How can I solve this challenging card counting problem?

Question

A deck of 52 cards is shuffled and evenly split amongst 13 people(4 cards each). What is the probability that one of them has exactly 2 aces and 2 others have exactly 1 ace each?

I feel like a good way to think about this would be to divide the number of ways to place 2 aces in one sequence of 4 cards and 1 ace in two sequences of 4 cards each by the number of ways to place 4 aces in 52 cards(52C4). I'm really confused about how to calculate the numerator though. Any insights on how to solve this? Would also appreciate general tips for solving more challenging counting problems.

Answer 1

There are 12 ways to choose 1 ace, 1 ace, and 2 aces (4!/1!1!2!). For a given choice, the probability that they end up at the first, second, and third player respectively, is 1/13 * 1/13 * 4/52*3/51. Now the is $12\cdot {11 \choose 2}$ ways to choose these 3 players.

You should work out the answer from this.

Answer 2

Let's implement the strategy you suggested. If we look at sequences of cards, there are $\binom{52}{4}$ ways to select positions for the four aces.

If we divide these sequences into $13$ blocks of four cards each, there are $13$ ways to select the block which will contain two aces, $\binom{4}{2}$ ways to select two positions in that block for those two aces, $\binom{12}{2}$ ways to select which two of the remaining $12$ blocks will contain one ace each, and four ways to select a position in each of those blocks for the ace in those blocks.

Hence, the probability that one person receives two of the four aces, while two other people each receive one ace when $13$ players each receive four cards is $$\frac{\dbinom{13}{1}\dbinom{4}{2}\dbinom{12}{2}\dbinom{4}{1}\dbinom{4}{1}}{\dbinom{52}{4}}$$

Answer 3

Another approach is to count the ways for exactly three players to get at least one ace.

There are $\binom{13}3$ ways to pick three players.

There are $\binom{48}{8}$ ways to choose the non-aces for those three players.

There are $\frac{40!}{4!^{10}}$ to divide the remaining cards amongst the other ten players.

So if $M$ is the number of ways to divide $12$ cards, including four aces, amongst three players, so that each player gets at least one ace, the resulting probability is:

$$\frac{M\binom{48}8\binom{13}3\frac{40!}{4!^{10}}}{\frac{52!}{4!^{13}}} =\frac{M4!^3\binom{48}{8}\binom{13}3}{52\cdot51\cdot\cdots\cdot 41} =\frac{M4!^3\binom{13}{3}}{52\cdot 51\cdot 50\cdot 49\cdot 8!}$$

There are a number of ways to compute $M.$ One ways is inclusion-exclusion, but it is probably easier to use a different technique for $M.$

Answer 4

There are $13$ groups of $4$ each

• Choose $3$ groups which will have the aces: $\binom{13}3$

• From them, choose group with two aces, and place: $\binom32\binom42$

• Place one ace each in the two other groups:$\binom41\binom41$

• Putting the pieces together, $$Pr = {\binom{13}3\binom32\binom42\binom41\binom41\over\binom{52}4} = \dfrac{82368}{270725}= \dfrac{6336}{20825} \approx 0.3042 $$

---

Another way is to choose the "double" card in $\binom42$ ways, and place the four cards sequentially in proper groups, thus $$\left[\binom42\frac{52}{52}\frac3{51}\right]\frac{48}{50}\frac{44}{49}$$ where $\frac{52}{52}$, of course, could just have been omitted.

Answer 5

First, let us find the probability that $A\spadesuit$ and $A\heartsuit$ are together, while the other two aces are separate.

• Each card can end up in one of $52$ equally likely positions. Given the location of $A\spadesuit$, there are $51$ remaining equally likely positions for $A\heartsuit$. There are $3$ of these locations which are in the same hand as $A\spadesuit$. Therefore, $P(\text{A$\spadesuit$ and A$\heartsuit$ in same hand})=1/17$.

• Next, we need $A\clubsuit$ to be in a different hand, conditional on the first bullet point. There are $50$ remaining positions for $A\clubsuit$. Two of these are in the same hand as $A\spadesuit$ and $A\heartsuit$, leaving $48$ valid positions, for a probability of $48/50$.

• Finally, a similar calculation shows that the conditional probability that $A\diamondsuit$ is in a different hand from all previous aces is $44/49$.

Multiplying these three probabilities together, we get the probability that $A\spadesuit$ and $A\heartsuit$ are together, while the other two aces are separate. There are $\binom42$ total ways to choose the two aces that are together, so we have to multiply the above by $\binom42$ to account for all possibilities. The final answer is $$ \binom{4}{2}\cdot \frac1{17}\cdot \frac{48}{50}\cdot \frac{44}{49}=\frac{6336}{20825}\approx 30.42\% $$