shuffle array in Go

As your list is just the integers from 1 to 25, you can use Perm :

list := rand.Perm(25)
for i, _ := range list {
    list[i]++
}

Note that using a permutation given by rand.Perm is an effective way to shuffle any array.

dest := make([]int, len(src))
perm := rand.Perm(len(src))
for i, v := range perm {
    dest[v] = src[i]
}

Since 1.10 Go includes an official Fisher-Yates shuffle function.

Documentation: pkg/math/rand/#Shuffle

math/rand: add Shuffle

Shuffle uses the Fisher-Yates algorithm.

Since this is new API, it affords us the opportunity to use a much faster Int31n implementation that mostly avoids division.

As a result, BenchmarkPerm30ViaShuffle is about 30% faster than BenchmarkPerm30, despite requiring a separate initialization loop and using function calls to swap elements.

See also the original CL 51891

First, as commented by shelll:

Do not forget to seed the random, or you will always get the same order.
For example rand.Seed(time.Now().UnixNano())

Example:

words := strings.Fields("ink runs from the corners of my mouth")
rand.Shuffle(len(words), func(i, j int) {
    words[i], words[j] = words[j], words[i]
})
fmt.Println(words)

dystroy's answer is perfectly reasonable, but it's also possible to shuffle without allocating any additional slices.

for i := range slice {
    j := rand.Intn(i + 1)
    slice[i], slice[j] = slice[j], slice[i]
}

See this Wikipedia article for more details on the algorithm. rand.Perm actually uses this algorithm internally as well.