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I'm aware that similar questions exist that ask how to solve the problem, but I have a specific question, given the following question/solution: Let $X_1, X_2, X_3$ be independent uniform $[0,1]$ random variables. What is the probability that we can form a triangle with three sticks of length $X_1, X_2, X_3$? Solution: By symmetry, any one of the sides could be the largest with equal probability, so $$ P(X_1, X_2, X_3 \text{form a triangle}) = 3 \int_0^1 \int_{x_2+x_3\geq x_1, x_2 \leq x_1, x_3 \leq x_1} dx_2dx_3dx_1$$ $$ = 3\int_0^1\frac{x_1^2}{2}dx_1 = \frac{1}{2}$$.

My confusion: I imagined that we would argue that "WLOG $X_1$ is the maximum" and not need to multiply by 3. My justification was that we want to find the probability that three numbers are valid sides of a triangle, and of those one of them is actually the largest -- whether we label this $X_1, X_2,$ or $X_3$ does not affect the probability. Why is this incorrect? How can I reason whether to use "WLOG any one could be the largest" or the symmetry argument for other similar problems?

Community
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Jess
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  • It's WLOG in terms of the labels, but not the distributions. By relabeling $X_1$ as the longest segment, you are changing its distribution from uniform to a beta-distribution. Then, there wouldn't be a $3$, but the integrand would change. – Michael Biro Dec 15 '16 at 04:16
  • Why does the distribution change? Is there a similar case where we would be able to argue WLOG we can relabel without changing the probability? Being able to see the example relative to another would help greatly. – Jess Dec 15 '16 at 05:21

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You make an additional assumption ($X_1$ is the longest segment) and this assumption you have to implement in your model… and your model influence your distribution.

Let's explain this on a much easier problem: Throw two dices and calc the probability that they are not equal (or equivalent: that one dice shows a greater value then the other).

It's well known and easy to show that the probability for that is $\frac{30}{36} = \frac{5}{6}$, because all throws are fine except two of a kind.

Now we take your ansatz: Both dices are not equal if one shows are greater value then another, wlog let dice number 1 be that dice.

If we assume the distribution wouldn't change we get a probability for that event of $\frac{15}{21} = \frac{5}{7}$

But obviously $\frac{5}{6} \not= \frac{5}{7}$

The reason is: In the first case you have 36 non-ordered 2-tupel and a discrete uniform distribution for them, in the second you get only 21 ordered 2-tupels. If you keep the discrete uniform distribution on that model you change the probability for each event. So you have to adapt the assumed distribution to keep your results true.

Gono
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