Broken Stick Puzzle

Passacaglia · 08-01-2008, 02:34 PM

A perfectly straight and very thin rod is marked at random at two points, then divided into three parts at those points. What is the probability that you can form a triangle from the three parts?

MikeVic · 08-01-2008, 02:35 PM

1 in 3.

st.cronin · 08-01-2008, 02:39 PM

QuikSand · 08-01-2008, 02:42 PM

Well, you have no triangle if the largest segment is longer than or equal to the sum of the other two. Pretty sure that would be the only case where you'd fail to make a triangle.

So, on a scale of 0-1, how often will you have a segment with length 0.5 or longer -- isn't this essentially the question? Anyone see something I'm missing?

gstelmack · 08-01-2008, 02:42 PM

100%. You're just going to take the two outer pieces and rotate them up until they meet. Take a piece of string, loop it, put three fingers in (for the three corners), and move the fingers around.

That's my story and I'm sticking to it.

gstelmack · 08-01-2008, 02:43 PM

Quote:

Originally Posted by QuikSand

Well, you have no triangle if the largest segment is longer than or equal to the sum of the other two. Pretty sure that would be the only case where you'd fail to make a triangle.

So, on a scale of 0-1, how often will you have a segment with length 0.5 or longer -- isn't this essentially the question? Anyone see something I'm missing?

I've changed my story. I'm with Quik.

st.cronin · 08-01-2008, 02:44 PM

ok seriously

Under what conditions will three lines NOT make a triangle? When 1 line is longer than the other two combined, right. So if both "cuts" are on the same half of the stick, no triangle.

My answer is 1 : 1, on the theory that the second cut is equally likely to be on one half of the stick as the other.

QuikSand · 08-01-2008, 02:59 PM

I don't know intuitively how to calculate this, but let's think of this road as reaching from 0 to 1. SO, brute force below...

First cut is a random place, let's put it at X. For our orientation purposes here, it's somewhere on the interval from 0 to 0.5.

So, the the places that create a SuperRod (0.5 or longer) include those positions between (X+0.5) and 1.

My shortcut guess here is since the median value for X here is 0.25, this probability is going to smooth out to 1 in 4. But I may well be missing something more complex in the distribution.

Thinking a bit more... As X approached 0.5, the chance of a SuperRod approaches zero. As X approaches 0, the chance of a SupeRod approaches 0.5. More support for the answer of 0.25.

gkb · 08-01-2008, 03:04 PM

Quote:

Originally Posted by QuikSand

I don't know intuitively how to calculate this, but let's think of this road as reaching from 0 to 1. SO, brute force below...

First cut is a random place, let's put it at X. For our orientation purposes here, it's somewhere on the interval from 0 to 0.5.

So, the the places that create a SuperRod (0.5 or longer) include those positions between (X+0.5) and 1.

My shortcut guess here is since the median value for X here is 0.25, this probability is going to smooth out to 1 in 4. But I may well be missing something more complex in the distribution.

Thinking a bit more... As X approached 0.5, the chance of a SuperRod approaches zero. As X approaches 0, the chance of a SupeRod approaches 0.5. More support for the answer of 0.25.

Did I miss a post about worthwhilemoney?

gkb · 08-01-2008, 03:07 PM

Dola...kind of found what I was looking for in one of your posts.

Quote:

In time, I want to use this site to identify this sort of available proposition — where we can play one market against another (or against some other comparable betting lines) to ensure a guaranteed yield. This will be one major element of a “beat the markets” strategy that I am advocating here.

And man is this off-topic for this thread. Sorry...carry on.

st.cronin · 08-01-2008, 03:09 PM

Quote:

Originally Posted by QuikSand

Thinking a bit more... As X approached 0.5, the chance of a SuperRod approaches zero. As X approaches 0, the chance of a SupeRod approaches 0.5. More support for the answer of 0.25.

I think this is correct, and the only real explanation needed.

QuikSand · 08-01-2008, 03:11 PM

Quote:

Originally Posted by gkb

Did I miss a post about worthwhilemoney?

No, haven't posted anything here, but feel free to drop by.

gstelmack · 08-01-2008, 03:14 PM

There's a 50% chance that the first cut makes this impossible (by being over the halfway mark already).

Out of the remaining 50% (where at least half the rod is available), there is a 50% chance the second cut will create a breaking rod.

