Mind the Gap: Cards
Imagine you have a deck of four cards, each uniquely numbered from 1 to 4. After shuffling the deck, the cards are laid out in a single row. What is the probability that each pair of adjacent cards in this sequence has a number difference greater than 1? Express this probability as a reduced fraction.
Answer
Total permutations: \(4!=24\). Allowed adjacent pairs are those with \(\lvert i-j\rvert>1\), namely \(\{1,3\},\{1,4\},\{2,4\}\) (you basically count 12 13 14, 23 24, 34). Then it turns out that the only favorable orderings are
Therefore,
Comment 1; Why the “complement” approach fails: You might be tempted to say “complement = the two monotone sequences with all gaps =1, so \(1-2/24=22/24\),” but that’s wrong because the complement of “all gaps greater than 1” is “at least one gap \(\le1\),” not “all gaps \(=1\).”