I play a game that involves standard decks of playing cards, and scales with the number of players. 4 players use two decks of standard playing cards, 6 players use three decks, etc.
One sequence of cards in the game is consecutive pairs of cards. So two 2's of hearts and two 3's of hearts is such a set of consecutive pairs.
I want to calculate the probability that a hand contains at least one consecutive pair. A hand is 27 cards. I thought the way to do this would be counting the ways to have a consecutive pair times the number of remaining combinations of 23 cards, divided by the total number of hands. So in a 3 deck game:
$$ {4 \choose 1}{12 \choose 1}{(52*3)-4 \choose 23} $$ The first term chooses a suit, the second the starting rank of the consecutive pair, and the last the remaining combinations of cards out of the total decks.
Then the denominator is just ${52*3 \choose 27}$
I think this is accurate, but am not sure if I am missing some other way to count the fact that there are multiple decks. What specifically confuses me - with more decks in play, I would expect this probability to increase, as there are more copies of each individual card. But using the above formula modified for 4 decks, the probability decreases - .03 for 3 decks with 27 card hands, and .009 for 4 decks with 27 card hands.