That lends credence to the 25% theory Quik put forth.

Maple Leafs · 08-01-2008, 03:15 PM

Quote:

Originally Posted by QuikSand

Thinking a bit more... As X approached 0.5, the chance of a SuperRod approaches zero.

Are you sure? If X is 0.4999, you still get a super-rod if the second cut is anywhere to the low side of that, no?

QuikSand · 08-01-2008, 03:16 PM

Quote:

Originally Posted by gstelmack

There's a 50% chance that the first cut makes this impossible (by being over the halfway mark already).

I don't quite get how this is true. Maybe I over-abbreviated with my range of 0 to 0.5?

Put it this way -- any spot on the "right" side of the bar is just the mirror image of another spot on the left side of the bar. Flip the bar around, if you prefer, to get thevalues back to a more easy-to-grasp 0-0.5 range. Agreed?

QuikSand · 08-01-2008, 03:17 PM

Quote:

Originally Posted by Maple Leafs

Are you sure? If X is 0.4999, you still get a super-rod if the second cut is anywhere to the low side of that, no?

Yeah, I was just going back to edit in -- I have isolated one way to do it, but that ignored the other. New post coming to add that in. Duh.

BrianD · 08-01-2008, 03:18 PM

That doesn't sound right. As X approaches 0, the chance of a SupeRod approaches 1.

st.cronin · 08-01-2008, 03:19 PM

Quote:

Originally Posted by Maple Leafs

Are you sure? If X is 0.4999, you still get a super-rod if the second cut is anywhere to the low side of that, no?

What I think Quik means is that, if the first cut cuts the rod in half, there is no chance that the second cut will not create a triangle. The closer the first cut is to cutting it in half, the smaller the spectrum of cuts that will make a triangle.

The closer the first cut is to being a 0 cut (cutting off an infitesimally small piece), the larger the spectrum of possible 2nd cuts for a triangle (actually I am not sure this is true - so I'm not quite convinced this is the right answer).

Thinking, sketching, brb.

QuikSand · 08-01-2008, 03:20 PM

I guess you also need to add in (to my previous thinking above) another way to make the SuperRod -- when the second cut is to the "left" of the first.

Thus, any value less than X as also creating a super-rod, which I'd guess in itself adds in another 0.25 chance.

I hate it when one of the obvious "dartboard" answers happens to be right... but that sure sounds like 0.5 in total.

For any value X, your two ways of getting a SuperRod are X and 1-(X+0.5)... and the sum of those two simplifies nicely to 0.5. Ugh.

st.cronin · 08-01-2008, 03:20 PM

Quote:

Originally Posted by BrianD

That doesn't sound right. As X approaches 0, the chance of a SupeRod approaches 1.

I think it approaches .5, actually.

st.cronin · 08-01-2008, 03:25 PM

The most logical way I can think of is like this:

Code:

______________________
.5        0

The .5 represents what happens if the first cut is infitesimally small - a cut on the very end gives a 50 percent chance of the second cut making a triangle impossible. The 0 represents what happens if the first cut is exactly halfway - it makes a "superrod" impossible. Assuming an even distribution, this gives us an average of .25. However, the 2nd cut is equally likely to be on the OTHER side of the midpoint, meaning we have to double that average to .50.

So I'm back to 1 : 1 as my answer.

gstelmack · 08-01-2008, 03:27 PM

Actually, 2 unique numbers, then comparing the three ranges you get.

BrianD · 08-01-2008, 03:35 PM

Quote:

Originally Posted by st.cronin

I think it approaches .5, actually.

Our ultimate conclusion is the same, but I'd argue with this point. Here is my example.

If I make the first cut at 0.1, where can I make the second cut to not leave anything greater than 0.5? My second cut has to be between 0.5 and 0.6. In this instance, I've only got 10% of the total rod in which to make my second cut.

If I make the first cut at 0.001, I'll only have 0.1% of the remaining rod which won't make a SupeRod. As X approaches 0, the odds of a SupeRod approach 1.

As X approahces 0.5, the odds of a SupeRod approach 0.5.

Now wait, does that mean a 75% chance of not making a triangle?

TCY Junkie · 08-01-2008, 03:36 PM

I would say slightly above .75 there will be a triangle. But I could definitely be wrong. Whats the answer pass

QuikSand · 08-01-2008, 03:55 PM

Okay, third layer, you're right Brian. First cut is at X, where X is between 0 and 0.5.

