Conditional probability that first card is spades given that second and third one are spades Attempt:
The original sample space is just the 52 cards. Now, Suppose the events that second and third card are spades have occured. We know now have a reduced sample space of $50$ cards which contains 2 spaces only, so
$$P( \text{first card spade} ) = \frac{2}{50} = \frac{1}{25} $$
but the correct answer should be 11/50 which I have no idea how they got it. What is the correct reasoning with this type of problems?

Added: Actually, thanks to the given hint I think I missunderstood.
So, basically we $\bf have$ 13 spades. Given the event that reduces
our sample space now we have $50$ cards and $13-2 = 11$ spades. Hence,
$$ P = \frac{11}{50} $$
You got the correct answer, though you didn't use conditional probability. Your method works here because, as you noted, knowing that the 2nd and 3rd cards are spades is basically equivalent to removing them from the deck before we picked the 1st card.

The way to do this with conditional probability would be to take the probability of both events -- i.e., the probability that the first three cards are all spades -- and divide by the probability of the observed event -- i.e., the probability that the second and third cards are both spades. We get

$$\dfrac{\frac{13}{52}\cdot \frac{12}{51}\cdot\frac{11}{50}}{\frac{13}{52}\cdot \frac{12}{51}} = \dfrac{11}{50}   $$
I think is goes like this:

$P(A):=$ 1st card a space; $P(B):=$ only 2nd &amp; 3rd cards are spades, first might be a spade

Event E is the questions noted above.

$$P(E)=P(A \cap B)/P(B)$$

$$P(A \cap B)=\text{all three cards are spades}=13/52*12/51*11/50$$

$$P(B)=\text{all three cards are spades + 2nd &amp; 3rd cards are spades, first might be a spade}$$

$$P(B)=\underbrace{13/52*12/51*11/50}_{\text{ all three are spades}} + \underbrace{39/52*13/51*/12/50}_{ \text{ first not spade, rest spades}}$$

Plug into $P(E)$ equation and you get 11/50!