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 bottomleft 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?
HintIf 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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Comments
tpg76
Jun 15, 2008
 I found this to be a very challenging and interesting teaser  good job! I was on the right track but didn't use the "sums", so it got complicated quickly and I didn't get the right answer. I should have had some "humble pie" (I did, anyway, in the end!) and used the hint.
Thanks! 
kunju
Jun 16, 2008
 hard but interesting 
sourdough
Jun 19, 2008
 Mathematically imperative that the answer must come true!
$D 
javaguru
Dec 08, 2008
 Like all probability problems, there's more than one way to get the answer. I used the principle that I need an even number of odd spins in order to move an even number of spaces. This means that there can be either 0, 2, 4 or 6 odd spins.
There are 6P2 = 15 ways to order either two evens and four odds or four odds and two evens.
This gives:
(1/3)^6 + 15x(1/3)^4x(2/3)^2 + 15x(1/3)^2x(2/3)^4 + (2/3)^6
= 1/729 + 60/729 + 240/729 + 64/729
= 365/729 
ron93
Nov 24, 2015
 Perhaps my solution is probably the weirdest of all, why ?, it's because I didn't use the hint at all. i was able to get the same answer anyway.
Let's say the lowerleft corner square is white in color. i observed that (1,1), (1,3), (2,2),(3,1),(3,3) all resulted in the same white color (First coordinate is up move and second is right move). If you have (1,2), (2,1), (2,3), and (3,2) you get opposite color which is black.
Therefore for ONE process only we have,
P(whitestarted with white) = 5(1/3)^2 = 5/9
And
P(blackstarted with white) = 4(1/3)^2 = 4/9
Because 1,2,3 are equally likely and they are independent events
Note that the equations above are symmetric.
That was only for the first process. Since we want to end up with white color after three processes. We want
WWW
BBW
WBW
BWW
It is easy to see that P(WWW) = (5/9)(5/9)(5/9) = 125
Now, finding P(BBW) is a bit trickier. The probability of getting B at first process is 4/9. Then to get black again on the next given that we had black first then we have
P(blackstarted with black) = 5/9
Then to get white, we can view the ending color of the second process as the new starting color for the third process
so that P(white black) = 4/9
Thus P(BBW) = (4/9)(5/9)(4/9) = 80/729
a similar logic goes for finding that
P(WBW) = 80/729
P(BWW) = 80/729
Therefore the total probability of having the last color as white also is
(125/729) + (80/729) + (80/729) + (80/729) =365/729 
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