Three ways to make a superrod:
-second cut is to the left of X (probability = x)
-second cut is to the right of X, but less than 0.5 (probability = (0.5-x))
-second cut is so far to the right that the space between the two cuts if greater than 0.5 (probability = [1-(X+0.5)] or (0.5-x)

So, the first two just sum to 0.5, and the other equals (0.5 minus a range from 0 to 0.5) -- meaning we're going to make a super-rod 75% of the time...meaning a triangle only 25%.

*sigh*

Pretty cumbersome path to a solution, but I suspect that's it.

BrianD · 08-01-2008, 04:00 PM

In my post above, I talked myself into NOT making a triangle 75% of the time. The best you can do it to make your first cut close to the 0.5 point. In that instance, you have a 50% chance of your second cut making the triangle. That is the best case. The farther away your first cut is from the 0.5 point, the less chance you have of making the triangle. If you make your first cut close to either end, you have close to a 0% chance of making the triangle. Because of all of this, I'd say the chance of being able to make the triangle is 25% and the chance of not being able to make the triangle is 75%.

jeheinz72 · 08-01-2008, 04:14 PM

Cheese.

RendeR · 08-01-2008, 04:16 PM

There is nothing in the poriginal post that requires this puzzle to force the pieces to meet at their ENDS.

The answer therefore is 100% you can MAKE a triangle out of any length of pieces, there may be overhanging material however.

If you want it to be 25% wich everyone is working out, then the original post must be edited to correctly stipulate the problem.

BrianD · 08-01-2008, 04:24 PM

Good point. We are making an assumption on our constraints.

Maple Leafs · 08-01-2008, 04:27 PM

Quote:

Originally Posted by RendeR

There is nothing in the poriginal post that requires this puzzle to force the pieces to meet at their ENDS.

(Shoots Render in the fucking head.)

So, where were we?

Logan · 08-01-2008, 04:35 PM

Quote:

Originally Posted by RendeR

There is nothing in the poriginal post that requires this puzzle to force the pieces to meet at their ENDS.

The answer therefore is 100% you can MAKE a triangle out of any length of pieces, there may be overhanging material however.

If you want it to be 25% wich everyone is working out, then the original post must be edited to correctly stipulate the problem.

That was the first thing I thought of but didn't want to ruin it for anybody else.

Crim · 08-01-2008, 05:31 PM

Quote:

Originally Posted by Maple Leafs

(Shoots Render in the fucking head.)

So, where were we?

lolz

rowech · 08-01-2008, 06:28 PM

25% is what I believe it to be. Call the three pieces x, y, and z. Here are the possibilities...

x+y greater than z (triangle), x+y less than z, x+y equals z
x+z greater than y (triangle), x+z less than y, x+z equals y
y+z greater than x (triangle), y+z less than x, y+z equals x

The two endpoints are selected which keeps a single piece
The same point is selected twice which gives only two pieces.
An endpoint and a nonendpoint are selected which also gives only two pieces.

3/12 = 25%

I believe that's how the problem is originally intended but I think the wording is wrong in the problem given. I don't think you are forced into three pieces.

TCY Junkie · 08-01-2008, 09:03 PM

The big part can not be the same length as the smaller ones combined or it would just be a line. Thats where I was going with my answer above. The small parts have to equal greater than 50 percent. The big part if half or more makes no triangles. So it should approach 75 percent but not get there. I got it backwards earlier with help from my sinus infection.

hoopsguy · 08-01-2008, 09:12 PM

Quote:

Originally Posted by RendeR

There is nothing in the poriginal post that requires this puzzle to force the pieces to meet at their ENDS.

The answer therefore is 100% you can MAKE a triangle out of any length of pieces, there may be overhanging material however.

If you want it to be 25% wich everyone is working out, then the original post must be edited to correctly stipulate the problem.

This is how I read the problem as well.

QuikSand · 03-07-2021, 01:49 PM

well, Pass, what's the dealio?

Toddzilla · 03-08-2021, 10:00 PM

The only way we CANNOT make a triangle is if any of the 3 pieces are equal to or larger than 1/2 the total length.

First cut needs to be less than halfway, anything larger than that and no triangle. That's basically a 50% shot.

