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!!!!!!!
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?
Official answers - back of the book!!!! Is it incorrect?
To calculate the probability of there being exactly one winning set of consecutive cards, we need to count the total number of possible outcomes and the number of favorable outcomes.
Total number of outcomes:
There are 52 cards in a standard deck, and we need to select 9 cards to form the 3 by 3 grid. This can be calculated using combinations. The total number of possible outcomes is given by:
C(52, 9) = 52! / (9! * (52-9)!) = 719,917,548.
Number of favorable outcomes:
To count the number of favorable outcomes, we need to consider the possible arrangements of consecutive cards in the grid. Let's break it down based on the type of consecutive cards:
1. Horizontal consecutive cards:
We can have 3 rows and select 3 consecutive cards in each row, or we can have 2 consecutive cards in one row and 4 consecutive cards in another row. There are C(3, 2) ways to choose the rows for 2 and 4 cards.
So, the number of favorable outcomes for horizontal consecutive cards is:
C(9, 3) * (C(3, 3) + C(3, 2) * C(6, 4)) = 280.
2. Vertical consecutive cards:
Similar to the horizontal case, we can have 3 columns and select 3 consecutive cards in each column, or we can have 2 consecutive cards in one column and 4 consecutive cards in another column. There are C(3, 2) ways to choose the columns for 2 and 4 cards.
So, the number of favorable outcomes for vertical consecutive cards is:
C(9, 3) * (C(3, 3) + C(3, 2) * C(6, 4)) = 280.
3. Diagonal consecutive cards:
We have two possible diagonal directions: from top-left to bottom-right, and from top-right to bottom-left. For each direction, we can either have 3 consecutive cards or 2 consecutive cards and 4 consecutive cards.
So, the number of favorable outcomes for diagonal consecutive cards is:
2 * (C(9, 3) * C(3, 3) + C(9, 3) * C(3, 2) * C(6, 4)) = 560.
Thus, the total number of favorable outcomes is 280 + 280 + 560 = 1,120.
Probability:
The probability of exactly one winning set of consecutive cards is given by the number of favorable outcomes divided by the total number of outcomes:
P(exactly one winning set of consecutive cards) = 1,120 / 719,917,548 ≈ 0.000001554.
So, the probability is approximately 0.0001554 or about 0.1004%.