Calculating Probability of Non-Capturing Rooks on a Chessboard

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Homework Help Overview

The problem involves calculating the probability that 8 rooks placed randomly on a chessboard do not threaten each other, meaning no row or column contains more than one rook. This falls under the subject area of combinatorial probability.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the total number of arrangements for placing rooks with and without restrictions. Questions arise regarding the validity of certain calculations, particularly the use of factorials in determining arrangements.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning the assumptions made about the arrangements of the rooks. Some guidance has been offered regarding the placement of rooks and the implications of restrictions on their positioning.

Contextual Notes

There is a noted confusion regarding the equivalence of the problem statements and the impact of rook color on capture possibilities, which may affect the interpretation of the problem.

fk378
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Homework Statement


If 8 rooks (castles) are randomly places on a chessboard, compute the probability that none of the rooks can capture any of the others. That is, compute the probability that no row or file contains more than one rook.


The Attempt at a Solution


I just started it by knowing there are 64 squares on a chessboard. If there are 8 rooks, then that leaves 56 empty blocks.

Where to go from here?
 
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How many different ways to put the rooks on the board without the no-capture restriction? How many ways with the restriction?
 
no-capture restriction: 8! ways
without the restriction: 56! ways
 
Not to give it away, but so you can check: P(no capture) [tex]\approx .911\times 10^{-5}[/tex].
 
I don't understand how that is? Can you explain?
 
Can you start by explaining why you think 56! is the number of ways to place them without restriction? That is certainly wrong.
 
First, place a rook in a random spot on the board. How many ways are there to do this? Next, figure out how many places are left on the board where the second rook can't take the first rook. Now, how many ways are there to place another rook on the board so that it can't take either of the first two. Continue this until you get to the last rook (there should only be one space left for that one). Finally, figure out how many possible ways there are to place the 8 rooks on the board with no restrictions.
 
2w7mm89.jpg


So you need one rook in one row.

[tex]P(A)=\frac{m}{n}[/tex]

8th row - the rook can move on 8! ways

Can you find the probability now?

Notice: you got 8 rooks so n=8 * ?
 
fk378 said:

Homework Statement


If 8 rooks (castles) are randomly places on a chessboard, compute the probability that none of the rooks can capture any of the others. That is, compute the probability that no row or file contains more than one rook.

The first sentence is not equivalent to the second. :-p
 
  • #10
epenguin said:
The first sentence is not equivalent to the second. :-p

Why not?
 
  • #11
Because capture possibility depends on the colour of the rooks. :smile:
 

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