Probability of Consecutive Cards I've got a deck of 36 cards... 6 are labeled with #1, 6 with #2, and so on up to 6 with #6. I'm drawing 3 cards from the deck & wondering what the probability is that I will end up with 2 or more consecutive cards (including wrap-arounds--meaning a #6 can wrap-around to a #1 and visa-versa).
I know that it's 66.7% WITH replacement as I actually drew the whole thing out on a spreadsheet with every possible outcome (and got 24/36), but I know that the number will increase WITHOUT replacement, and I'm just not sure how to come up with the formula to figure it out.
I came up with this formula... 12/35 + (24/35 x 12/34)+(16/35 x 12/34) = 74.6%, but that seems way too high as I can't see an 8% jump just because I didn't replace 2 cards.
Is there anybody out there who can help me?
 A: Let's count the number of hands which don't contain any consecutive cards.
If all the cards are distinct and there are no consecutive cards, then we have only two possible hands to consider, and they are $\{1,3,5\},\{2,4,6\}$; there are ${2 \choose 1}{6 \choose 1}^3$ ways to get these hands from our deck of $36$. Now if our hand of three cards contains only two distinct numbers and no consecutive cards (like, for example, the hand $\{3,1,1\}$), then we have ${6 \choose 1}{3 \choose 1}{6 \choose 1}{6 \choose 2}$ hands of this type. Since there are ${6 \choose 1}{6 \choose 3}$ hands of three cards that contain only one number, the probability that your hard contains at least two consecutive cards is $$1-\frac{{2 \choose 1}{6 \choose 1}^3+{6 \choose 1}{3 \choose 1}{6 \choose 1}{6 \choose 2}+{6 \choose 1}{6 \choose 3}}{{36 \choose 3}}\approx 0.696$$
A: Jacobi invert always invert. I am going to find the probability of not finding any consecutive cards then I will do 1-p.
1st draw. it doesn't matter, for practical purpouses let's say you draw a 1
2nd draw. you have to draw either a 1, 3, 4 or 5 (q=23/35).
3rd draw.
if 2nd was 1 -> you have to draw either a 1, 3, 4 or 5 (r=22/34) => $22/34*5/35$
if 2nd was 3 -> you have to draw either a 1, 3 or 5 (r=16/34) => $16/34*6/35$
if 2nd was 4 -> you have to draw either a 1 or 4 (r=10/34) => $10/34*6/35$
if 2nd was 5 -> you have to draw either a 1, 3 or 5 (r=16/34) => $16/34*6/35$
Then your probability is $1-22/34*5/35-16/34*6/35-10/34*6/35-16/34*6/35=414/595$