Second cut needs to be BOTH (a) big enough so 1st plus 2nd is more than

50% of the total length, and (b) smaller than 50% of the total original length.

So there is a "sweet spot" in the middle of the rod after the first cut where the second cut has to be made - and it's exactly equal to the size of the first cut.

We have a 50% chance of the first cut providing a valid path forward, and the second cut is 50% of that.

25%

QuikSand · 03-09-2021, 08:06 AM

Quote:

Originally Posted by Toddzilla

First cut needs to be less than halfway, anything larger than that and no triangle. That's basically a 50% shot.

25% may be the right answer, but I'd quarrel with the logic above. By definition, any first cut is equal to or less than 50% away from one end, right?

NobodyHere · 03-09-2021, 10:18 AM

I just discovered this puzzle.

I find that if you were to graph the odds of making a triangle based on the position of the first cut, it would resemble ,well, a triangle. Getting triangle if your first cut is at either end is virtually impossible. As your first cut approaches the middle the percentage goes up to 50 in a linear fashion.

So I get 25% too.

Vince, Pt. II · 03-09-2021, 10:38 AM

Quote:

Originally Posted by QuikSand

25% may be the right answer, but I'd quarrel with the logic above. By definition, any first cut is equal to or less than 50% away from one end, right?

Question of perspective, right? If you imagine the stick is in a fixed position and define "first cut" as the cut closest to a fixed side of your fixed stick, this logic holds.

Unless I'm missing something?

QuikSand · 03-10-2021, 07:22 AM

I don't see how any first cut can rule out an eventual triangle, other than the precise 0.5 location, which is infinitesimally unlikely. Any other cut by definition creates two sides, one shorter than 0.5 and one longer than 0.5.

AnalBumCover · 03-10-2021, 09:04 AM

Guys, I think we're missing the puzzle entirely. There is no "first cut" or "second cut". The original post clearly says the rod is marked at two points, and THEN divided to three parts at those points. Because of this, the sizes of the pieces are already determined before any cuts are made.

AnalBumCover · 03-10-2021, 09:06 AM

dola

That being said, we can change our discussion to read "first mark" and "second mark" and still come to the same conclusion.

Carry on.

albionmoonlight · 03-10-2021, 09:54 AM

It's a cool puzzle.

It does seem strong that the answer is .25

Hard to formalize, but it makes sense in an intuitive way

Vince, Pt. II · 03-10-2021, 11:53 AM

Quote:

Originally Posted by QuikSand

I don't see how any first cut can rule out an eventual triangle, other than the precise 0.5 location, which is infinitesimally unlikely. Any other cut by definition creates two sides, one shorter than 0.5 and one longer than 0.5.

I absolutely understand what you are saying, and agree. I think the other perspective is misconstruing the word "first" to mean "left-most." With two cuts, imagining a horizontal stick means there will always be a left and a right cut. If you call the left cut the "first" cut, and it happens to be to the right of the midpoint, you have a super rod, and therefore can't have a triangle.

Maple Leafs · 03-10-2021, 06:12 PM

Quote:

Originally Posted by Maple Leafs

(Shoots Render in the fucking head.)

Jesus dude, calm down.

Toddzilla · 03-10-2021, 06:44 PM

So let's approach it like this......

Imagine the stick has a mark in the very middle, point m

The first mark on the stick is on one side of m. That's what we're calling piece A.

Half A is 1/2 the stick with a mark on it and Half B is 1/2 the stick with no mark.

A second mark is made.

If the second mark is on Half A, we cannot make a triangle (as determined a few ways above.)

The chances of a second mark being on the same Half as the first mark is....50%.

If the second mark is on Half B, then we MAY be able to have a triangle.

So the chances we can make a triangle are demonstrably LESS than 50%.

Imagine the stick now a line with 5 points: The "beginning" of the stick, the first mark, the middle, the second mark, and the "end" of the stick.

In order for us to be able to make a triangle, the length of the stick between the first mark and second mark must be less than 1/2 the length of the stick.

Another way to say that is this: The distance between the "beginning" of the stick and the first mark must be GREATER THAN the distance between the middle of the stick and the second mark. Otherwise the distance between the first and second marks is greater than 1/2 the length of the rod, which is a no-go.

Now it's a matter or probabilities and limits, which I'll try not to overdo.

