perplexus.info :: Probability : Checkerboard Chance Conclusion

Checkerboard Chance Conclusion (Posted on 2022-02-07)

Terence has a checkerboard with 10 rows and 10 columns, a single checker at the bottom-left corner, 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 the 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.

He repeats this process twice more, thus spinning the spinner for a total of six times from the outset.

Determine the probability that the checker will land on the same color as the square on which it started.

No Solution Yet Submitted by K Sengupta

Rating: 5.0000 (1 votes)

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Solution

| Comment 2 of 9 |

I already read tomarken's solution but mine is a bit more explicit.

Note the problem is equivalent to making 6 moves each with a 1/3 chance of the color staying the same and 2/3 of switching.

(x(n),y(n)) = color distribution probability with n spins and x the starting color.

(x(n+1),y(n+1))=(x(n)+2y(n),2x(n)+y(n))/3

Which leads to the sequence

n=0 (1,0)

n=1 (1,2)/3

n=2 (5,4)/9

n=3 (13,14)/27

n=4 (41,40)/81

n=5 (121,122)/243

n=6 (365, 364)/729

And the general formula (1 + (-1/3)^n)/2

Posted by Jer on 2022-02-07 11:00:46

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