Symmetry and Geometric Probability

Many probability questions can be answered without a calculation, because the setup treats several outcomes the same way. If nothing in the problem tells two outcomes apart, they have the same probability. Swap the labels and every number in the problem stays the same, so the probabilities stay the same too.

The cleanest case is a sequence of i.i.d. continuous draws $X_1, \dots, X_n$. Ties have probability zero, and permuting the indices leaves the joint distribution unchanged, so each of the $n!$ orderings is equally likely. The chance that $X_7$ is the largest of seven draws is $1/7$, and the chance that $X_1 < X_2 < X_3$ is $1/3! = 1/6$. The common distribution doesn't matter: uniform, normal, or daily stock returns modeled as i.i.d. all give the same answers.

Dice need one extra step because ties can happen. Roll three dice. What is the probability that the faces come out strictly increasing? First the faces must all differ: $P(\text{all distinct}) = \frac{6 \cdot 5 \cdot 4}{216} = \frac{120}{216}$. Given three distinct faces, the six orders are equally likely and exactly one of them is increasing:

A direct count agrees. Choose which three faces appear, $\binom{6}{3} = 20$ ways, and each choice gives exactly one increasing sequence. With $k$ dice the answer is $\binom{6}{k}/6^k$.

Ties also catch people on the simplest question. For two dice, symmetry gives $P(D_1 > D_2) = P(D_1 < D_2)$, and neither one is $\frac{1}{2}$. A tie has probability $\frac{6}{36}$, so each strict inequality has probability $\frac{1}{2}\left(1 - \frac{1}{6}\right) = \frac{5}{12}$.