No. There are (52 × 51 × 50) / 6, or 22,100, ways that three different cards can be chosen from a deck of 52.

If we remove the 12 picture cards, then 40 cards remain, and the number of ways of choosing three cards from a 40-card deck is (40 × 39 × 38) / 6, or 9,880. Call this group of 9,880 three-card sets Set A.

Since there are 22,100 three-card sets in a normal deck, and 9,880 three-card sets that don’t contain a picture card, there must be 22,100 – 9,880 = 12,220 sets of three cards in an ordinary deck that include at least one picture card. Call these 12,220 three-card sets Set B.

Now, after we’ve shuffled the deck and divided it into three heaps, the set of three topmost cards belongs to either Set A (i.e., none of them is a picture card) or Set B (i.e., at least one of them is a picture card).

So the probability that none of the three topmost cards is a picture card is 9,880 / 22,100, or about 44.7 percent, and the probability that at least one of the three is a picture card is 12,220 / 22,100, or about 55.3 percent. So the odds are roughly 55 to 45 in the stranger’s favor.

(From Owen O’Shea, The Call of the Primes, 2016.)