In fact, the "correct" definition of Littlewood-Richardson coefficients shows a surprising $S_3$-symmetry among all the indices $\lambda,\mu,\nu$. See http://arxiv.org/abs/0704.0817.

A further example related to symmetric functions is the symmetry between the area and bounce statistics of Dyck paths. See for instance Chapter 3 of http://www.math.upenn.edu/~jhaglund/books/qtcat.pdf. No combinatorial proof of symmetry is known.

There are many enumeration problems with "hidden symmetry." For instance, what is the probability that 1 and 2 are in the same cycle of a (uniform) random permutation of $1,2,\dots,n$? More interesting, suppose that I shuffle an ordinary deck of 26 red cards and 26 black cards. I turn the cards face up one at a time. At any point before the last card is dealt, you can guess that the next card is red. What strategy maximizes the probability of guessing correctly? The surprising answer is that all strategies have a probability of 1/2 of success! There is a very elegant way to see this.

In fact, the "correct" definition of Littlewood-Richardson coefficients shows a surprising $S_3$-symmetry among all the indices $\lambda,\mu,\nu$. See Thomas and Yong - An $S_3$-symmetric Littlewood–Richardson rule.

A further example related to symmetric functions is the symmetry between the area and bounce statistics of Dyck paths. See for instance Chapter 3 of Haglund - The $q, t$-Catalan numbers and the space of diagonal harmonics. No combinatorial proof of symmetry is known.

There are many enumeration problems with "hidden symmetry." For instance, what is the probability that 1 and 2 are in the same cycle of a (uniform) random permutation of $1,2,\dots,n$? More interesting, suppose that I shuffle an ordinary deck of 26 red cards and 26 black cards. I turn the cards face up one at a time. At any point before the last card is dealt, you can guess that the next card is red. What strategy maximizes the probability of guessing correctly? The surprising answer is that all strategies have a probability of 1/2 of success! There is a very elegant way to see this.

In fact, the "correct" definition of Littlewood-Richardson coefficients shows a surprising $S_3$-symmetry among all the indices $\lambda,\mu,\nu$. See http://arxiv.org/pdf/0704.0817v1.pdf.

A further example related to symmetric functions is the symmetry between the area and bounce statistics of Dyck paths. See for instance Chapter 3 of http://www.math.upenn.edu/~jhaglund/books/qtcat.pdf. No combinatorial proof of symmetry is known.

There are many enumeration problems with "hidden symmetry." For instance, what is the probability that 1 and 2 are in the same cycle of a (uniform) random permutation of $1,2,\dots,n$? More interesting, suppose that I shuffle an ordinary deck of 26 red cards and 26 black cards. I turn the cards face up one at a time. At any point before the last card is dealt, you can guess that the next card is red. What strategy maximizes the probability of guessing correctly? The surprising answer is that all strategies have a probability of 1/2 of success! There is a very elegant way to see this.

In fact, the "correct" definition of Littlewood-Richardson coefficients shows a surprising $S_3$-symmetry among all the indices $\lambda,\mu,\nu$. See http://arxiv.org/abs/0704.0817.

A further example related to symmetric functions is the symmetry between the area and bounce statistics of Dyck paths. See for instance Chapter 3 of http://www.math.upenn.edu/~jhaglund/books/qtcat.pdf. No combinatorial proof of symmetry is known.

There are many enumeration problems with "hidden symmetry." For instance, what is the probability that 1 and 2 are in the same cycle of a (uniform) random permutation of $1,2,\dots,n$? More interesting, suppose that I shuffle an ordinary deck of 26 red cards and 26 black cards. I turn the cards face up one at a time. At any point before the last card is dealt, you can guess that the next card is red. What strategy maximizes the probability of guessing correctly? The surprising answer is that all strategies have a probability of 1/2 of success! There is a very elegant way to see this.