Could someone explain to me this solution : (a) all 13 spades end up in the same hand? ...

Question

Suppose we deal four 13-card bridge hands from an ordinary 52-card deck. What is the probability that

(a) all 13 spades end up in the same hand?

Solution:

$${4\choose1}{13\choose13}{39\choose 13 \text{ } 13 \text{ } 13} / {52\choose13 \text{ } 13 \text{ } 13 \text{ } 13}$$

I believe we are giving $13$ cards to four hands. So the sample space is $52C13$

I don't get the top part

Answer 1

The denominator $\dbinom {52}{13,13,13,13}$ is the count of ways to deal 52 cards into four hands of 13 each.   This is a multinomial coefficient, and is an extension from the same concept as the binomial coefficient.$$\dbinom{52}{13,13,13,13}=\dfrac{52!}{13!~13!~13!~13!}={^{52}\mathrm C_{13}}{^{39}\mathrm C_{13}}{^{26}\mathrm C_{13}}{^{13}\mathrm C_{13}}$$

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The numerator $\dbinom{4}{1}\dbinom{13}{13}\dbinom{39}{13,13,13}$ is the count of ways for selecting one from four hands to place 13 from 13 spades, and dealing the remaining 39 cards into thirteen cards each for the remaining three hands.

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I believe we are giving $13$ cards to four hands. So the sample space is $^{52}\rm C_{13}$

Why now, this is not wrong.   Notice that there can be cancelation of common factors $$\dfrac{\dbinom{4}{1}\dbinom{13}{13}\dbinom{39}{13,13,13}}{\dbinom {52}{13,13,13,13}}~=~\dfrac{\dbinom 4 1 \dbinom {13}{13}}{\dbinom{52}{13}}$$

The RHS may be viewed as the probability for sorting the deck so all 13 spades are sorted to be deal to one from the four hands, when the 13 spades may be sorted fairly among the 52 places in the deck.   That is where your sample space comes in.

So it is just a different view of the same task.

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Of course, $\binom{13}{13}$ equals one, so that too may vanish.   Although it is useful to leave it in to show the logic behind the expression.

Answer 2

Four hands the thirteen spades could end up in x 1 way to choose the cards for that hand x multinomial coefficient for distributing the other 39 cards into the other 3 hands, divided by multinomial coefficient for all the ways to distribute 52 cards among all 4 hands.

Answer 3

Another way to tackle the problem is to consider just the spades being distributed.

The first spade can go to any of the $52$ "slots", but then the remaining $12$ must go into the same hand, thus$\quad\frac{52}{52}\cdot\frac{12}{51}\cdot\frac{11}{50}\cdot\frac{10}{49} .....\frac{1}{40}$

How the remaining cards are distributed is now immaterial.