In a 52 card deck, what is the probability of getting a run of exactly 3 cards of the same suit in a 4-card hand?
So, treat aces (A) as a 1 so that A, 2, 3, 4 is a 4-card run (not a 3-card run), while J, Q, K, A is a 3-card run.
Valid examples are:
- 9♣️, 10♣️, J♣️, 5♥️
- 8♣️, 9♣️, 10♣️, 3♣️
- 2♣️, 3♣️, 4♣️, K♦️
Invalid examples are
- A♣️, 2♣️, 3♣️, 4♣️
- 8♣️, 9♣️, 10♣️, J♣️
My solution so far is:
$$x = \frac{(_4C_1 × 11) × ??? }{52C4}$$
I have recognized that there are 11 possible positions for a 3-run arrangement in any give suit, as such, I used 4C1 × 11 to represent all the possible 3-run arrangements in a deck of 52 cards.
However, I'm not too sure how the 4th card should be dealt as. For any of the remaining 49 cards to be picked, the 4th card must NOT turn the 3-run into a 4-run.
For the 4th card, I have concluded 2 cases:
Case 1: if the 3-run is on the edge of a suit (A♣️-2♣️-3♣️), then there's only 1 card that the 4th card cannot be (4♣️).
Case 2: if the 3-run is in the middle of a suit (3♣️-4♣️-5♣️), then there are 2 cards that the 4th card cannot be (2♣️, 6♣️).
I have tried approaching this problem from many different perspectives, and I have not been able to come to any firm conclusions on the part regarding the # of ways of picking the 4th card. If anyone has any ideas, please feel free to share and please keep in mind that this is a Combinations problem, NOT a Permutations problem.
Your cases are right, you just need to add them up. There are $4\cdot2$ instances of case $1$ and $4\cdot9$ instances of case $2$, so the probability is
$$ \frac{4\cdot2\cdot48+4\cdot9\cdot47}{\binom{52}4}=\frac{2076}{270725}\approx0.77\%\;. $$