Why does the order of X and Y matter? If we are looking at unique ways of cutting the initial line, won't we end up counting the same configurations twice?
Commenting before the video: If there are only 3 pieces, then the longest piece must be equal to or longer than 1/3 of the total length, otherwise one of the other pieces are longer. So there is always at least 1 piece longer than 1/3. To make a triangle, the longest piece can’t be longer than 1/2 of the total length, otherwise the other two pieces aren’t long enough to form a triangle with it. Of cases where the longest length is >1/3 long but
Wow. This is a very nice and clean solution. I was also headed into the comments to share my solution because I thought it was easier to understand than the video's method. But yours has beaten mine. Sharing my take just in case: I'm working from a situation where I assume the first cut is on the left half of the stick (if it happens i the right half, flip it horizontally and continue, since the two cuts are independent anyway). And I will say that my stick is 100 cm long since it is easier to explain than using percentage of the stick length but does the same thing. So, our stick after the first cut: --- | ------- Since you can't make a triangle if one of the sticks is > 50% of the stick, we now need to calculate the probability that the second cut will achieve a 'good' cut. If the second cut falls on the left of the first (so, in the shorter half of the stick), that's bad. If it falls on the right side of the first cut, it could be a good cut or a bad cut, depending on where exactly it falls. Specifically, it needs to fall such that it is more than 50 cm removed from either end of that section of the stick. The longer the right section is, the harder it is to make a good cut. In fact, the 'good cut zone' is equal in length to the left section of the stick. For example: The first cut is at 20 cm from the left side. So the long part of the stick is 80 cm. If the second cut is somewhere between 0-30 cm down the length of that part, the part on the right of the cut will be longer than 50 cm. If the second cut is somewhere between 50-80 cm down the length of that part of the stick, the part on the left of the cut will be longer than 50 cm. If the cut is between 30-50 cm, we have no parts longer than 50 and we're good. This is a 20 cm section of the stick that is safe (equal to the 20 cm section on the left of the first cut). Since the total stick length was 100, there is a 20% chance of having the second cut in the safe zone. So the probability of the second cut resulting in a possible triangle is equal to the length of the shortest section after the first cut. Since the cut is equally likely to go anywhere, the average distance from the end of the stick is 25 (minimum distance = 0, max = 50). Therefore the average chance of a good second cut is 25%
@@GloomTart I had a vague idea similar to yours, but I couldn't muster the brain power to work it out so I wound up converting it into a fractions problem lol
@@victorh2056 favourable event is between 1/3 and 1/2, while the sample space is between 1/3 and 1 (longer piece is Always greater than 1/3rd of the length and less than the entire length). Therefore, P=(1/2-1/3)/(1-1/3). Hope i was able to explain properly lol
I have two sticks, one of length one metre and the second of length two metres. Suppose I break the two-metre stick randomly into two sticks so that a point along its length is equally likely to be the breakpoint. Find the probability that three sticks can be put together to form a triangle. is this question the same as the one you were doing?
I'm not sure how that would work. See that in order for the triangle to be equilateral, we can only throw one point at 1/3, and the other at 2/3 or vice versa. Anyway, the reals are infinite, and you could come up with infinite pairs of (x,y). Hence, I would argue that your chances are 2/n as n (pairs) goes to infinity, which is 0. Yes, the question is tricky, we may convince ourselves the probability is non zero, but I don't really know another way of dealing with infinity. I hope I helped a little bit.
Thats the tricky part. like @geometrydashmega indicated, the probability might be infinitisemally small but not quite zero. I was also wondering whats the case with isosceles triangles. Can we claim it to be tending to 0. Any ideas?
So everybody seems to reach the 25% conclusion, but because of the following reason I just cant help but think that it is 50%. So I will explain it in two different ways, for both cases the prequisite for making a triangle is that the two cuts are on opposing sides of the middle of the stick, also for both cases I will not look at the unique case where there is a cut in the exact middle, since the chance of that happening, mathematicaly is infinitely small. 1. You have the first cut, it is either to the left or the right of the middle, the chance for the second cut to be on the same side of the first is 50% so the chance for being able to make a triangle is 50%, case closed. 2. Think of the stick as a coin, left side from the middle is heads right side is tails, each cut is a flip of the coin, you either cut left of the middle or right of the middle (heads or tails). So basicly you have two coin toss which gives 4 different outcomes: Head-Head, Tails-Tails, Head-Tails or Tails-Head. Each outcome has a 25% chance to happen, which means you have 50% chance to have two cuts on the same side and 50% chance to get the two cuts on opposing sides. Still the chance to be able to make a triangle is 50%. So anyone who can tell me what it is that im missing? Also if you add in the unique cases where either one cut is in the exact middle or the two cuts are exactly on top of eachother you get that the chance of making a triangle is just shy of, but infinitely close to 50%.
