All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
Back to throne (Posted on 2015-05-21) Difficulty: 2 of 5
Imagine a chess King placed in the central square C(5,5) of a 9x9 “checkerboard”. King’s step consists of his displacement to the neighboring square in one of the 8 directions.

What is the probability that his 4-step random walk terminates at square C?

See The Solution Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution One way to solution | Comment 1 of 3
First consider everywhere the King can get in 2 moves and the number of ways to get there.

1 2 3 2 1
2 2 4 2 2
3 4 8 4 3
2 2 4 2 2
1 2 3 2 1

Note: the 8 is on the starting square.

Now from any of these squares there are the same number of ways to get back to the center in 2 moves, so we can just square all these numbers and add them up.

The sum is 216.  There are 8^4 = 4096 four-move sequences so

216/4096 = 27/512 = .052734375

Incidentally you could use an ordinary chessboard for this problem.  The starting board really only needs to be 7x7, since if the first three moves bring you that far out, the fourth can't bring you back.

  Posted by Jer on 2015-05-21 09:40:07
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information