# Drawing 3 cards from a deck

3 cards are drawn from a standard deck of 52 cards without replacement. Let $X_1$ be a random variable that is 1 if the 1st card drawn is a King and 0 otherwise $X_2$ be a random variable that is 1 if the 2nd card drawn is a king and 0 otherwise. What is the probability that $X_2 = 1$?

My thinking is that $P(X_2 = 1) = P(X_2 = 1|X_1 = 1) + P(X_2 = 1|X_1 = 0)$, so $P(X_2 = 1) = \frac{\frac{4}{52} \times \frac{3}{51}}{\frac{4}{52}} + \frac{\frac{48}{52} \times \frac{4}{51}}{\frac{48}{52}} = \frac{7}{51}$.

Is this right way of thinking, or should the answer just be $\frac{4}{52}$ by symmetry?

You made a mistake in the first equation of not multiplying $P(X_1=1)$ and $P(X_1=0)$, it should be $$P(X_2=1)=P(X_1=1)P(X_2=1|X_1=1)+P(X_1=0)P(X_2=1|X_1=0)$$ and your answer is indeed $\frac{4}{52}$. (This equation is the Law of Total Probability)

Note: The calculation is $$P(X_2=1)=\frac{1}{13}\cdot \frac{3}{51}+\frac{12}{13}\cdot\frac{4}{51}=\frac{1}{13}$$

The utmost right way of thinking is: there are $52$ equiprobable candidates for the second card drawn and $4$ of them are kings, so the probability that it will turn out to be a king is $\frac4{52}=\frac1{13}$.

You come to the same solution by practicizing the Law of Total Probability (as has been shown in the answer of Lazy Lee), but there is no reason at all to reach out for more complicated solutions.

Also symmetry (mentioned by yourself) can be missed.

Keep things easy!

The probability of the first card being a King is 4/52, or 1/13. Should the first card had been a King, the probability of the second card being a King is now 3/51, or 1/17. However, if the first card wasn't a King, then the odds that the second card is a King is 4/51.

If you sum the two scenarios, applying a product of their condition, you get the following:

$P(X_2) = 1/13 * 1/17 + 12/13 * 4/51$

$= 1/221 + 48/663$

$= 1/221 + 16/221$

$= 17/221$

$= 1/13$

Edit:

Fixed fraction error.