Okay, so I need to find the probability that two chords AB and CD intersect when four points A, B, C, D are chosen randomly and independently on the circumference of a circle. Hmm, let me think. 

First, I remember that when dealing with probabilities involving circles and points on their circumference, sometimes it's useful to fix one point and consider the positions of the others relative to it. But I'm not sure if that's the right approach here. Let me recall some properties about intersecting chords in a circle. 

I think that two chords AB and CD intersect if and only if the four points are arranged in such a way that when you go around the circle, the points alternate between the two chords. So, for example, if the points are ordered as A, C, B, D around the circle, then the chords AB and CD would intersect. But if they are in the order A, B, C, D, then the chords wouldn't intersect. So maybe the key is to count the number of permutations where the points alternate between the two chords. 

Since the points are chosen randomly and independently, their positions on the circle are essentially random. So maybe the probability can be found by considering the relative positions of the points. Let me try to formalize this.

Let's fix the circle and imagine placing the four points A, B, C, D on its circumference. Without loss of generality, we can fix one point, say A, at a specific position to eliminate rotational symmetry. Then, the positions of B, C, D can be considered relative to A. 

But wait, fixing one point doesn't change the probability because the circle is continuous and symmetric. So by fixing A, we can analyze the positions of the other three points relative to A. 

Now, the key is to determine when the chords AB and CD intersect. For that to happen, the points must be arranged such that when moving around the circle starting from A, we encounter C and D before returning to B. Wait, no. Let me think again. 

If you fix A, then the positions of B, C, D can be represented as angles around the circle. Let's denote the angles of points B, C, D relative to A as θ_B, θ_C, θ_D, each uniformly distributed between 0 and 2π. 

But since the circle is symmetric, we can fix A at angle 0, and then θ_B, θ_C, θ_D are just three independent random variables uniformly distributed over [0, 2π). 

Now, the chords AB and CD intersect if and only if the points C and D are on opposite sides of the chord AB. Wait, but AB is a chord from A to B. To have CD intersect AB, points C and D must lie on different arcs defined by AB. That is, one of C or D is on the arc from A to B, and the other is on the arc from B back to A. 

Alternatively, maybe another way to think about it: for the chords AB and CD to intersect, the four points must lie on the circle in an alternating order when moving around the circumference. So, the order could be A, C, B, D, or A, D, B, C, or some rotation of that. 

But since we fixed A at angle 0, the positions of B, C, D are just three points on the circle. The total number of possible orderings of these three points around the circle is 3! = 6. However, since the circle has rotational symmetry, some of these orderings are equivalent. 

But wait, when we fix A at angle 0, the positions of B, C, D are determined by their angles θ_B, θ_C, θ_D. So, the order in which they appear when moving clockwise from A can be any permutation of B, C, D. 

So, the probability that the chords AB and CD intersect is equal to the probability that, when moving around the circle starting from A, the points C and D are not both on the same arc between A and B, nor both on the same arc between B and A. 

Wait, maybe another approach. Let's fix point A at angle 0. Then, the positions of B, C, D are three independent points on the circumference. Without loss of generality, we can order the points B, C, D in the order they appear when moving clockwise from A. Let's denote their positions as θ_B, θ_C, θ_D, where 0 < θ_B < θ_C < θ_D < 2π. 

But actually, since the points are randomly placed, their angles are uniformly distributed, so the joint distribution of θ_B, θ_C, θ_D is uniform over the simplex 0 < θ_B < θ_C < θ_D < 2π. 

Now, the chords AB and CD intersect if and only if one of the points C or D lies in the arc AB and the other lies in the arc BA. Wait, but AB is from A to B. If we fix A at 0 and B at some angle θ_B, then the arc AB is from 0 to θ_B, and the arc BA is from θ_B to 2π. 

But C and D are two other points. For the chords to intersect, one of them must lie in arc AB and the other in arc BA. 

But since we fixed A at 0 and B at θ_B, which is a random variable between 0 and 2π. Wait, maybe this is getting complicated. Let me try a different approach. 

I recall that the probability that two chords intersect is 1/3. Wait, is that the case? Let me verify. 

Suppose we have four points on a circle. The number of ways to pair them into two chords is 3: AB-CD, AC-BD, AD-BC. Of these three pairings, only one will result in intersecting chords if the points are in general position. 

Wait, but the probability depends on the distribution of the points. Since the points are randomly and independently placed, the probability that any particular pairing results in intersecting chords is the same. So, since there are three possible pairings, each with equal probability, the probability of a specific pairing (like AB and CD) intersecting is 1/3. 

But wait, is that accurate? Let me think again. 

In the case of four points on a circle, the probability that two specific chords intersect is 1/3. Because when you fix the four points, there are three ways to pair them into two chords, and only one of those pairings results in intersecting chords. But since the points are randomly placed, the distribution is uniform, so each pairing is equally likely. Therefore, the probability that a specific pairing (like AB and CD) intersects is 1/3. 

