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25 replies to this topic

#1 unreality

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Posted 20 March 2008 - 08:46 PM

You have a 7x7 board that looks like this:

7by7board.jpg

There are white spaces and black spaces on the board, as shown. You have a ruler three spaces long that are you going to put on the board... a random space is picked out of the 49 spaces (either random 7 to determine the row, and then random 7 to determine the column, or just random 49 to determine the square number, whatever, somehow you randomly determine, with equal chances, a square from the board). This random square on the board is where you start the ruler, and you place it facing any of the four orthogonal directions (left, right, up, down), randomly (ie, random out of 4). Obviously if it's in the corner it's only random out of 2 directions, or 3 directions if it's a side piece. (Remember, the ruler is 3 spaces long)

1. What is the probability that the ruler's end will fall on a white space?
2. What is the probability that the ruler's end will fall on a black space?
3. What is the probability that the ruler's end will land on a color that is the same color as the space under the ruler's beginning? (ie, one end is on black, the other end is on black, or one is on white and the other end is on white)
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#2 EventHorizon

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Posted 20 March 2008 - 10:10 PM

just a couple clarification questions...

1. whenever the end would fall outside the board, you can't choose that orientation....right? So one square away diagonally from a corner square (a gray square in all cases) only has 2 valid orientations to choose from just like the corner, etc.

2. is the ruler starting from the center of the chosen square or covering it completely? In other words, once the ruler is down, would 3 or 4 squares have the ruler covering part or all of them? I would assume only 3, but wanted to make sure.
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#3 toddpeak

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Posted 20 March 2008 - 10:22 PM

One more clarification:

The "ruler's end" means the one that wasn't randomly chosen, right? Or do we need to take into account the 1/49 chance of each square at the beginning? What I'm asking is, do we count both ends or just the "second" end after the "first" one's location has been randomly chosen.

Oops, never mind. I see the difference between "end" and "beginning" now, my bad.

Edited by toddpeak, 20 March 2008 - 10:22 PM.

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#4 EventHorizon

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Posted 20 March 2008 - 10:36 PM

Spoiler for answers...

Edited by EventHorizon, 20 March 2008 - 10:39 PM.

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#5 unreality

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Posted 20 March 2008 - 11:41 PM

EventHorizon has the wrong answers ;D except for #3, which is correct

Clarification:

yes the ruler covers 3 spaces, the "starting space", the "middle space" and the "end space"

the "starting space" of the ruler has 1/49 chance to land on each of the squares. In other words, the starting space of the ruler is at a random space on the board

the "end space" is a space two over from that, in one of the 4 orthogonal directions, but RANDOMLY. If the ruler would end up going off the board in one direction, it cant go in that direction. For example, if the starting space is in the upper left corner, it's equally random choices are down or right since there's no room to go up or left. And remember, like I hinted in my post, the ruler has THREE spaces and is 3 long, so "corner spaces" aren't just the 4 corner spaces, there are 16 corner spaces. And the "side spaces" that would only have 3 directions aren't just the few along each side, they go out 1 as well. Just like EventHorizon said:

whenever the end would fall outside the board, you can't choose that orientation....right? So one square away diagonally from a corner square (a gray square in all cases) only has 2 valid orientations to choose from just like the corner, etc.


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#6 itachi-san

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Posted 21 March 2008 - 12:45 AM

For #2, the probability of landing on a black square is 12/49.

Reasoning: take each 4 square black section. 1 square (the outer-most) has 2 possibilities, 2 squares have 3 and 1 square (the inner-most) has 4.

Added together, this makes 12 * 4 groups of black squares = 48.

And the probability for the ruler is 1/49 for the first square * 1/4 for where it lands (only 4 possible places) = 196

48/196 = 12/49
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#7 unreality

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Posted 21 March 2008 - 12:53 AM

Nope, because multiple squares can connect to multiple black spaces.

It seems everyone is having a harder time than I thought! :D

I'll post my answers, and leave you guys to figure out the method still (I'll post my method too, but after a few other people have tried)

Spoiler for my answers...

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#8 itachi-san

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Posted 21 March 2008 - 12:53 AM

#1 is 23/49 by the same reasoning:

There are 12 white squares that have 2 possibilities (at the corners)
There are 16 white squares that have 3 possibilities (by the centers of each boundary line)
There are 5 white squares that have 4 possibilties (in the center)

=24 + 48 + 20 = 92 possibilities

divided by 196 (49 possibilities for the beginning of the ruler * 4 possibilities for the end)

=92/196

=23/49
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#9 unreality

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Posted 21 March 2008 - 12:55 AM

I think you posted before seeing my post above yours ;D
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#10 itachi-san

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Posted 21 March 2008 - 12:55 AM

what does:

multiple squares can connect to multiple black spaces

mean?

Oh i get it! Makes more sense if you say 1 square can connect to multiple black spaces

Edited by itachi-san, 21 March 2008 - 12:58 AM.

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