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+2
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Question:

8 rooks are randomly placed on different squares of a chessboard. A rook is said to attack all of the squares in its row and its column.

Compute the probability that every square is occupied or attacked by at least  rook.

P.S.

I think I may have figured out the answer, which according to my calculations is (2*8^8 - 8!)/(64*63*62*61*60*59*58*57) from the Inclusion-Exclusion principle that I used in finding it. Yet I hope someone can verify my answer's correctness since I am not sure if it is accurate.

Also, I posted this last night but didn't got any responses so I hope someone can help me cuz I need the help asap(It's due today)

Jun 29, 2022

#1
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Also, I posted this last night but didn't got any responses so I hope someone can help me cuz I need the help asap(It's due today)

You’re an idiot waiting until the last day to ask for help for a question this complex and laborious.

--. .-

Jun 29, 2022
#2
+2440
-1

Is this really your calculation? It looks like you copied it from somewhere.

If it is yours, then how did you formulate your calculation?

It is unlikely what you presented is the correct probablity.

Re-Present your calculation as an equation with the work product: delineate your calculations showing how the Inclusion-Exclusion principle applies to your calculation. Specifically, define the included set, the excluded set, and the union of these two sets.

If you do this, then it should be easy for someone to verify your solution, without deriving and solving this. (As an analogy, it’s easy to verify the primes of large numbers; it’s not so easy to find the primes in the first place.)

If you do decide to do this, use LaTex. Reading math in ASCII slop is often irritating.

GA

--. .-

GingerAle  Jun 29, 2022
#3
+118604
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I think it is just

$$\displaystyle \frac{8!}{64C8}$$

Jun 29, 2022
edited by Melody  Jun 29, 2022
edited by Melody  Jun 29, 2022
edited by Melody  Jun 29, 2022
#4
+2440
+1

There is no casual solution to this problem; though the solution process is not ultra complex.

The use of the Inclusion-Exclusion principle will accelerate the process. Like usual, its application requires careful and (sometimes) laborious attention to avoid over-counting successes and failures.

The question resolves to a binary state of either an arrangement of eight (8) rooks occupying or attacking every square, or it does not. Analyzing the counts of unoccupied and un-attacked squares is not requires for this question.

(Permutations of the rooks among themselves are not necessary, as this will be done for both success and failure counts and will factor out.)

Setup:

Vertical format.

Starting with an 8 X 8 chessboard place eight rooks on the top row (row 1) and observe that every square is either occupied or attacked by a rook. Moving the rooks vertically (row-wise) still allows for every square to be either occupied or attacked by a rook. There are (8^8) unique positions when moving the rooks vertically.

This same sequence is repeatable in a horizontal format.

Place the eight rooks in the left column (Column 1) observe that every square is either occupied or attacked by a rook. Moving the rooks horizontally (column-wise) still allows for every square to be either occupied or attacked by a rook. There are (8^8) unique positions when moving the rooks vertically. HOWEVER, in this horizontal format, there are two (2) cases where the positions of the rooks identically match the positions in the vertical format. These two (2) cases occur when the rooks align diagonal positions across the board.  This occurs twice: from the top-left to the bottom-right, and from the bottom-left to the top-right.  These two (2) cases are exclusions and need to be subtracted from the total. So at this point, the subtotal for success counts are (8^8) + [(8^8) – (2)].

At this point it’s (apparently) discernable that for every square to be either occupied or attacked by a rook, every column or every row must have at least one rook. Removing all rooks from a column leaves some (not all) row squares on that column un-attacked.  Like-wise, removing all rooks from a row leaves some (not all) column squares on that row un-attacked.

If the above is true, then these counts (8^8) + [(8^8) – (2)] are the only successes.

Probability of a random arrangement of eight (8) rooks to occupy or attack every square on a chessboard:

$$\rho_{(s)} = \dfrac {(2*8^8)-2} {\binom {64}{8}} \approx 0.75809\%$$

GA

--. .-

GingerAle  Jun 29, 2022
#5
+118604
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Thanks Ginger.

My problem is that I do not understand the question.

(And at present I am too unenthused to try and work it out from what you have said.)

Maybe I will become more enthused later...

Melody  Jun 29, 2022
edited by Melody  Jun 29, 2022
#6
+118604
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Hi Ginger,

I finally got a chess board out and looked at the problem.  I understand it now.

I think that the 8 rooks must be placed 1 in every column which is  8^8

or

one in every row which is 8^8

But I have to subtract all the ones that are in both every row and every column (becasue they have been added twice.)

This number is 8!

So I am now pretty certain that the answer is :

$$\displaystyle \frac{2*8^8-8!}{64C8}= \frac{33 554 432-40320}{4426165368} = \frac{33514112}{4426165368}\approx 0.75718\%$$

Jul 2, 2022
#7
+2440
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Hi Melody,

But I have to subtract all the ones that are in both every row and every column (becasue they have been added twice.)This number is 8!

You’ve subtracted (8!) from the total successful arrangements. That is a lot duplication that is NOT indicated in the positional analysis of the rooks. The only time there is a rook in every row and every column is when they are positioned diagonally across the board.  Other than this, one or more rows and/or one more columns are unoccupied, but not unattacked.

As I described above, this diagonal arraignment occurs twice: once, descending from the top right to the bottom left, and again from the bottom left to the top right. These diagonal arraignments are the ONLY two arrangements that occur in both the vertical movement progression and the horizontal movement progression.  There are no other duplicates, so where does the (8!) come from?

Appended to this post are images depicting the chessboard with eight rooks in horizontal and vertical formats, demonstrating their respective movements on the chessboard and points of interest.

If I am missing (8!-2) arrangements where these two formats intersect, then I have royally rooked myself. I would really like to see them. Can you graphically demonstrate where some of these duplications occur?

GA

--. .-

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*****

GingerAle  Jul 4, 2022
#8
+118604
+1

Hi Ginger,

Sorry about the late response, Ive been in flood the last couple of days and have had not electricity of internet.

It is just luck that i saw thisat all.

-----------------------------------

Every row has a rook

and every column has a rook

So it has been counted twice.

There are 8! such combinations that have been counted twice.

So I am stickig by my answer.

Jul 7, 2022
#9
+2440
-1

Hi Melody,

OMG! I saw two trees but missed the whole bloody forest.

[You’re in a flood, but I’m the one that’s all wet...]

Until I saw your graph, I had tunnel vision. I didn’t really do a (full) positional analysis. I only compared the positions of the two sets for each simultaneous sequence.   The two diagonals are the only two that occur simultaneously in the 8! counts.

It’s Spooky Quantum Dumbness at a distance and I walked right into it, face first, because of quantum blindness.

Lancelot Link would throw banana peels and coconuts at me for that.

Very Cool! Thank you.

GA

--. .-

GingerAle  Jul 7, 2022
#10
+118604
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LOL Ginger,

It is easy to wear blinkers with probabilitiy questions :)

But yea, the forest was hiding behind two trees.  I like that :)

Melody  Jul 7, 2022