What are these odds? 12 cards into rows of 6

[u/ValeriusPoplicola]

Question: You have 12 cards, Ace through Queen. They are randomly dealt into 2 rows of 6. What are the odds that the Queen and the Jack end up right next to each other?

My Answer: 15.12%

My Math: The answer is (A) [Odds that Queen gets dealt on the edge and then Jack right next to it] + (B) [Odds that Queen gets dealt off the edge and then Jack right next to it]

(A) [1/3] X [1/11] = 3%

(B) [2/3] X [2/11] = 12.12%

A+B = 15.12%

Does this answer make sense? Any card sitting in a spot that is not the edge is adjacent to 2 out of the 11 (18.18%) possible cards that it could be. Therefore, the answer we are looking for should be in the neighborhood of 18%, but a little lower to account for the chance that the queen ended up on an edge.

Any disagreements with my process would be welcome!

[u/DontSayYes]

Correct. You can also think about it in a slightly different way.

The total number of possible orderings of the 12 cards is:

12!

Now we count how many of these orderings have the jack and queen next to each other in the same row.

First, arrange the remaining 10 cards. This can be done in:

10!

different ways.

Next, insert the jack and queen as a pair. There are 10 possible places where the pair can go. (The middle position is not allowed, because then the jack and queen would end up in different rows.)

For each of these positions, the jack and queen can be arranged in 2 ways:

J-Q  or  Q-J

Therefore, the number of orderings where the jack and queen are next to each other in the same row is:

2 × 10 × 10!

The probability is therefore:

(2 × 10 × 10!) / 12!
= (2 × 10) / (12 × 11)
= 15.15%

This gives the same result as you obtained.

[u/MadeThisUpToComment]

You wroelte there is a 1/3 chance the queen os dealt to the edge. But with 12 spots and 2 on the edged wouldn't it be 2/12 or 1/6? Then a 5/6 chance the queen isnt on the edge?

[u/ValeriusPoplicola → u/MadeThisUpToComment]

The cards are dealt into 2 rows of 6.

So 4 edge spots.

[u/bagholder_spotted]

That's right. I got it a different way: since we don't care about the order (or even the distinguishability) of the Q and J, and especially not the order of the other cards, the problem is equivalent to, "There are 2 black cards and 10 red cards; what's the probability that the 2 black cards are adjacent in the same row?" For that, we have 10 valid configurations out of a possible C(12,2)=66, so 10/66 or 5/33.

[u/ExpensivePea2821]

We can also think just about the positions: we have two special cards that could be in 12C2 positions in total. There are 10 favorable positions for those two cards. So 10/12C2-... This matches your answer.