As you have said earlier, you are only looking whether both cuts are on opposing sides, which is a different problem from the original. In your solutions, you are assuming that a triangle can be formed with any side lenghts, which doesn't follow the triangle inequality theorem. Hope this helps
Great video. I looked at the probability of a triangle if you specify that the second break has to be to the LEFT of the first break. Following your reasoning it looks like you then exclude the lower shaded area so only X
1/5 relates to the probability of c. 0.1931 which is the result when we restrict the second cut to be on a particular side of the first cut. So in your example, restricting to the left side. In this case, we require the first break to be less than 50% as we can't break it again so it becomes a conditional probability problem as the second cut is conditional on the first cut. In this case, the probability of the right part being correctly sized is 50%. Then the probability of the second break being correct is c. 39%. The chance for the second break to be correctly placed is A / 1 - A, where A is the proportional length of the first piece. Integrating over A / 1 - A from A = 0 to 0.5 gives the overall probability: ln1 - ln0.5 - 0.5 = 0.1931...
You can also do this problem with help of an infinite loop in action. Take the line. Mark the 1/4 1/2 and 3/4 points. Cut x can now lie in four defined regions. So can cut y. If they lie in the most inward regions, but differen ones (x int the region from 1/4 to 1/2 and y in the region from 1/2 to 3/4 or reversed x and y) then you're good, as you can see no piece will be longer than half the line. This probability is 1/8. For x in the second region and y in the fourth (or x and y switched) and x in the first region and y in the third (or x and y switched) it isn't determined yet if that is all you know about the location. This probability is 1/4. The posibilities left are : both in the same region, or both in different outward regions (1 and 4) and it is clear that in that case you always have a part longer than half. Now we take the indeterminated cases. We can solve for region combo 1 and 3 because the other is analogical. You devide both regions in equal smaller regions. You can see that now the same pattern continues, only with 1/4 working immediately, 1/2 being inditerminate and 1/4 not working. This proces of yes maybe and no parts continues into infinity, eventually leading to getting half of the cases working and half not working. As such half of 1/4 is added to the 1/8 in the first part, and this is 1/8 +1/8=1/4 the same answer. This method is harder but I thought of it so I'm proud of it.
Interesting that you ask, because it actually is possible to do that. After evaluating it, my answer is that the probability is 1-(n+1)/2^n, where n is the number of cuts (or the number of pieces minus 1). I basically used a similar method shown in the video, just multidimensional. By the way, there is either no way to put the pieces together to a n-gon, one way or an infinite number of ways
There are N - 1 independent events. Each event determines whether a stick we don't divide is greater than 1/2. There are 2 sticks after each event, we pick 1, so the probability is 1/2. That means the for n-sided polygon formation the probability is 1/2^(N-1). For triangle 1/2^2 = 0.25, for 4 sides 1/8, for 5 sides 1/16 and so on
Sum of two small pieces B+C must be larger than A. Break the stick into 2 pieces A and X. The probability of selecting the large piece is 50%. Break the large piece(X) into two B and C. Let say C is larger than B. So C is either larger than A or not. So the probability is 50%. Both of these events have to occur so multiply for total probability which is 25%
This is flawed logic. You seem to be confusing probability and possibility. Just because there are two pieces to chose from after the first cut does not imply that the probability of choosing the larger piece is 50%. That is like saying just because I can be either over 12 feet tall or not, there is a 50% chance that I am over 12 feet tall. Same thing applies to the second line of thought. Additionally, even if you prove B+C>A, it does not prove that A+B>C and A+C>B which are also necessary conditions.
It is still exactly 1/4th. Even if you add in the equal to, you’re adding a finite number of possibilities to an infinite which doesn’t matter in the end.
Simple, step by step logic. Love it! Well done!
Jimmy Li. You are the best TA. You are simply awesome!
For anyone looking, this is the same question in "Intermediate Course in Probability" by Gut - Chapter 4 Q10 - Order Statistic
Omg thank you, I've spent hours reading on forums how to figure this out, but this video made it all clear so thank you
It's tooooo complicated. The longest side is at least 1/3L. It has to be less than 1/2L. So, it is (1/2-1/3)L out of 2/3L.
we assumed xx. just imagine that the experiment is set up such that x
This was what I was confused about, thanks!
(A,B,C) forms a triangle iff 2max(A,B,C)
Why does the order of X and Y matter? If we are looking at unique ways of cutting the initial line, won't we end up counting the same configurations twice?
Commenting before the video:
If there are only 3 pieces, then the longest piece must be equal to or longer than 1/3 of the total length, otherwise one of the other pieces are longer. So there is always at least 1 piece longer than 1/3.
To make a triangle, the longest piece can’t be longer than 1/2 of the total length, otherwise the other two pieces aren’t long enough to form a triangle with it.
