Probability of 4th card drawn to be a king? Problem: Out of a pack of 52 playing cards, 3 cards are drawn one after the other without replacement. If a fourth card is drawn, what is the probability that it is a king?
One of my friends has given this as the solution. 
`The probability that the 4th card is a king when the cards are drawn one by   one without replacement is same as the probability that the fourth card obtained  is a king when the 4 cards are drawn simultaneously.
This is the same as the probability that the first card drawn is a king when the four cards are drawn simultaneously, because when all the cards are drawn simultaneously, order doesn't matter. This probability is given by 4 C 1/ 52 C 1, hence the required probability is 4/52 = 1/13.`
But I don't think it is correct. Since chances of getting King in 4th trial will depend on the number of kings drawn in the first 3 draws? Am I correct?
So, solution would be
(4/52)(48/51)(47/50)(3/49) + (4/52)(3/51)(48/50)(2/49) + (4/52)(3/51)(2/50)*(1/49)
Is my solution correct? If not why not? If yes, then is there a faster way to get to the answer?
 A: A priori the $k^{\rm th}$ card of the deck is a king with probability ${1\over13}$, whatever $k\in[52]$. Since we are not told the values of the first three cards drawn this probability is still the same, namely ${1\over13}$, at the moment we are asked the question about the fourth card.
A: For the first 3 cards, there are 4 cases possible:
${ }$
CASE 1: If the first 3 cards doesn't contain any king, then $P_1 = \frac{48}{52}.\frac{47}{51}.\frac{46}{50}.\frac{4}{49}$
CASE 2: If the first 3 cards contain exactly 1 king, then $P_2 = 3.\frac{4}{52}.\frac{48}{51}\frac{47}{50}.\frac{3}{49}$
CASE 3: If the first 3 cards contain exactly 2 kings, then $P_3 = 3.\frac{4}{52}.\frac{3}{51}\frac{48}{50}.\frac{2}{49}$
CASE 4: If the all the first 3 cards are kings, then $P_4 = \frac{4}{52}.\frac{3}{51}.\frac{2}{50}.\frac{1}{49}$
${ }$
where $P_i$ is the probability of $4^{th}$ card being a King in $i^{th}$ case.
${ }$
$$P(Fourth\ card\ is\ King) = P_1 + P_2 + P_3 + P_4 = \frac{499800}{6497400} = \frac{4}{52} = 0.07692$$
A: No number answer: 
shuffle the cards, splay them out:
Have your friend pull any card and
re-insert it 4th from the top. 
Now before you deal out the cards, ask yourself, "what is the probability
it will be one of the four kings out of a deck
of fifty-two" ? 
A: As kings are nor specially favoured, the probability that one comes fourth (or indeed anywhere in sequence) is the same as that for any other denomination. 13 denominations, total probability 1 (some card must appear), required probability 1/13.
A: As Christian has answered, one way to think of this experiment would be to have you and your friend together pick all $4$ cards instantaneously. The fourth card is equally likely to be one of four kings from $52$ cards. Hence, the unconditional probability is $4/52$.
