Odds in a New Jersey lottery 
New Jersey has a six-way combination lottery in which you pick three digits, each between 0 and 9, and you win if the digits the lottery picks
  are the same as yours in any of the six possible orders. For a dollar bet, a winner receives \$45.50. Two possible strategies are (a) to pick three different digits, or (b) to pick the same digit three times. Let $X$ be the winnings for a
  single bet.
Find the probability distribution of $X$ and its mean for both strategies.

I understand that for part (b), $P(X=45.50)=\left(\frac1{10}\right)^3=0.001$, and $P(X=0)=0.999$.
For part (a), isn't the working supposed to be like $P(X=45.50)=\frac1{10}\times\frac19\times\frac18=0.00138$? The answer in my book says $P(X=45.50)=0.006,P(X=0)=0.994$ – how do I get this answer?
 A: You forgot the multiple orders in which your numbers may have occured. If you picked the digits $a$, $b$ and $c$, there are $3!=6$ possible outcomes of the lottery that might help you: ${(a,b,c), (a,c,b), (b,a,c), (b,c,a), (c,a,b), (c,b,a)}$. Every other of the $10 ^ 3 = 1000$ outcomes won't help you.
If you can oversee the possible outcomes and the lucky outcomes, you can calculate your chances by:
$P("lucky") = \frac{\#lucky}{\#possible}$
So your chance of winning $45.50$ is:
$P(45.50) = \frac{6}{1000} = 0.006$
$P(0)=1-P(45.50) = 0.994$
You can also look at it like that:


*

*For the first digit, there are $3$ digits that might help you ($a$,$b$,$c$) and 7 that are bad for you. So you have a $\frac{3}{10}$ chance that you get a good one.

*For the next digit, there are only $2$ usefull digit left, so your chance to get that right is $\frac{2}{10}$

*Last digit, only one digit can help you: $\frac{1}{10}$


So you chance that you get lucky for all $3$ digits is $P(45.50) = \frac{3}{10} \cdot \frac{3}{10} \cdot \frac{3}{10} = 0.006$
A: If you pick three different digits, like 571, there are six different numbers the lottery machine can draw out of 1000 such that you win the \$45.50. The chosen digits do not affect the chance, so $P(X=45.50)$ for strategy A is simply $\frac6{1000}=0.006$, and $P(X=0)$ is the complement, or $0.994$.
A: if you choose 3 numbers the same, then on each of the 3 draws you have a 1/10 chance of the correct one coming out - making 
(1/10) x (1/10) x (1/10) = 1/1000
if you choose 3 different digits, you have 3/10 chance on first draw, if you succeed it goes down to 2/10 on the second one (because the first digit that came out is then used up) then 1/10
(3/10) x (2/10) x(1/10) = 6 / 1000 = 3/500
for A
P(X = 45.50) = .006 
P(X = 0) = .994
E(X) = .273
A: The strategy of picking $3$ different digits doesn't mean that the sample-space changes from $\color\red{\text{all possible permutations of $3$ digits}}$ into $\color\green{\text{all possible permutations of $3$ different digits}}$.
So the total number of permutations remains $10^3=1000$.
When picking a permutation of $3$ different digits, we cover $3!=6$ possible permutations.
Hence the probability of winning is $\frac{6}{1000}=0.006$.