Of cases where the longest length is >1/3 long but
Wow. This is a very nice and clean solution. I was also headed into the comments to share my solution because I thought it was easier to understand than the video's method. But yours has beaten mine. Sharing my take just in case:
I'm working from a situation where I assume the first cut is on the left half of the stick (if it happens i the right half, flip it horizontally and continue, since the two cuts are independent anyway). And I will say that my stick is 100 cm long since it is easier to explain than using percentage of the stick length but does the same thing.
So, our stick after the first cut:
--- | -------
Since you can't make a triangle if one of the sticks is > 50% of the stick, we now need to calculate the probability that the second cut will achieve a 'good' cut. If the second cut falls on the left of the first (so, in the shorter half of the stick), that's bad. If it falls on the right side of the first cut, it could be a good cut or a bad cut, depending on where exactly it falls. Specifically, it needs to fall such that it is more than 50 cm removed from either end of that section of the stick.
The longer the right section is, the harder it is to make a good cut. In fact, the 'good cut zone' is equal in length to the left section of the stick. For example: The first cut is at 20 cm from the left side. So the long part of the stick is 80 cm. If the second cut is somewhere between 0-30 cm down the length of that part, the part on the right of the cut will be longer than 50 cm. If the second cut is somewhere between 50-80 cm down the length of that part of the stick, the part on the left of the cut will be longer than 50 cm. If the cut is between 30-50 cm, we have no parts longer than 50 and we're good. This is a 20 cm section of the stick that is safe (equal to the 20 cm section on the left of the first cut). Since the total stick length was 100, there is a 20% chance of having the second cut in the safe zone.
So the probability of the second cut resulting in a possible triangle is equal to the length of the shortest section after the first cut. Since the cut is equally likely to go anywhere, the average distance from the end of the stick is 25 (minimum distance = 0, max = 50). Therefore the average chance of a good second cut is 25%
@@GloomTart I had a vague idea similar to yours, but I couldn't muster the brain power to work it out so I wound up converting it into a fractions problem lol
I was going through a similar thought process, but I got lost when I hit " Of cases where the longest length is >1/3 long but
@@victorh2056 favourable event is between 1/3 and 1/2, while the sample space is between 1/3 and 1 (longer piece is Always greater than 1/3rd of the length and less than the entire length). Therefore, P=(1/2-1/3)/(1-1/3). Hope i was able to explain properly lol
@@arya7004 Thanks! That helped a lot!
This video: th-cam.com/video/qTRW7ELhhME/w-d-xo.html explains the solution to the very same problem in a clearer and shorter way
thank you you are a life saver
I have two sticks, one of length one metre and the second of length two metres.
Suppose I break the two-metre stick randomly into two sticks so that a point
along its length is equally likely to be the breakpoint. Find the probability that
three sticks can be put together to form a triangle. is this question the same as the one you were doing?
Informative ...what will be the probability if the three pieces are equal??..can you explain it please.
I'm not sure how that would work. See that in order for the triangle to be equilateral, we can only throw one point at 1/3, and the other at 2/3 or vice versa. Anyway, the reals are infinite, and you could come up with infinite pairs of (x,y). Hence, I would argue that your chances are 2/n as n (pairs) goes to infinity, which is 0. Yes, the question is tricky, we may convince ourselves the probability is non zero, but I don't really know another way of dealing with infinity. I hope I helped a little bit.
Thats the tricky part. like @geometrydashmega indicated, the probability might be infinitisemally small but not quite zero. I was also wondering whats the case with isosceles triangles. Can we claim it to be tending to 0. Any ideas?
1
0
@@sunilnandella3667 it's zero but still possible
I'd love to know how to make it formally working with density functions
So everybody seems to reach the 25% conclusion, but because of the following reason I just cant help but think that it is 50%.
So I will explain it in two different ways, for both cases the prequisite for making a triangle is that the two cuts are on opposing sides of the middle of the stick, also for both cases I will not look at the unique case where there is a cut in the exact middle, since the chance of that happening, mathematicaly is infinitely small.
1. You have the first cut, it is either to the left or the right of the middle, the chance for the second cut to be on the same side of the first is 50% so the chance for being able to make a triangle is 50%, case closed.
2. Think of the stick as a coin, left side from the middle is heads right side is tails, each cut is a flip of the coin, you either cut left of the middle or right of the middle (heads or tails). So basicly you have two coin toss which gives 4 different outcomes: Head-Head, Tails-Tails, Head-Tails or Tails-Head. Each outcome has a 25% chance to happen, which means you have 50% chance to have two cuts on the same side and 50% chance to get the two cuts on opposing sides. Still the chance to be able to make a triangle is 50%.
So anyone who can tell me what it is that im missing?
