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### #11

Posted 21 March 2008 - 12:57 AM

according to you, the probability of the end space of the ruler being on white was: 23/49

according to you, the probability of the end space of the ruler being on black was: 12/49

those add to 35/49... so what other color is there on the board? lol

### #12

Posted 21 March 2008 - 01:01 AM

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

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)

~~~~

you asked me what "multiple spaces can connect to multiple black squares" means... it means that for a black space, there can be multiple spaces that connect to that color... and in the spaces that connect to that color can also connect to other spaces of both colors (some of them just one though)

What I did, my method, is to generalize them into groups (corner groups, side groups, middle square) and figure out as if each square was the

*starting space*and what other spaces it can connect to. I did the chances of connecting to a white. For example, if one space can go to blacks in two directions, but white if its orientation becomes straight ahead, then it has a 1/3 of ending a white, etc, and I did that for all the spaces (though using symmetry to my advantage to only have to do a quarter of all the squares) and then figuring it out. Then the chance to land on a black is 1-w, where w is the chance to land on a white. I then double-checked the black answer of course.

### #13

Posted 21 March 2008 - 01:02 AM

### #14

Posted 21 March 2008 - 01:05 AM

### #15

Posted 21 March 2008 - 01:07 AM

as for you not being sold, your solution is clearly fallacious somewhere, as it totals to 35/49. There are only two colors on the board ;D

I will try to re-explain WHY it's fallacious:

you say x amount of spaces go to a white space... well most of those spaces can also connect to other white spaces, and black spaces too. You're counting backwards from each space... I suggest thinking forward from each space.

For example, each corner space as a starting space has a 1/1 chance of the other end of the ruler landing on white. The spaces next to the corner spaces have a 1/2 chance of landing on white, the other 1/2 is inward, to a black. etc. Soon, after more people have tried, I'll show my exact methods for each three problems

### #16

Posted 21 March 2008 - 01:22 AM

it does need to add up to 49... if the chances to end on white + the chances to fall on black are less than 49, it means that there are chances to fall on something OTHER than black or white.Black plus white doesn't have to add up to 49 because the ruler is not the same as just picking squares.

But enough of that, you saw the error of your solution cuz you edited your other post

anyways, you wanted my method, I'll give my method for solving #1:

### #17

Posted 21 March 2008 - 01:32 AM

### #18

Posted 21 March 2008 - 01:52 AM

I see you edited your other post ;D you saw the problem with your idea then?

it does need to add up to 49... if the chances to end on white + the chances to fall on black are less than 49, it means that there are chances to fall on something OTHER than black or white.

But enough of that, you saw the error of your solution cuz you edited your other post

anyways, you wanted my method, I'll give my method for solving #1:Spoiler for method for #1

**Edited by EventHorizon, 21 March 2008 - 02:01 AM.**

### #19

Posted 21 March 2008 - 02:03 AM

I think I got it, let me know where I might have gone wrong.

Spoiler for My attempt

I just wrote a small Perl program to go through all possibilities and got the same answer I got above...I feel very confident this is the correct solution.

for($x=0;$x<7;$x++){for ($y=0;$y<7;$y++){if (($x==1||$x==2||$x==4||$x==5)&&($y==1||$y==2||$y==4||$y==5)){$box[$x][$y]=1;}else{$box[$x][$y]=0;}print $box[$x][$y];}print "\n";}$total=0;$black=0;$white=0;$same=0;print "\n";for($x=0;$x<7;$x++){for ($y=0;$y<7;$y++){if($x>1){$total++;if ($box[$x-2][$y]==1){$black++;}else{$white++;}if ($box[$x-2][$y]==$box[$x][$y]){$same++;}}if($y<5){$total++;if ($box[$x][$y+2]==1){$black++;}else{$white++;}if ($box[$x][$y+2]==$box[$x][$y]){$same++;}}if($x<5){$total++;if ($box[$x+2][$y]==1){$black++;}else{$white++;}if ($box[$x+2][$y]==$box[$x][$y]){$same++;}}if($y>1){$total++;if ($box[$x][$y-2]==1){$black++;}else{$white++;}if ($box[$x][$y-2]==$box[$x][$y]){$same++;}}}}print "$white \/ $total\n";print "$black \/ $total\n";print "$same \/ $total\n";

### #20

Posted 21 March 2008 - 02:13 AM

I just wrote a small Perl program to go through all possibilities and got the same answer I got above...I feel very confident this is the correct solution.

for($x=0;$x<7;$x++){for ($y=0;$y<7;$y++){if (($x==1||$x==2||$x==4||$x==5)&&($y==1||$y==2||$y==4||$y==5)){$box[$x][$y]=1;}else{$box[$x][$y]=0;}print $box[$x][$y];}print "\n";}$total=0;$black=0;$white=0;$same=0;print "\n";for($x=0;$x<7;$x++){for ($y=0;$y<7;$y++){if($x>1){$total++;if ($box[$x-2][$y]==1){$black++;}else{$white++;}if ($box[$x-2][$y]==$box[$x][$y]){$same++;}}if($y<5){$total++;if ($box[$x][$y+2]==1){$black++;}else{$white++;}if ($box[$x][$y+2]==$box[$x][$y]){$same++;}}if($x<5){$total++;if ($box[$x+2][$y]==1){$black++;}else{$white++;}if ($box[$x+2][$y]==$box[$x][$y]){$same++;}}if($y>1){$total++;if ($box[$x][$y-2]==1){$black++;}else{$white++;}if ($box[$x][$y-2]==$box[$x][$y]){$same++;}}}}print "$white \/ $total\n";print "$black \/ $total\n";print "$same \/ $total\n";

that would probably be right if each orientation possibility had equal probability....but that is not the case. Each of the 49 squares have equal probability. After a square is chosen, then each of it's orientation possibilities have equal probability.

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