Checkerboard Chances
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Timothy has a checkerboard with ten rows and ten columns and a spinner with the numbers 1, 2, and 3. There is an equal chance of each number being spun. He first spins the spinner and moves a checker from the bottom-left corner up the number of spaces spun. He then spins the spinner again, this time moving the checker right the number of spaces spun. If he repeats this process twice more, what are the chances the checker will land on the same color as the square on which it started?
Hint
If in total, the checker moves up an even number of spaces and to the right an even number of spaces it will land on the same color it started on. If it moves up and to the right an odd number of spaces it will also land on the same color it started on.
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Answer
Answer: 365/729
If the checker moves in total up an even number of spaces and to the right an even number of spaces, it will land on the same color it started on. If it moves up an odd number of spaces and to the right an odd number of spaces, it will land on the same color it started on.
The spinner is numbered 1, 2, and 3. The combinations of three of those numbers whose sum is even are the following: (2, 2, 2), (2, 1, 1), (2, 3, 3), (2, 1, 3). These three spins represent either the total move up (three moves up) or the total move to the right (three moves to the right).
Since the numbers can be spun in any order, the total number of permutations is 1+3+3+6=13. Therefore, the probability that three spins will total an even number is 13/27 (27 is the total number of ways the spinner could have been spun). The probability that three spins will total an odd number is 1-(13/27)=14/27 since a number is either even or odd.
The probability that both sets of spins will be even is (13/27)^2. The probability that both will be odd is (14/27)^2. The probability that both will be either even or odd is (13/27)^2+(14/27)^2=365/729
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