BrainDen.com - Brain Teasers
• 0

## Question

You have a 7x7 board that looks like this:

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)

## 25 answers to this question

• 0

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.

##### Share on other sites
• 0

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

##### Share on other sites
• 0

13 squares, if chosen, will have a 100% chance to have the end on white

16 have a 2/3 chance

16 have a 1/2 chance

and 4 have a 1/3 chance

A weighted average gives 33/49 chance to have the end on a white square.

Subtracting from 1 gives 16/49 chance for the end to land on a black square.

For 9 squares, if chosen, the end will have a 100% chance to be the same color.

8 have a 2/3 chance

16 have a 1/2 chance

12 have a 1/3 chance

and 4 have no chance.

A weighted average gives 26.3333333/49 or 79/147 chance for the end and beginning to end on the same square.

This uses the assumptions I made in the clarification questions.

Edited by EventHorizon

##### Share on other sites
• 0

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.

##### Share on other sites
• 0

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

##### Share on other sites
• 0

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

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)

1. 328/441

2. 113/441

3. 79/147

##### Share on other sites
• 0

#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

##### Share on other sites
• 0

I think you posted before seeing my post above yours ;D

##### Share on other sites
• 0

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

##### Share on other sites
• 0

So, see my post between your two posts for the answers, and in the meantime I'll show your fault:

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

##### Share on other sites
• 0

I'll repost part of it:

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)