On Half A, there is equal probability of the first mark being anywhere on Half A.

If the first mark is VERY CLOSE to the "beginning" of the stick, the odds the second mark is CLOSER to the halfway mark is VERY SMALL. As the first mark gets closer and closer to the "beginning" of the stick, those odds approach 0%

If the first mark is VERY FAR AWAY from the "beginning" of the stick, the odds the second mark is CLOSER to the halfway mark is VERY LARGE. As the first mark gets further and further away from the "beginning" of the stick (i.e. as it approaches the middle), those odds approach 100%.

The average of 0% and 100% is 50%

and

50% (the chances the second mark is closer to the middle than the first mark is to the "beginning" of the stick) of 50% (the chances the second mark is not on the same half as the first mark) is 25%

08-01-2008, 02:39 PM	#3
st.cronin General Manager Join Date: Oct 2004 Location: New Mexico	2/3 __________________ co-commish: bb-bbcf.net knives out

08-01-2008, 02:42 PM	#5
gstelmack Pro Starter Join Date: Oct 2000 Location: Cary, NC	100%. You're just going to take the two outer pieces and rotate them up until they meet. Take a piece of string, loop it, put three fingers in (for the three corners), and move the fingers around. That's my story and I'm sticking to it. __________________ -- Greg -- Author of various FOF utilities

08-01-2008, 02:44 PM	#7
st.cronin General Manager Join Date: Oct 2004 Location: New Mexico	ok seriously Under what conditions will three lines NOT make a triangle? When 1 line is longer than the other two combined, right. So if both "cuts" are on the same half of the stick, no triangle. My answer is 1 : 1, on the theory that the second cut is equally likely to be on one half of the stick as the other. __________________ co-commish: bb-bbcf.net knives out Last edited by st.cronin : 08-01-2008 at 02:49 PM.

08-01-2008, 03:14 PM	#13
gstelmack Pro Starter Join Date: Oct 2000 Location: Cary, NC	There's a 50% chance that the first cut makes this impossible (by being over the halfway mark already). Out of the remaining 50% (where at least half the rod is available), there is a 50% chance the second cut will create a breaking rod. That lends credence to the 25% theory Quik put forth. __________________ -- Greg -- Author of various FOF utilities

08-01-2008, 03:20 PM	#19
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	I guess you also need to add in (to my previous thinking above) another way to make the SuperRod -- when the second cut is to the "left" of the first. Thus, any value less than X as also creating a super-rod, which I'd guess in itself adds in another 0.25 chance. I hate it when one of the obvious "dartboard" answers happens to be right... but that sure sounds like 0.5 in total. For any value X, your two ways of getting a SuperRod are X and 1-(X+0.5)... and the sum of those two simplifies nicely to 0.5. Ugh. Last edited by QuikSand : 08-01-2008 at 03:22 PM.

08-01-2008, 02:35 PM	#2
MikeVic Head Coach Join Date: Mar 2003 Location: Hometown of Canada	1 in 3.

08-01-2008, 02:59 PM	#8
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	I don't know intuitively how to calculate this, but let's think of this road as reaching from 0 to 1. SO, brute force below... First cut is a random place, let's put it at X. For our orientation purposes here, it's somewhere on the interval from 0 to 0.5. So, the the places that create a SuperRod (0.5 or longer) include those positions between (X+0.5) and 1. My shortcut guess here is since the median value for X here is 0.25, this probability is going to smooth out to 1 in 4. But I may well be missing something more complex in the distribution. Thinking a bit more... As X approached 0.5, the chance of a SuperRod approaches zero. As X approaches 0, the chance of a SupeRod approaches 0.5. More support for the answer of 0.25.

08-01-2008, 03:18 PM	#17
BrianD Grizzled Veteran Join Date: Jan 2005 Location: Appleton, WI	That doesn't sound right. As X approaches 0, the chance of a SupeRod approaches 1.

08-01-2008, 03:25 PM	#21
st.cronin General Manager Join Date: Oct 2004 Location: New Mexico	The most logical way I can think of is like this: Code: ______________________ .5 0 The .5 represents what happens if the first cut is infitesimally small - a cut on the very end gives a 50 percent chance of the second cut making a triangle impossible. The 0 represents what happens if the first cut is exactly halfway - it makes a "superrod" impossible. Assuming an even distribution, this gives us an average of .25. However, the 2nd cut is equally likely to be on the OTHER side of the midpoint, meaning we have to double that average to .50. So I'm back to 1 : 1 as my answer. __________________ co-commish: bb-bbcf.net knives out Last edited by st.cronin : 08-01-2008 at 03:26 PM.

