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Help me out with my math exam Help me out with my math exam

09-22-2014 , 04:06 AM
So I had my math exam today about probability. The last exercise was this:

If pieces A and B are put randomly on a chess board, what's the probability that the squares they are in touch each other.

I assumed that the touching means that the squares contact each other diagonally or that they are next to each other.

My solution:

(4/64)*(3/64)+(24/64)*(5/64)+(36/64)*(8/64)=0,10253...

It really bothers my that did I calculate this correctly, please help me
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09-22-2014 , 09:47 AM
If that were my exam, I wouldn't assume diagonals count, I'd ask the professor.

P(touch orthogonally) = 8 * 7 * 2 / C(64,2)
(8 rows, 7 ways to touch horizontally on each row. Same with the columns.)

P(touch diagonally) = (7*2) + 4(1+2+3+4+5+6) = [14 + 4*C(7,2)] / C(64,2)
(2 main diagonals of length 8, then 4 diagonals of lengths 2 to 7.)
Sum from 1 to 6 = C(7,2) and there might be a way to get C(7,2) by different reasoning, which I'll think about.

P(touch either way) = sum of those = 5/48 or 10.41666...%
Help me out with my math exam Quote
09-22-2014 , 10:11 AM
Can anyone confirm that the suggested solution is right?
Help me out with my math exam Quote
09-22-2014 , 10:28 AM
Someone will confirm/deny it, but meanwhile, decide if you agree with my reasoning. If you don't, state your objections.

Also explain the reasoning behind what you tried. Our answers are close, so you might have only made a small mistake and I can point it out.
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09-22-2014 , 03:17 PM
Quote:
Originally Posted by errorpoker
So I had my math exam today about probability. The last exercise was this:

If pieces A and B are put randomly on a chess board, what's the probability that the squares they are in touch each other.

I assumed that the touching means that the squares contact each other diagonally or that they are next to each other.

My solution:

(4/64)*(3/64)+(24/64)*(5/64)+(36/64)*(8/64)=0,10253...

It really bothers my that did I calculate this correctly, please help me
This approach is fine except the 2nd piece has only 63 squares available and not 64, so it's (4/64)*(3/63)+(24/64)*(5/63)+(36/64)*(8/63) ≈ 10.4167%

or more simply: (4*3 + 24*5 + 36*8)/P(64,2) ≈ 10.4167%

Last edited by R Gibert; 09-22-2014 at 03:37 PM.
Help me out with my math exam Quote
09-22-2014 , 03:33 PM
Quote:
Originally Posted by heehaww
If that were my exam, I wouldn't assume diagonals count, I'd ask the professor.

P(touch orthogonally) = 8 * 7 * 2 / C(64,2)
(8 rows, 7 ways to touch horizontally on each row. Same with the columns.)

P(touch diagonally) = (7*2) + 4(1+2+3+4+5+6) = [14 + 4*C(7,2)] / C(64,2)
(2 main diagonals of length 8, then 4 diagonals of lengths 2 to 7.)
Sum from 1 to 6 = C(7,2) and there might be a way to get C(7,2) by different reasoning, which I'll think about.

P(touch either way) = sum of those = 5/48 or 10.41666...%
There is no reason to be dividing by C(64,2) "as you go along." Count the combos 1st and divide once is simpler.

Your 2nd paragraph is needlessly complicated. With a diagonal pairing argument so that you always get seven way for pieces to be touching e.g. a1-h8 diag with the a8 square, b1-h7 diag with
the a7b8 diag, etc. you would then count 7*7*2 for 98 combos.

The answer is then (8*7*2 + 7*7*2)/C(64,2) ≈ 10.4167%
Help me out with my math exam Quote
09-22-2014 , 03:57 PM
Quote:
Originally Posted by R Gibert
There is no reason to be dividing by C(64,2) "as you go along." Count the combos 1st and divide once is simpler.
I did that so as to answer both possible interpretations at once.
Quote:
With a diagonal pairing argument so that you always get seven way for pieces to be touching e.g. a1-h8 diag with the a8 square, b1-h7 diag with
the a7b8 diag, etc. you would then count 7*7*2 for 98 combos.
Nice, I agree that's a little better. I did what first came to mind, which wasn't much more complicated since I know choose-2's by heart: 14+4(21) vs 7*7*2

I hope this exam was already taken like OP said, and that we didn't just help him cheat on an online exam.
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09-22-2014 , 04:24 PM
Quote:
Originally Posted by heehaww
I hope this exam was already taken like OP said, and that we didn't just help him cheat on an online exam.
That's doesn't matter. If true, a person who BSs his way through life is creating a hell of his own making. I not going to worry about it.
Help me out with my math exam Quote
09-22-2014 , 05:08 PM
Quote:
Originally Posted by heehaww
I hope this exam was already taken like OP said, and that we didn't just help him cheat on an online exam.
Ye I did it today, didn't cheat just curious to know what's the answer, since we get them back only after a few weeks.

Quote:
Originally Posted by R Gibert
This approach is fine except the 2nd piece has only 63 squares available and not 64, so it's (4/64)*(3/63)+(24/64)*(5/63)+(36/64)*(8/63) ≈ 10.4167%

or more simply: (4*3 + 24*5 + 36*8)/P(64,2) ≈ 10.4167%
Why does the 2nd piece only have 63 squares available, I kinda thought that you would just randomly put the piece anywhere so it could technically go in the same square as the 1st piece. I did not count that as "touching" though
Help me out with my math exam Quote
09-22-2014 , 06:40 PM
Quote:
Originally Posted by errorpoker
Ye I did it today, didn't cheat just curious to know what's the answer, since we get them back only after a few weeks.



Why does the 2nd piece only have 63 squares available, I kinda thought that you would just randomly put the piece anywhere so it could technically go in the same square as the 1st piece. I did not count that as "touching" though
Your professor will mark your answer wrong I'm afraid. Perhaps you will get partial credit for producing an answer that was otherwise on the right track. Some professors might. Good luck.

When you ask him/her for the partial credit, make sure you've practiced your pathetic look in the mirror. Every little bit can help. In my own case, I would not need to practice. I'm a "natural"
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09-22-2014 , 06:56 PM
That's another reason the question was poorly worded. There are 4 valid interpretations of the question, and hence 4 equally correct answers. The best play would have been to email your prof immediately after taking the exam (assuming it was online) and explain why that question was vague. A math prof should know better. In math, everything has to be well-defined and answers should have 0 element of subjectivity.
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09-23-2014 , 07:22 PM
Here's another simple approach.

Consider only columns and sweep from left to right. For any column, the top square touches 3 others not previously counted – right, down diagonal and below. The next 6 squares touch 4 others, up diagonal, right, down diagonal and below. The bottom square touches just 2 others, up diagonal and right. So, each of 7 columns touch 3+6*4+2 = 29 squares for a total 203 touches. Now the last column has no column to the right but each of 7 squares has a touching square on the bottom, so total touches is 210

Prob = 210/C(64,2) = 10.42%.
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