# If you draw 26 cards from 52 cards, what is the probability that you get 4 kings?

The 52 cards of a standard playing card deck are randomly distributed to two persons: 26 cards to each person. Find the probability that the first person receives all four Kings. Note: The 52 cards include four Kings.

I had this question in my probability exam and my answer was $$\frac{ {4 \choose 4} . {48 \choose 22}}{52 \choose 26}$$

However the teaching assistant's answer was $$(\frac{1}{2})^4 = \frac{1}{16}$$ as each card has a probability $$\frac{1}{2}$$ to go to either of the 2 persons

• The probability of drawing 4 kings is the number of combinations of drawing 4 kings divided by the total number of combinations of drawing 26 cards out of 52, which is 48C22 divided by 52C26 Commented Apr 2, 2020 at 3:04
• The TA does not consider the condition that each player must receive $26$ cards. Commented Apr 2, 2020 at 3:04
• The two answers are very nearly equal, but yours is the correct one. Commented Apr 2, 2020 at 3:04
• You could also say $\frac{\binom{26}{4}}{\binom{52}{4}}$, which is equal to your answer. Commented Apr 2, 2020 at 3:08

As Angina Sing noted, your TA is assuming that the $$4$$ events are independent, but they are not. If we correct his calculation, we arrive at the same answer you did. The probability that the first player is dealt the King of Spades is $$\frac{26}{52}$$. Once he has the King of Spades, what is the probability that he also receives the King of Hearts? There are $$25$$ spots remaining in his hand, and $$51$$ spots overall, so the probability that he is dealt both Aces is $$\frac{26}{52}\cdot\frac{25}{51}$$. (If you have learned about conditional probability, this is simply the fact that $$\Pr(S\cap H)=\Pr(S)\Pr(H|S)$$.) Continuing in this manner, we get that the probability that he gets all $$4$$ Aces is $$\frac{26\cdot25\cdot24\cdot23}{52\cdot51\cdot50\cdot49}$$ the same answer you got.
• @YuvrajSingh... The argument above explains why they are not independent. The probability that any card goes to the first player is $\frac12.$ If all these events were independent, the probability that he gets all the cards would be $\left(\frac12\right)^{52}$, but actually it's $0$ by the conditions of the problem. Commented Apr 2, 2020 at 14:51