrupt with a very large probability. However, once in a lifetime (not even of a human but of the universe) you win an unspeakably large amount of money, so large you can't even imagine.
In the light of this observation, a reasonable person would never play this game only once. It is worth playing only if you can play it indefinitely, while you have access to unlimited borrowing. What will happen is that you will keep playing for billions of years, accumulating an enormous debt using your infinite line of credit. But after a very long time, you will win so much money that is sufficient for you to pay off this large debt and still purchase the whole world. As @IanColey put it, this is because the chances of winning so much money are very, very tiny, but the payoffs associated with these very, very tiny probabilities are much, much, much more enormous than the probabilities are tiny.
Best Answer
First of all, the Pareto distribution is defined on non-negative real numbers. Thus it doesn't really make sense to force it to the interval $ (0, 1) $.
Suppose, however, you knew the wealth of two people: someone at the 90th percentile and the 99.90th percentile. Then, if you assume that wealth is truly Pareto-distributed, you could take a method-of-moments approach.
Letting $ A $ be the wealth of the 90th percentile person, and $ B $ the wealth of the 99.90th percentile person, and $ F(x) $ be the cumulative distribution function of a Pareto($x_m, \alpha$) random variable, we have that
$$ F(x) = 1 - (\frac{x_m}{x})^\alpha \iff F^{-1}(y) = \frac{x_m}{(1 - y)^{1/\alpha}} \iff x_m = x(1-y)^{1/\alpha} $$
From two statements that you provided, we have that the person with wealth $ B $ has greater wealth than 78% of the population, and that the person with wealth $ A $ has greater wealth than 22% of the population. You could then estimate the Pareto distribution parameters by solving the equations
$$ x_m = A \cdot (1-0.22)^{1/\alpha} $$ $$ x_m = B \cdot (1-0.78)^{1/\alpha} $$
If you had the full dataset, however, you could certainly perform MLE (@Clarinetist's suggestion). You might get better theoretical guarantees that way.