# Probability of a shuffled deck having two adjacent cards that are same suit and numerically adjacent (with A-K looping)

The question is: what is the probability that a shuffled deck will have two position-adjacent cards that are of the same suit and numerically adjacent (with A being considered adjacent to both 2 and K)? Any pair of such cards anywhere in the deck satisfies the requirement.

I thought I understood how to tackle this problem, which a buddy came up with (for fun), but my logic produces a clearly-wrong result. Here is my incorrect reasoning:

We can tackle this by counting the number of decks that satisfy the requirements. This should be:

(51 positions in the deck where a match can occur)(52 choices for the first card in the match)(2 choices for the second card in the match)(50 possibilities for the first other spot in the deck)(49 possibilities for the second other spot) . . . (2 possibilities for the next-to-last other spot in the deck)(1 possibility for the last spot in the deck) = 2 * 52!

However, there are only 52! shuffled decks possible, so there cannot be more that satisfy the requirements. What have I done wrong?

(Also, a Monte Carlo simulation for this question gives about 87% chance of having such a deck.)

• The possibility one position pair has two numerically adjacent cards is not quite independent of the other positions having another pair, but if it was then the probability of at least one case would be $1-\left(1- \frac{2}{51}\right)^{51} \approx 0.870007$, close to both $e^{-2}$ and your simulation. – Henry Mar 30 '16 at 7:27
• Wow. That's great! Can you, or somebody else, walk me through how that answer works? I'm not experienced enough in probabilities for it to make sense to me. (Which is why I was trying to tackle this this through combinatorics, not probabilities.) – Dave Hirsch Mar 30 '16 at 7:43

• considering a particular position pair, the probability the second card is numerically adjacent to the first is $\frac{2}{51}$ since of the $51$ cards not identical to the first, two are numerically adjacent
• considering a particular position pair, the probability the cards are not numerically adjacent is $1-\frac{2}{51}$ as the complement of the previous calculation
• assuming independence (almost but not quite true) between position pairs, the probability none of the $51$ position pairs have pair of numerically adjacent cards is $\left(1-\frac{2}{51}\right)^{51}$, multiplying the probabilities of independent events
• assuming independence (almost but not quite true) between position pairs, the probability at least one of the $51$ position pairs has pair of numerically adjacent cards is $1 - \left(1-\frac{2}{51}\right)^{51}$, which is about $0.87$