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 is the total number of cards in the deck, of which are of rank A and are of rank B. (For example, in the original problem, and ; in the second variant, and .) Then the probability of at least one A/B adjacency is

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

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