08-01-2008, 03:27 PM	#22
gstelmack Pro Starter Join Date: Oct 2000 Location: Cary, NC	Actually, 2 unique numbers, then comparing the three ranges you get. __________________ -- Greg -- Author of various FOF utilities

08-01-2008, 03:36 PM	#24
TCY Junkie College Prospect Join Date: Jan 2007 Location: TX	I would say slightly above .75 there will be a triangle. But I could definitely be wrong. Whats the answer pass __________________ I try to open things I probably have no chance of opening.

08-01-2008, 03:55 PM	#25
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	Okay, third layer, you're right Brian. First cut is at X, where X is between 0 and 0.5. Three ways to make a superrod: -second cut is to the left of X (probability = x) -second cut is to the right of X, but less than 0.5 (probability = (0.5-x)) -second cut is so far to the right that the space between the two cuts if greater than 0.5 (probability = [1-(X+0.5)] or (0.5-x) So, the first two just sum to 0.5, and the other equals (0.5 minus a range from 0 to 0.5) -- meaning we're going to make a super-rod 75% of the time...meaning a triangle only 25%. sigh Pretty cumbersome path to a solution, but I suspect that's it. Last edited by QuikSand : 08-01-2008 at 10:33 PM. Reason: math was right, but corrected 75%/25% conclusion

08-01-2008, 04:00 PM	#26
BrianD Grizzled Veteran Join Date: Jan 2005 Location: Appleton, WI	In my post above, I talked myself into NOT making a triangle 75% of the time. The best you can do it to make your first cut close to the 0.5 point. In that instance, you have a 50% chance of your second cut making the triangle. That is the best case. The farther away your first cut is from the 0.5 point, the less chance you have of making the triangle. If you make your first cut close to either end, you have close to a 0% chance of making the triangle. Because of all of this, I'd say the chance of being able to make the triangle is 25% and the chance of not being able to make the triangle is 75%.

08-01-2008, 04:16 PM	#28
RendeR Coordinator Join Date: Aug 2001 Location: Buffalo, NY	There is nothing in the poriginal post that requires this puzzle to force the pieces to meet at their ENDS. The answer therefore is 100% you can MAKE a triangle out of any length of pieces, there may be overhanging material however. If you want it to be 25% wich everyone is working out, then the original post must be edited to correctly stipulate the problem. __________________ http://wotlabs.net/s...8/signature.png http://wotlabs.net/sig_dark/na/banichi18/signature.png

08-01-2008, 04:24 PM	#29
BrianD Grizzled Veteran Join Date: Jan 2005 Location: Appleton, WI	Good point. We are making an assumption on our constraints.

08-01-2008, 06:28 PM	#33
rowech Pro Starter Join Date: Feb 2006	25% is what I believe it to be. Call the three pieces x, y, and z. Here are the possibilities... x+y greater than z (triangle), x+y less than z, x+y equals z x+z greater than y (triangle), x+z less than y, x+z equals y y+z greater than x (triangle), y+z less than x, y+z equals x The two endpoints are selected which keeps a single piece The same point is selected twice which gives only two pieces. An endpoint and a nonendpoint are selected which also gives only two pieces. 3/12 = 25% I believe that's how the problem is originally intended but I think the wording is wrong in the problem given. I don't think you are forced into three pieces. Last edited by rowech : 08-01-2008 at 06:31 PM.

08-01-2008, 09:03 PM	#34
TCY Junkie College Prospect Join Date: Jan 2007 Location: TX	The big part can not be the same length as the smaller ones combined or it would just be a line. Thats where I was going with my answer above. The small parts have to equal greater than 50 percent. The big part if half or more makes no triangles. So it should approach 75 percent but not get there. I got it backwards earlier with help from my sinus infection. __________________ I try to open things I probably have no chance of opening.

03-07-2021, 01:49 PM	#36
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	well, Pass, what's the dealio?

