Brainteaser with 6 random variables and combinations

Question

Six points are drawn subsequently from a uniform distribution. The first two points are marked blue, the second marked green and the last two red on the real line. What is the probability of having the 2 blue points, the 2 red points and the two green points adjacent? (i.e. the points have order RRGGBB or BBGGRR etc.)

My initial attempt was to try an integral like:

$$ 3 \int_{0}^{1} \int_{x}^{1} \int_{y}^{1} \int_{z}^{1} \int_{r}^{1} \int_{q}^{1} dp dq dr dz dy dx $$

Where the limits if the integration are determined by the previously drawn colours. So if we say that the first colour X can be drawn anywhere between 0 and 1, the next colour must be drawn from the smaller distribution between X and 1, and the next colour after than between the second colour and 1 and so on.

This integral gives $ \frac{93}{720}$, which I don't think is correct.

Instead a much more intuitive way to approach the question is to just think of the combinations of ways in which the colours can be arranged. There are $6!$ arrangements, of which $3! * 2^{3}$ give the order we require, which gives a probability of $\frac{1}{15}$.

If the second way of approaching the problem is correct, there must be an analogue to solving the problem with integrals - can anybody see find this second method?

Answer 1

The combinatorial way of thinking is ok, we first make a choice of an uncolored $x=(x_1,x_2,x_3,x_4,x_5,x_6)\in I^6$, where $I=[0,1]$, then we put the colors on the components. But ok, the OP asks for a way to see this result by using integrals. Here is this way. (Done by computation, and done by seing it is the same way as the combinatorial way.)

---

Let $A$ be the set of all $x\in I^6$ so that the intervals

• between $x_1$ and $x_2$,
• between $x_3$ and $x_4$,
• between $x_5$ and $x_6$,

are disjoint. Let $1_A$ be its characteristic function, one on $A$, zero else.

The needed probability $P$ is: $$ \begin{aligned} P &= \iint_{(x_1,x_2)\in I^2} dx_1\; dx_2 \iint_{(x_3,x_4)\in I^2} dx_3\; dx_4 \iint_{(x_5,x_6)\in I^2} 1_A(x)\; dx_5\; dx_6 \\ &= 2^3 \iint_{\substack{(x_1,x_2)\in I^2\\x_1<x_2}} dx_1\; dx_2 \iint_{\substack{(x_3,x_4)\in I^2\\x_3<x_4}} dx_3\; dx_4 \iint_{\substack{(x_5,x_6)\in I^2\\x_5<x_6}} 1_A(x)\; dx_5\; dx_6 \\ &= 2^3 \iint_{\substack{(x_1,x_2)\in I^2\\x_1<x_2}} \iint_{ \substack{(x_3,x_4)\in I^2 \\x_3<x_4 \\x_1<x_2<x_3<x_4\text{ or }x_3<x_4<x_1<x_2 } } dx_1\; dx_2 dx_3\; dx_4 \iint_{\substack{(x_5,x_6)\in I^2\\x_5<x_6}} 1_A(x)\; dx_5\; dx_6 \\ &= 2^3\cdot 2 \iint_{\substack{(x_1,x_2)\in I^2\\x_1<x_2}} \iint_{ \substack{(x_3,x_4)\in I^2 \\x_3<x_4 \\x_1<x_2<x_3<x_4 } } dx_1\; dx_2 dx_3\; dx_4 \iint_{\substack{(x_5,x_6)\in I^2\\x_5<x_6}} 1_A(x)\; dx_5\; dx_6 \\ &= 2^3\cdot 2 \iint_{\substack{(x_1,x_2)\in I^2\\x_1<x_2}} \iint_{ \substack{(x_3,x_4)\in I^2 \\x_3<x_4 \\x_1<x_2<x_3<x_4 } } dx_1\; dx_2\; dx_3\; dx_4 \iint_{\substack{(x_5,x_6)\in I^2\\x_5<x_6 \\ \text{ and } \\(\ x_5<x_6<x_1<x_2<x_3<x_4\text{ or} \\ x_1<x_2<x_5<x_6<x_3<x_4\text{ or} \\ x_1<x_2<x_3<x_4<x_5<x_6\ ) }} dx_5\; dx_6 \\[3mm] &\qquad\text{ and now we can compute or (simpler) further use the symmetry:} \\[3mm] &\qquad\text{ The Computation:} \\[3mm] P &= 2^3\cdot 2 \iint_{\substack{(x_1,x_2)\in I^2\\x_1<x_2}} \iint_{ \substack{(x_3,x_4)\in I^2 \\x_3<x_4 \\x_1<x_2<x_3<x_4 } } dx_1\; dx_2\; dx_3\; dx_4 \\ &\qquad\qquad\cdot \frac 12 (\ (x_1-0)^2+(x_3-x_2)^2+(1-x_4)^2\ ) \\ &= 2^3 \iiint_{0<x_1<x_2<x_3<1}(x_1-0)^2(1-x_3)\;dx_1\; dx_2\;dx_3 \\&\qquad +2^3\iiint_{0<x_1<x_2<x_3<1}(x_3-x_2)^2(1-x_3)\;dx_1\; dx_2\;dx_3 \\&\qquad\qquad +2^3\iiint_{0<x_2<x_3<x_4<1}(1-x_4)^2 x_2\; dx_2\; dx_3\;dx_4 \\ &= 2^3 \iint_{0<x_1<x_3<1}(x_1-0)^2(1-x_3)(x_3-x_1)\;dx_1\;dx_3 \\&\qquad +2^3\iint_{0<x_2<x_3<1}(x_3-x_2)^2(1-x_3)x_2\; dx_2\;dx_3 \\&\qquad\qquad +2^3\iint_{0<x_2<x_4<1}(1-x_4)^2 x_2(x_4-x_2)\; dx_2\;dx_4 \\ &=2^3\left(\frac 1{360}+\frac 1{360}+\frac 1{360}\right)=\frac 1{15}\ . \\[3mm] &\qquad\text{ Using the symmetry, from the point we started the computation:} \\[3mm] P&= 2^3\cdot 2 \iint_{\substack{(x_1,x_2)\in I^2\\x_1<x_2}} \iint_{ \substack{(x_3,x_4)\in I^2 \\x_3<x_4 \\x_1<x_2<x_3<x_4 } } dx_1\; dx_2\; dx_3\; dx_4 \iint_{\substack{(x_5,x_6)\in I^2\\x_5<x_6 \\ \text{ and } \\(\ x_5<x_6<x_1<x_2<x_3<x_4\text{ or} \\ x_1<x_2<x_5<x_6<x_3<x_4\text{ or} \\ x_1<x_2<x_3<x_4<x_5<x_6\ ) }} dx_5\; dx_6 \\ &= 2^3\cdot 2\cdot 3 \iint_{\substack{(x_1,x_2)\in I^2\\x_1<x_2}} \iint_{ \substack{(x_3,x_4)\in I^2 \\x_3<x_4 } } \iint_{\substack{(x_5,x_6)\in I^2\\x_5<x_6 \\ \text{ and } \\ x_1<x_2<x_3<x_4<x_5<x_6 }} dx_1\; dx_2\; dx_3\; dx_4\; dx_5\; dx_6 \\ &=2^3\cdot 2\cdot 3\cdot\frac 1{6!}=\frac 1{15}\ . \end{aligned} $$