## Pick any two cards

Let’s play a game: thoroughly shuffle a standard pack of 52 playing cards.  Now name any two card ranks, e.g., seven and jack.  If a seven and a jack happen to appear next to each other in the deck, you pay me a dollar, otherwise I pay you a dollar.  Should you be willing to play this game?  How likely do you think it is that I will win?

This game is discussed in a Grey Matters blog post, which in turn refers to a “Scam School” video describing the game as a bar trick.  I think it’s interesting because it’s not intuitive– it doesn’t seem very likely at all that an arbitrary pre-selected pair of ranks, just 4 cards each, should appear adjacent in a random shuffle of the 52 total cards in the deck.

But it’s also interesting because it’s not as likely as the Scam School video seems to suggest.  It turns out to be slightly worse than a coin flip, where the probability that two given card ranks appear adjacent is 284622747/585307450, or about 0.486.  Not very useful as a bar bet, I guess.

So, to improve the odds a bit, consider the following slight variation: shuffle the deck, and name any single card rank that is not a face card (i.e., not a jack, queen, or king).  If a card of the named rank appears next to a face card somewhere in the deck, you pay me a dollar.  Now what is the probability that I win?  It turns out this is a much better bet, with a probability of about 0.886 of such an adjacency.

Finally, we don’t have to resort to simulation as in the Grey Matters post.  It’s a nice puzzle to compute the probability in general, where $n=a+b+c$ is the total number of cards in the deck, of which $a$ are of rank A and $b$ are of rank B.  (For example, in the original problem, $n=52$ and $a=b=4$; in the second variant, $a=4$ and $b=12$.)  Then the probability of at least one A/B adjacency is $1-\frac{1}{{n \choose a,b,c}} \sum_{k=0}^{a-1} {a-1 \choose k} {n-2a+k \choose b} ({c-1 \choose a-k} + 2{c-1 \choose a-k-1} + {c-1 \choose a-k-2})$

the derivation of which I think is a nice puzzle in itself.

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