Probability question about cards being dealt and probability of getting a particular suit

Question

In this card game, the entire standard deck of $52$ cards is dealt out to $4$ players. What is the probability that

(a) one of the players receives all $13$ spades?

(b) each player receives $1$ Queen?

Answer 1

There are $52!$ ways to order the $52$ cards, after which you give the first $13$ to player $1$, the next $13$ to player $2$ and so on. There are four possible ways to select a player, $13!$ ways to order the spades and $39!$ ways to order the remaining cards. The probability of a player receiving all spades, thus equals:

$$\frac{4 \cdot 13! \cdot 39!}{52!} \approx 6.3 \cdot 10^{-12}$$

There are $4!$ ways to assign one queen to each player, and another $13^4$ to select the position of each queen in the players' decks. Once these have been placed, there are $48!$ ways to order the remaining cards. The probability of each player receiving one queen, thus equals:

$$\frac{4! \cdot 13^4 \cdot 48!}{52!} \approx 0.1055$$

Answer 2

(a) For this question you can consider a equivalent setup, which is easier to work with. Suppose you have a bucket with 52 marbles, consisting of 13 green marbles (representing the spades) and 39 blue marbles (representing the not-spades). What is the probability of picking 13 green marbles after picking randomly from the bucket 13 times?

Lets start from the beginning. The first time you randomly select a marble from the bucket you have a probability of $\frac{13}{52}$ to select a green marble. Say you were successfull and got a green marble, what is the probability you get another green one? Of course it is $\frac{12}{51}$. You can repeat that untill you find that the probability of randomly getting all green marbles is $$ P(13 \text{ green marbles}) = \frac{13 \cdot 12 \cdot ... \cdot 1}{52 \cdot 51 \cdot ... \cdot 40} = \frac{13! \cdot 39!}{52!} \approx 1.57 \cdot 10^{-12}$$

Another way to approach this problem would be to wonder how many different combinations of cards you can be dealt. There is only one combination with 13 spades. Thus $$P(13 \text{ spades}) = \frac{1}{{52}\choose{13}}$$

We thus know the probability of one person getting all 13 spades. Since all 4 have equal probability of getting all 13 spades the correct answer is

$$P(\text{one player recieves all 13 spades}) = 4P(\text{13 spades}) \approx 6.3 \cdot 10^{-12}$$

Can you now find the answer on the second question (b)?