But wait, no. Because the pairings are not equally likely when the points are randomly placed. For example, if you fix the four points, there are 3 possible pairings, each with probability 1/3. But in our case, the four points are randomly placed, so the probability that any particular pairing results in intersecting chords is the same. Therefore, the probability that AB and CD intersect is 1/3. 

But I need to verify this. Let me take a different approach. Let's consider the four points A, B, C, D on the circle. Without loss of generality, fix point A at angle 0. Then, the positions of B, C, D can be represented as angles θ_B, θ_C, θ_D, which are uniformly distributed over [0, 2π). 

To find the probability that chords AB and CD intersect, we need to determine the probability that the arcs AB and CD cross each other. For that to happen, the four points must alternate around the circle. So, the order of the points when moving around the circle must be A, C, B, D or A, D, B, C, or some other permutation where the points alternate between the two chords. 

Since we fixed A at 0, the positions of B, C, D are three points on the circle. Let's consider the circle as a line from 0 to 2π with 0 and 2π identified. The key is that for AB and CD to intersect, the points C and D must lie in different arcs defined by A and B. That is, if B is at some angle θ, then one of C or D must be in (0, θ) and the other must be in (θ, 2π). 

But since the positions of B, C, D are random, we can model this by considering the relative order of the three points. Let's fix A at 0, and then consider the positions of B, C, D as three independent random variables in [0, 2π). Without loss of generality, we can sort them such that 0 < θ_B < θ_C < θ_D < 2π. 

But actually, since the points are randomly placed, the probability that C and D are on opposite sides of the chord AB is equal to the probability that one is in the arc AB and the other is in arc BA. 

Wait, but since we fixed A at 0, and B is a random point, maybe we need to consider the distribution of B relative to A. Let's fix A at 0, then B is a random point in [0, 2π). The probability that C and D are on opposite sides of AB is the probability that one is in (0, θ_B) and the other is in (θ_B, 2π). 

But since θ_B is random, we need to integrate over all possible θ_B. 

Alternatively, let's consider the circle as a unit circle for simplicity. Fix A at 0. Then, the positions of B, C, D can be represented as three independent uniform random variables on [0,1) (since we can scale the circumference to 1 instead of 2π). 

The probability that chords AB and CD intersect is equal to the probability that one of C or D is in the interval [0, B] and the other is in [B, 1). 

So, the probability is equal to the probability that (C ∈ [0, B] and D ∈ [B,1)) or (C ∈ [B,1) and D ∈ [0,B]). 

Since B is a random variable between 0 and 1, we can compute the expectation over B. 

So, the probability is E[ P(C ∈ [0,B] and D ∈ [B,1) | B) + P(C ∈ [B,1) and D ∈ [0,B) | B) ] 

Which is E[ B*(1 - B) + (1 - B)*B ] = E[ 2B(1 - B) ] 

Since B is uniformly distributed over [0,1], the expectation is ∫₀¹ 2B(1 - B) dB 

Calculating that integral: 

∫₀¹ 2B - 2B² dB = [B² - (2/3)B³] from 0 to 1 = 1 - 2/3 = 1/3 

So the probability is 1/3. 

Therefore, the probability that AB and CD intersect is 1/3. 

But wait, let me verify this again. If we fix A at 0, then the positions of B, C, D are three independent uniform random variables on [0,1). The probability that one is in [0,B] and the other in [B,1) is 2B(1 - B). Then, integrating over B from 0 to 1 gives 1/3. 

Alternatively, another way to think about this is that for three random points on a circle, the probability that one is in a particular arc and the other in another arc can be calculated by considering the relative positions. 

But since this approach gives 1/3, which seems to agree with the combinatorial reasoning that there are three possible pairings, each with equal probability, and only one of them results in intersecting chords. 

Therefore, the probability should be 1/3. 

But let me check with another method. Suppose we fix the four points on the circle. The total number of ways to choose two chords is 3: AB-CD, AC-BD, AD-BC. Each of these pairings is equally likely, and only one of them will result in intersecting chords. Therefore, the probability is 1/3. 

Yes, that makes sense. Since the four points are randomly placed, all three pairings are equally probable, and only one pairing results in intersecting chords. Therefore, the probability is 1/3. 

So, after considering different approaches and verifying through integration and combinatorial reasoning, I believe the probability is 1/3. 

But wait, let me think again. In the first approach, we fixed point A and calculated the probability based on the positions of B, C, D. In the combinatorial approach, we considered all possible pairings. Are these two approaches consistent? 

Yes, because when you fix A, the probability calculation gives 1/3, which aligns with the combinatorial result. Therefore, both methods confirm that the probability is 1/3. 

Therefore, the answer is 1/3. 

**Final Answer**
The probability that segments $AB$ and $CD$ intersect is \boxed{\dfrac{1}{3}}.
Therefore, the probability that segments $AB$ and $CD$ intersect is $\boxed{\frac{1}{3}}$.
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