the 1st eleven cards in a suite can start a 3-card consecutive set
each card can start 16 con. sets ... 4 suites each for the 2nd and 3rd cards
4 suites * 11 cards per suite * 16 sets per card = 704 con.sets
there are 52C3 possible 3-card sets ... 22100 sets
... probability of a set being consecutive ... 704 / 22100 = .03186
... probability of a set not being consecutive ... 1 - .03186 = 0.96814
there are 8 possible sets on the grid
... the probability of exactly one being consecutive is
... 8 * .96814^7 * .03186
doesn't seem to agree with the "answer" ... where did it come from?
A game involves making a 3 by 3 grid with nine cards from a standard deck. You win if three cards in a row (horizontally, vertically, or diagonally) are the same denomination or are consecutive (in any order).
What is the probability that there is exactly one winning set of consecutive cards?
Ans - about 0.1004
Explanation of process required!!!!!!!
2 answers
Official answers - back of the book!!!! Is it incorrect?