Induced probability of card drawn from deck The problem:
Let a card be selected from an ordinary deck of playing cards. The outcome $c$ is one of these $52$ cards. Let $X(c) = 4$ if $c$ is an ace, let $X(c) = 3$ if $c$ is a king, let $X(c)=2$ if $c$ is a queen, let $X(c)=1$ if $c$ is a jack, and let $X(c)=0$ otherwise. Suppose that $P$ assigns a probability of $\frac{1}{52}$ to each outcome $c$. Describe the induced probability $P_X(D)$ on the space $\mathcal{D} = \{ 0, 1, 2, 3, 4 \}$.
My attempt:
This problem should be relatively simple: presumably, we find the probability mass function by finding $$\sum_{d_j \in D} p_x(d_j), \,\,\,\, D \subset \mathcal{D}$$
But in order to do this, we need to construct some table, and I'm not sure what it should be. What's even stranger is the book gives five answers, those being $9/13$, $1/13$, $1/13$, $1/13$, and $1/13$.
If I could just figure out how to create this table and figure out why they have five answers I'd have it solved really fast.
 A: $p(d_x)$ is given as being $\frac1{52}$ for every card, so we just have to count, for each element of $\mathcal{D}$, how many cards there are that $X$ sends to $\mathcal{D}$, and divide that by 52. Since there are five elments of $\mathcal{D}$, there will be five such answers.
If D=0, then we're looking for cards that are not Jack, Queen, King, or Ace. There are 36 such cards, so we get $\frac{36}{52}=\frac9{13}$. For the rest, there are four cards, giving $\frac4{52}=\frac1{13}$
A: There are 5 possible outcomes which I would think of as Ace, King, Queen, Jack, and "all other cards".
There is a $\frac 1 {13}$ chance of drawing an Ace; the same for King, Queen, Jack respectively and finally a $\frac 9 {13}$ chance of drawing another rank.
So those are your five probabilities, and expressed in terms of X (the value assigned to those ranks) those should be written in the order $\frac 9 {13}, \frac 1 {13}, \frac 1 {13}, \frac 1 {13}, \frac 1 {13}$ because the space is described in the order 0,1,2,3,4 points.
