# What is the probability that a randomly chosen $10$-card hand has exactly three three- of-a-kinds (and no four-of-a-kinds)?

This is my attempt:

For the first three-of-a-kind:

There are $${13}\choose{1}$$ options for the three cards alike and $${4}\choose{3}$$ for the suits

For the second three-of-a-kind:

There are $${12}\choose{1}$$ options for the three cards alike and $${4}\choose{3}$$ for the suits

Similarly for the third: $${11}\choose{3}$$ $$\times$$ $${4}\choose{3}$$

Finally, for the remaining card, there are $${43}\choose{1}$$ options for the remaining card and $${4}\choose{1}$$ for the suit.

So there are $$\displaystyle\frac{C(13,1)\times C(4,3)\times C(12,1)\times C(4,3) \times C(11,1)\times C(4,3)\times C(43,1)\times C(4,1)}{C(52,10)}$$

Can someone please check and verify? If this is wrong, can someone help me through this problem.

• Your system counts $3334445556$ as different from $4443335556$, even if all suits are the same in both cases. Sep 25 '18 at 2:33
• do i have to account for this by dividing by $2$? @ZubinMukerjee Sep 25 '18 at 2:33
• How many times does it count each hand? What exactly do you have to divide by? Sep 25 '18 at 2:34
• since $3334445556, 3335554446, 4443335556, 4445553336, 5553334446,5554443336$ are all the same...would i have to divide by 6? @ZubinMukerjee Sep 25 '18 at 2:41
• Yes, I think you're overcounting by a factor of $6$. There are also two other errors in the post - you count the suit of the final singleton card after already choosing it (if you choose 1 card from a set of 43, then that card's suit is already chosen), and you also choose the 4th card without considering the restriction that 4 of a kind is not allowed. You have counted $3334445553$ and similar hands that shouldn't count. Sep 25 '18 at 2:44

We'll divide the number of successful hands by the total number of hands.

The total number of hands is $$\binom{52}{10}$$

The number of successful hands is the number of ways to choose $$3$$ card values that will be repeated $$3$$ times, then $$1$$ card value that will occur once, then the excluded suit for each of the sets of $$3$$, and finally the suit of the singleton card:

$$\binom{13}{3}\binom{10}{1}\binom{4}{1}^3\binom{4}{1}$$

Our final probability is

$$\displaystyle\frac{\displaystyle\binom{13}{3} \cdot 10 \cdot 4^4\,\,}{\displaystyle\binom{52}{10}}$$

• so ${13}\choose{3}$ is the value of the three cards alike, ${10}\choose{1}$ is the value of the single card, $({4}\choose{1})^3$ is the suits of the three cards alike, and ${4}\choose{1}$ for the singleton card? Sep 25 '18 at 2:43
• $\binom{13}{3}$ is choosing the values of all three sets of three cards : for the example $33344445556$, it would be choosing $3,4,5$ out of the $13$ choices of card value. Then $\binom{4}{3}^3$ is choosing the suits of all nine cards that are parts of sets of three. Side note: $$\binom{4}{3}=\binom{4}{1} = 4$$ corresponds to the fact that you can also choose the excluded suit for the sets of three, which is just a bit easier to think about. Sep 25 '18 at 2:47