What is the probability of at least one instance of two “Kings” being next to each other in a well-shuffled deck of cards?
When you see the term “at least one”, you either estimate the chance of 1, 2, 3, and 4 and add them up, or you calculate the probability of 0 appearance and subtract it from 1. We will do the latter.
The situation in which no King cards are next to each other can be obtained by arranging non-king cards in a row and placing the Kings in between. It is like considering the 52-4 = 48 non-Kings as the bars and 4 Kings are the stars. For each arrangement of the bars, these stars can be placed at 49 locations (from the left of the first bar to the right of the last bar)
*| | * | ….| * | * |
The arrangement of 4 cards in 49 places is 49P4 per each line of other cards. Since there are 48! arrangements possible, the total number becomes 49P4 x 48! If you divide this quantity by the maximum arrangements, 52! we get the required probability.
P(0 Kings) = (49P4 x 48!)/52! = ([49!/(49-4)!] x 48!)/52!
P(at least 1) = 1 – ([49!/(49-4)!] x 48!)/52! = 0.217
How do we verify this is correct? Let’s perform this exercise a million times; that is next.