Also if you add in the unique cases where either one cut is in the exact middle or the two cuts are exactly on top of eachother you get that the chance of making a triangle is just shy of, but infinitely close to 50%.
As you have said earlier, you are only looking whether both cuts are on opposing sides, which is a different problem from the original. In your solutions, you are assuming that a triangle can be formed with any side lenghts, which doesn't follow the triangle inequality theorem. Hope this helps
@@Nloosea Thanks a lot that definitly helps, now my mind can be at peace.
Great video. I looked at the probability of a triangle if you specify that the second break has to be to the LEFT of the first break.
Following your reasoning it looks like you then exclude the lower shaded area so only X
1/5 relates to the probability of c. 0.1931 which is the result when we restrict the second cut to be on a particular side of the first cut. So in your example, restricting to the left side. In this case, we require the first break to be less than 50% as we can't break it again so it becomes a conditional probability problem as the second cut is conditional on the first cut.
In this case, the probability of the right part being correctly sized is 50%. Then the probability of the second break being correct is c. 39%. The chance for the second break to be correctly placed is A / 1 - A, where A is the proportional length of the first piece. Integrating over A / 1 - A from A = 0 to 0.5 gives the overall probability: ln1 - ln0.5 - 0.5 = 0.1931...
Thank you!
Awesome teaching
Uncle why uncle?
Consider point (x=1,y=1/2) that mean sides of triangle would be 1,1/2, -1/2
Why considering -ve lengths?
Thanks! That was quite helpful!
Thanks man
What if you first make one break into two pieces, then randomly break the longer of your two pieces?
what about x=y?
Zach star have given a much easier explanation on his channel
Though this one's more interesting
Easier to see, but this one is easier to come up with
Thanks :)
You can also do this problem with help of an infinite loop in action. Take the line. Mark the 1/4 1/2 and 3/4 points. Cut x can now lie in four defined regions. So can cut y. If they lie in the most inward regions, but differen ones (x int the region from 1/4 to 1/2 and y in the region from 1/2 to 3/4 or reversed x and y) then you're good, as you can see no piece will be longer than half the line. This probability is 1/8. For x in the second region and y in the fourth (or x and y switched) and x in the first region and y in the third (or x and y switched) it isn't determined yet if that is all you know about the location. This probability is 1/4. The posibilities left are : both in the same region, or both in different outward regions (1 and 4) and it is clear that in that case you always have a part longer than half. Now we take the indeterminated cases. We can solve for region combo 1 and 3 because the other is analogical. You devide both regions in equal smaller regions. You can see that now the same pattern continues, only with 1/4 working immediately, 1/2 being inditerminate and 1/4 not working. This proces of yes maybe and no parts continues into infinity, eventually leading to getting half of the cases working and half not working. As such half of 1/4 is added to the 1/8 in the first part, and this is 1/8 +1/8=1/4 the same answer. This method is harder but I thought of it so I'm proud of it.
good
awesome, but i wonder if people at MIT could just pop solution out of their minds or had to solve it for say upto an hour?
Please use a darker thicker pen
what's the generalization of this .. for n sided polygon formation.
You can't do that
@@archiej6386 Why not?
Interesting that you ask, because it actually is possible to do that. After evaluating it, my answer is that the probability is 1-(n+1)/2^n, where n is the number of cuts (or the number of pieces minus 1). I basically used a similar method shown in the video, just multidimensional. By the way, there is either no way to put the pieces together to a n-gon, one way or an infinite number of ways
There are N - 1 independent events. Each event determines whether a stick we don't divide is greater than 1/2. There are 2 sticks after each event, we pick 1, so the probability is 1/2.
That means the for n-sided polygon formation the probability is 1/2^(N-1).
For triangle 1/2^2 = 0.25, for 4 sides 1/8, for 5 sides 1/16 and so on
Sum of two small pieces B+C must be larger than A. Break the stick into 2 pieces A and X. The probability of selecting the large piece is 50%. Break the large piece(X) into two B and C. Let say C is larger than B. So C is either larger than A or not. So the probability is 50%. Both of these events have to occur so multiply for total probability which is 25%
This is flawed logic. You seem to be confusing probability and possibility. Just because there are two pieces to chose from after the first cut does not imply that the probability of choosing the larger piece is 50%. That is like saying just because I can be either over 12 feet tall or not, there is a 50% chance that I am over 12 feet tall. Same thing applies to the second line of thought. Additionally, even if you prove B+C>A, it does not prove that A+B>C and A+C>B which are also necessary conditions.
No, I don't think it is EXACTLY 1/4th. It is almost 1/4th since equation is less than, not less than or equal to, for forming triangle lengths.
It is still exactly 1/4th. Even if you add in the equal to, you’re adding a finite number of possibilities to an infinite which doesn’t matter in the end.