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.

  • 1
    $\begingroup$ Your system counts $3334445556$ as different from $4443335556$, even if all suits are the same in both cases. $\endgroup$ Sep 25 '18 at 2:33
  • $\begingroup$ do i have to account for this by dividing by $2$? @ZubinMukerjee $\endgroup$
    – rover2
    Sep 25 '18 at 2:33
  • $\begingroup$ How many times does it count each hand? What exactly do you have to divide by? $\endgroup$ Sep 25 '18 at 2:34
  • $\begingroup$ since $3334445556, 3335554446, 4443335556, 4445553336, 5553334446,5554443336$ are all the same...would i have to divide by 6? @ZubinMukerjee $\endgroup$
    – rover2
    Sep 25 '18 at 2:41
  • $\begingroup$ 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. $\endgroup$ 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:


Our final probability is

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

  • $\begingroup$ 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? $\endgroup$
    – rover2
    Sep 25 '18 at 2:43
  • 1
    $\begingroup$ $\binom{13}{3}$ 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. $\endgroup$ Sep 25 '18 at 2:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.