Cfrgtkky

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?

"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 $3/48cdot 2/47$ 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 $52-4=48$, $5cdot4-1=19$ 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
$$p_1=underbrace{frac{19}{48}}_{text{prob. dealer has a card $leq 5$}}cdotunderbrace{frac{18}{47}}_{text{prob. other player has another card $leq 5$}}cdot underbrace{frac{17}{46}}_{text{prob. you get a card $leq 5$}}.$$

2) Only $1$ of the unknown cards is $leq 5$s and you draw a card $leq 5$: in this case the probability is
$$p_2=2cdot frac{19}{48}cdot frac{29}{47}cdot frac{18}{46}.$$

3) None of the unknown cards is $leq 5$s and you draw a card $leq 5$: in this case the probability is
$$p_3= frac{29}{48}cdot frac{28}{47}cdotfrac{19}{46}.$$

Can you take it from here?

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 $19/48$.