4

I was asked this question by a student at my college, and I answered it like this:

Since Bridge is played with 4 players, and there are 4 suits per deck of 52 cards, and assuming the deck is a fair, properly shuffled deck of cards, then the probability of 1 player getting 13 of the same suit is $\dfrac {\left( \begin{matrix} 13\\ 13\end{matrix} \right) \left( \begin{matrix} 39\\ 0\end{matrix} \right) } {\left( \begin{matrix} 52\\ 13\end{matrix} \right) }\times4$ This simplifies down to $\dfrac {4} {( \begin{matrix} 52\\ 13\end{matrix} ) }$. But since the question is about the probability of all 4 players getting 13 cards of the same suit, then, since there are $4!$ ways of assigning suits to players, then, isn't the answer $\dfrac {4} {( \begin{matrix} 52\\ 13\end{matrix} ) }\times \dfrac {3} {( \begin{matrix} 39\\ 13\end{matrix} ) }\times \dfrac {2} {( \begin{matrix} 26\\ 13\end{matrix} ) }\times \dfrac {1} {( \begin{matrix} 13\\ 13\end{matrix} ) }$ ?

I would like to know if I answered correctly or if there's another way of looking at this.

Thanks!

Jabernet
  • 1,724

4 Answers4

5

There are $4!$ different ways to deal the cards so that each player gets all the cards of one suit. (This is the same number of ways to distribute 4 objects (suits) to the 4 players.)

The number of ways to deal out the cards is $\displaystyle{52\choose 13}{39\choose 13}{26\choose 13}{13\choose 13}$. You choose 13 cards to give to the first player, 13 to the second, etc.

Using the rule for probability, you get an answer of $\displaystyle\frac{4!}{{52\choose 13}{39\choose 13}{26\choose 13}{13\choose 13}}$, which is what you got.

Christopher Carl Heckman
  • 4,458
2

$\color{green}\checkmark$ Yes, that is entirely correct.

There are $4!$ ways to arrange 4 suits among the players. There are $\binom{52}{13,13,13,13}$ ways to deal the 52 cards, 13 to each of 4 players. The probability is then

$$\frac{4!}{\dbinom{52}{13,13,13,13}}$$

Which is what you have.

Graham Kemp
  • 129,094
2

If you like permutations, you can get the same answer more compactly as

$$\frac{4!}{52!/(13!)^4}$$

The term $(13!)^4$ is to remove the permutations within each player's cards.

true blue anil
  • 41,295
-1

The first card does not matter. The second card has to be one of the 39 of a different suit of the 51 still left. Then, one of the 26 of the 2 other suits of the 50 still left. finally, one of the 13 the 49 that at left.

Then the pattern is that with each card, the bottom number goes down by one and the top number is 12 for four time, 11 for four times, 10 for four times etc.

So the bottom number is 51!, that is "51 factoral."

The top number is 39 X 26 X 13 X (12 to the fourth) times (11 to the fourth) etc.

I do not have a calculator. Figure it out.

Ken Moore
  • 1