Find the probability that the card number 18 is the second jack that you deal. 
You deal from a well-shuffled $52$-card deck, one card at a time. Find the probability that the card number 18 is the second jack that you deal. Include at least $4$ digits after the decimal point in your answer.

I have tried this numerous different ways, but I cant seem to get it. 
I've tried 
$$\frac{\dbinom{48}{16} \cdot \dfrac{4}{52}}{\dbinom{52}{16} \cdot \dfrac{3}{35}}$$ 
and I don't know why this doesn't work. 
 A: If the $18$th card is the second Jack you deal, then there must be one Jack and 16 non-Jacks among the first $17$ cards, then a second Jack at the $18$th card.  The probability of obtaining one of the four Jacks and $16$ of the $48$ non-Jacks in the first $17$ deals is 
$$\frac{\dbinom{4}{1}\dbinom{48}{16}}{\dbinom{52}{17}}$$
The probability of dealing one of the three remaining jacks from among the $52 - 17 = 35$ remaining cards is $3/35$.  Hence, the probability that the second Jack you deal occurs on the $18$th deal is 
$$\frac{\dbinom{4}{1}\dbinom{48}{16}}{\dbinom{52}{17}} \cdot \frac{3}{35}$$
A: There are $17$ ways to choose when the first Jack is drawn. 
$$17\tag{1}$$
Given this, there are $4$ ways to choose which Jack that is. 
$$4 \tag{2}$$
There are $16$ non-Jack cards drawn before the second Jack, and there are $48$ choices for the first of these cards, $47$ for the second, and so on ...
$$48\,\cdot\, 47\, \cdot\, 46\,\, \cdots \,\, 34 \,\cdot \,33 \tag{3}$$
Finally there are three choices for which Jack is drawn second:
$$3 \tag{4}$$

The number of successful draws is the product of values $(1)$ through $(4)$ above.
The total number of possible draws of $18$ cards in order is:
$$52 \, \cdot \, 51 \,\cdot \, 50 \,\,\cdots\,\,36 \,\cdot\, 35$$

The final probability is the number of successful possibilities, divided by the total number of possibilities:
$$\displaystyle\frac{17 \,\cdot\, 4 \,\cdot\,\left(\,48\,\cdot\, 47\, \cdot\, 46\,\, \cdots \,\, 34 \,\cdot \,33\,\right)\, \cdot \, 3 }{52 \, \cdot \, 51 \,\cdot \, 50 \,\,\cdots\,\,36 \,\cdot\, 35}$$
A: The first $17$ cards can be a single jack ($1$ from $4$) plus any $16$ from $48$. The $18$th card can be one of $3$ jacks from the $35$ remaining cards. So I get: $$4\cdot \frac{\binom{48}{16}}{\binom{52}{17}}\cdot \frac{3}{35} = 0.03523$$