03-08-2021, 10:00 PM	#37
Toddzilla Pro Starter Join Date: Jan 2001 Location: Burke, VA	The only way we CANNOT make a triangle is if any of the 3 pieces are equal to or larger than 1/2 the total length. First cut needs to be less than halfway, anything larger than that and no triangle. That's basically a 50% shot. Second cut needs to be BOTH (a) big enough so 1st plus 2nd is more than 50% of the total length, and (b) smaller than 50% of the total original length. So there is a "sweet spot" in the middle of the rod after the first cut where the second cut has to be made - and it's exactly equal to the size of the first cut. We have a 50% chance of the first cut providing a valid path forward, and the second cut is 50% of that. 25%

03-09-2021, 10:18 AM	#39
NobodyHere Grizzled Veteran Join Date: Nov 2013	I just discovered this puzzle. I find that if you were to graph the odds of making a triangle based on the position of the first cut, it would resemble ,well, a triangle. Getting triangle if your first cut is at either end is virtually impossible. As your first cut approaches the middle the percentage goes up to 50 in a linear fashion. So I get 25% too. __________________ "I am God's prophet, and I need an attorney"

03-10-2021, 07:22 AM	#41
QuikSand lolzcat Join Date: Oct 2000 Location: Annapolis, Md	I don't see how any first cut can rule out an eventual triangle, other than the precise 0.5 location, which is infinitesimally unlikely. Any other cut by definition creates two sides, one shorter than 0.5 and one longer than 0.5.

03-10-2021, 09:04 AM	#42
AnalBumCover College Starter Join Date: Nov 2002 Location: La Mirada, CA	Guys, I think we're missing the puzzle entirely. There is no "first cut" or "second cut". The original post clearly says the rod is marked at two points, and THEN divided to three parts at those points. Because of this, the sizes of the pieces are already determined before any cuts are made. __________________ ABC's Game Giveaway list

03-10-2021, 09:06 AM	#43
AnalBumCover College Starter Join Date: Nov 2002 Location: La Mirada, CA	dola That being said, we can change our discussion to read "first mark" and "second mark" and still come to the same conclusion. Carry on. __________________ ABC's Game Giveaway list

03-10-2021, 09:54 AM	#44
albionmoonlight Head Coach Join Date: Oct 2000 Location: North Carolina	It's a cool puzzle. It does seem strong that the answer is .25 Hard to formalize, but it makes sense in an intuitive way

03-10-2021, 06:44 PM	#47
Toddzilla Pro Starter Join Date: Jan 2001 Location: Burke, VA	So let's approach it like this...... Imagine the stick has a mark in the very middle, point m The first mark on the stick is on one side of m. That's what we're calling piece A. Half A is 1/2 the stick with a mark on it and Half B is 1/2 the stick with no mark. A second mark is made. If the second mark is on Half A, we cannot make a triangle (as determined a few ways above.) The chances of a second mark being on the same Half as the first mark is....50%. If the second mark is on Half B, then we MAY be able to have a triangle. So the chances we can make a triangle are demonstrably LESS than 50%. Imagine the stick now a line with 5 points: The "beginning" of the stick, the first mark, the middle, the second mark, and the "end" of the stick. In order for us to be able to make a triangle, the length of the stick between the first mark and second mark must be less than 1/2 the length of the stick. Another way to say that is this: The distance between the "beginning" of the stick and the first mark must be GREATER THAN the distance between the middle of the stick and the second mark. Otherwise the distance between the first and second marks is greater than 1/2 the length of the rod, which is a no-go. Now it's a matter or probabilities and limits, which I'll try not to overdo. On Half A, there is equal probability of the first mark being anywhere on Half A. If the first mark is VERY CLOSE to the "beginning" of the stick, the odds the second mark is CLOSER to the halfway mark is VERY SMALL. As the first mark gets closer and closer to the "beginning" of the stick, those odds approach 0% If the first mark is VERY FAR AWAY from the "beginning" of the stick, the odds the second mark is CLOSER to the halfway mark is VERY LARGE. As the first mark gets further and further away from the "beginning" of the stick (i.e. as it approaches the middle), those odds approach 100%. The average of 0% and 100% is 50% and 50% (the chances the second mark is closer to the middle than the first mark is to the "beginning" of the stick) of 50% (the chances the second mark is not on the same half as the first mark) is 25% Last edited by Toddzilla : 03-10-2021 at 06:45 PM.

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