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## Checkerboard Chances

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

 Puzzle ID: #42219 Fun: (2.32) Difficulty: (2.89) Category: Probability Submitted By: billy314

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?

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 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(white|started with white) = 5(1/3)^2 = 5/9 And P(black|started 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(black|started 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