Blackjack: probability of being dealt a card of value less than or equal to 5 given this scenario? There is only one deck of cards in play which has just been reshuffled. Six cards are dealt -- two to you, two to another player, and two to the dealer. You have a king and a six, the other player is showing a five; and the dealer is showing a seven. You are the first to decide whether to hit or stay. If you hit, what is the probability of being dealt a card with a value less than or equal to 5 (aces included)?
My idea:
$P$(dealt a card $\leq 5)$ = $P(5)$ + $P(4)$ + $P(3)$ + $P(2)$ + $P(ace)$.
To find $P(5)$, there are 3 cases: 
1) The 2 unknown cards are 5s $\implies$ $P(5)$ = $\frac{1}{46}$.  
2) 1 unknown card is a 5 $\implies$ $P(5)$ = $\frac{2}{46}$ + $\frac{2}{46}$ = $\frac{4}{46}$, since there are two ways this can happen.
3) None of the unknown cards is a 5 $\implies$ $P(5)$ = $\frac{3}{46}$.
Adding these up, $P(5) = \frac{8}{46}.$
To find $P(4)$, we consider:
1) 2 unknowns are 4s, so $P(4) = \frac{2}{46}$.
2) 1 unknown is a 4, so $P(4) = \frac{3}{46} + \frac{3}{46} = \frac{6}{46}$, since there are two ways this can happen.
3) None is a 4, so $P(4)$ = $\frac{4}{46}$.
So $P(4)$ = $\frac{12}{46}.$
Now, $P(3)$, $P(2)$, and $P(ace)$ are the same as $P(4)$. So our final answer is the sum of these results, so $\frac{8}{46} + 4(\frac{12}{46}) = \frac{56}{46}$... which obviously doesn't make sense. Where am I going wrong?
 A: "The 2 unknown cards are $5$s ⟹ $P(5) = 1/46$" is not correct.
Actually $1/46$ is the conditional probability that you got a $5$
after you know that the 2 unknown cards are $5$s. In order to obtain the correct probability you should multiply $1/46$ by $3/48\cdot 2/47$ i.e. the probability that the 2 unknown cards are $5$s.
This is a hint which follows your approach (although there is a shorter way to find the required probability). 
The unknown cards are $52-4=48$, $5\cdot4-1=19$ are $\leq 5$ and $48-19=29$ are $>5$. We consider $3$ distinct cases.
1) The $2$ unknown cards are $\leq 5$s and you draw a card $\leq 5$: in this case the probability is
$$p_1=\underbrace{\frac{19}{48}}_{\text{prob. dealer has a card $\leq 5$}}\cdot\underbrace{\frac{18}{47}}_{\text{prob. other player has another card $\leq 5$}}\cdot \underbrace{\frac{17}{46}}_{\text{prob. you get a card $\leq 5$}}.$$
2) Only $1$ of the unknown cards is $\leq 5$s and you draw a card $\leq 5$: in this case the probability is
$$p_2=2\cdot \frac{19}{48}\cdot \frac{29}{47}\cdot \frac{18}{46}.$$
3) None of the unknown cards is $\leq 5$s and you draw a card $\leq 5$: in this case the probability is
$$p_3= \frac{29}{48}\cdot \frac{28}{47}\cdot\frac{19}{46}.$$
Can you take it from here?
A: There are 48 unknown cards, including three 5's and four each of four denominations below 5, for a total of 19 good cards for you if you hit. Each of the 48 is equally likely as the card you get,  so the probability of a good hit is $19/48$. 