1. 328/441

2. 113/441

3. 79/147

~~~~

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.

##### Share on other sites
• 0

I'm not really sold on this, can you give an example showing this multiple connection? or can you show how you got your answers?

##### Share on other sites
• 0

I did the opposite. I took a square and looked at the possibility of how many positions could land on it. I'm still thinking I'm right. Black plus white doesn't have to add up to 49 because the ruler is not the same as just picking squares.

##### Share on other sites
• 0

Like I said, I will show my method (even though I just explained it mostly in the post above yours) after more people have tried

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

##### Share on other sites
• 0

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

Black plus white doesn't have to add up to 49 because the ruler is not the same as just picking squares.

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:

Each chance is the chance that that space (the starting space) will cause the ruler's end space to fall on white:

you arrive at 328/441

##### Share on other sites
• 0

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

first calculate the total possible spaces.

16 with 2 possible orientations(corner spots).

24 with 3 possible orientations(side spots).

9 with 4 possible orientations(middle spots).

giving a total of 16*2+24*3+9*4=140

since there are less black spots than white I will count how many ways you can end on a black spot in one of the 4 black groups and multiply by 4 to get the total. This works because of symetry. Using the bottom right square as a refference.

bottom right = 2 possible ways

bottom left = 3 possible ways

top right = 3 possible ways

top left = 4 possible ways

(2+3+3+4)*4 = 48 ways to land on black.

this means that 48/140 land on black and 92/140 land on white.

to figure out how many ways to land on the same color I first tackled landing on a black spot.

using the same technique as above...

bottom right = 0 possible ways.

bottom left = 1 possible way.

top right = 1 possible way.

top left = 2 possible ways.

so black to black = 4*4 = 16.

for white it will be more difficult.

corner = 2 ways * 4 corners = 8

next to corner = 1 possible way * 8 spots = 8

next one out = 2 possible ways * 8 spots = 16

next one out = 3 possible ways * 4 spots = 12

next one in = 1 possible way * 4 spots = 4

next one in = 2 possible ways * 4 spots = 8

center spot = 4 possible ways *1 spot= 4

so white to white = 60

52+16= 76 /140

1:92/140

2:48/140

3:76/140

I will make a program later to validate these results. If anyone can see why I would be wrong let me know.

##### Share on other sites
• 0
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:

Each chance is the chance that that space (the starting space) will cause the ruler's end space to fall on white:

you arrive at 328/441

hmm...those are the same numbers I had in the grid. I don't think I took the weighted average wrong......

(13/49)*1 + (16/49) *(2/3) + (16/49) * (1/2) + (4/49) * (1/3) = 33/49.

How are you averaging those numbers together to get 328/441?

Edited by EventHorizon

##### Share on other sites
• 0
I think I got it, let me know where I might have gone wrong.

first calculate the total possible spaces.

16 with 2 possible orientations(corner spots).

24 with 3 possible orientations(side spots).

9 with 4 possible orientations(middle spots).

giving a total of 16*2+24*3+9*4=140

since there are less black spots than white I will count how many ways you can end on a black spot in one of the 4 black groups and multiply by 4 to get the total. This works because of symetry. Using the bottom right square as a refference.

bottom right = 2 possible ways

bottom left = 3 possible ways

top right = 3 possible ways

top left = 4 possible ways

(2+3+3+4)*4 = 48 ways to land on black.

this means that 48/140 land on black and 92/140 land on white.

to figure out how many ways to land on the same color I first tackled landing on a black spot.

using the same technique as above...

bottom right = 0 possible ways.

bottom left = 1 possible way.

top right = 1 possible way.

top left = 2 possible ways.

so black to black = 4*4 = 16.

for white it will be more difficult.

corner = 2 ways * 4 corners = 8

next to corner = 1 possible way * 8 spots = 8

next one out = 2 possible ways * 8 spots = 16

next one out = 3 possible ways * 4 spots = 12

next one in = 1 possible way * 4 spots = 4

next one in = 2 possible ways * 4 spots = 8

center spot = 4 possible ways *1 spot= 4

so white to white = 60

52+16= 76 /140

1:92/140

2:48/140

3:76/140

I will make a program later to validate these results. If anyone can see why I would be wrong let me know.

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";
[/codebox]```

##### Share on other sites
• 0
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";
[/codebox]```

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.

##### Share on other sites
• 0
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.

What I meant was that each valid square orientation pair does not have equal probability.....just thought I would fix my horrible wording....

##### Share on other sites
• 0

hmm...those are the same numbers I had in the grid. I don't think I took the weighted average wrong......

(13/49)*1 + (16/49) *(2/3) + (16/49) * (1/2) + (4/49) * (1/3) = 33/49.

How are you averaging those numbers together to get 328/441?

Ah yes I'm sorry... I screwed up. I accidentally thought the 4 corners were all 1 instead of two 1's and two 1/2's, so I added on extra small amount which had to make me multiply 49 out to 441 to get a whole number numerator. Gah. I screwed up, I don't know what I was thinking. I had the numbers right but I misread my own handwriting when I was figuring out the averages. lol. To see exactly what I did wrong, pretend that the 1/2's adjacent the corners were 1's instead, that's what I read from my handwriting as I calculated the numbers (I did it all in my head), so yeah that came out to 328/441 (441 is 49*9). But yeah woops, sorry

1. 33/49

2. 16/49

3. 79/147

and the interesting thing about #1 and #2:

Hmph, maybe my problem wasn't so good after all, OR this is a coincidence, OR it's a unique property, but 33/49 is the chance that the first part of the ruler will fall on a white (ie, there are 33 white spaces) and 16/49 is for black (there are 16 black spaces on the board). What do you think, EventHorizon?

##### Share on other sites
• 0
and the interesting thing about #1 and #2:

Hmph, maybe my problem wasn't so good after all, OR this is a coincidence, OR it's a unique property, but 33/49 is the chance that the first part of the ruler will fall on a white (ie, there are 33 white spaces) and 16/49 is for black (there are 16 black spaces on the board). What do you think, EventHorizon?

That is interesting...it makes me think that perhaps there is a very simple lateral thinking way of solving this puzzle.

I can't quite seem to figure out what it would be yet. I'll think about it more tomorrow...I should really work on a paper...but have been avoiding it a little too much. It seems I would much rather solve puzzles and watch my team lose (grr!) than work on my paper.

##### Share on other sites
• 0

Each chance is the chance that that space (the starting space) will cause the ruler's end space to fall on white:

you arrive at 328/441

I think these probabilities apply to a ruler with length 2, not length 3.

Hint: Imagine the ruler begins in the center of a square.

##### Share on other sites
• 0

bonanova: Yes, 2 spaces long if you want to get picky, if it's from the center of one square to the center of the other, but we practically mean that it covers 3 spaces, the starting space, middle space, and end space. If it started at the middle of each square it would make it 2 spaces long then. If it started at the ends it would make it 3. It starts at the ends ;D and did you see that my 328/441 was incorrect, the chances are 33/49 for white and 16/49 for black (and 79/147 to land on the same color)

EventHorizon: yes that's what I was thinking. Maybe it's due to the fact that for a space that can reach less spaces, less spaces can reach that space too. I dunno, I'll think about it some more ;D good luck on your paper!

##### Share on other sites
• 0

still haven't thought of it...

## Create an account

Register a new account