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 Trading Places 2 (Posted on 2006-05-30)
n white markers are placed on the first n squares of a row of 2n+1 squares.
There is a space of 1 square and then n black markers.
White markers can only move right.
Black markers can only move left.
Markers can move forward one square, or can jump over a marker of either colour if there is an empty square to land on.
Markers are not removed from the board if jumped.
You DO NOT have to alternate moving black and white markers.

a) Find an algorithm to solve this puzzle.
b) How many moves does it take to complete?
c) If you make random moves what is the probability of completion?

 No Solution Yet Submitted by Sir Percivale Rating: 4.0000 (1 votes)

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 Formula for Part C Comment 3 of 3 |
(In reply to A start to part C by tomarken)

I don't have a proof or anything, but I think the probability of successfully completing this game using random moves is:

1 / [3^(2n-4)][2^(n^2-2n+4)]

for any n > 1.  That's ugly, I know.

This might clean it up a bit:

Let x = 2n-4
and y = n^2 - 2n + 4

Then the probability is:

1/(3^x)(2^y)

Anyway, it confirms the numbers I got earlier.  Now the real trick - can someone confirm if this is correct?

 Posted by tomarken on 2006-05-30 14:24:10

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