Go rand.Intn same number/value

For the functions in the rand package to work you have to set a 'Seed' value. This has to be a good random value as decided by the user because - as per https://golang.org/pkg/math/rand/#Rand.Seed this is the value golang uses to set the system to a deterministic state first to then generate a number based on that value.

For the sample code to work, you can try

func main() {
    rand.Seed(time.Now().UnixNano())
    fmt.Println("My favorite number is ", rand.Intn(10))
}

time.Now().UnixNano can give an arbitrary(like) number as the value is in 'one thousand-millionth of a second'

2 reasons:

1. You have to initalize the global Source used by rand.Intn() and other functions of the rand package using rand.Seed(). For example:

   rand.Seed(time.Now().UnixNano())
   

   See possible duplicate of Difficulty with Go Rand package.
   Quoting from package doc of rand:

   Top-level functions, such as Float64 and Int, use a default shared Source that produces a deterministic sequence of values each time a program is run. Use the Seed function to initialize the default Source if different behavior is required for each run.

2. The Tour runs examples on the Go Playground which caches its output.
   See details at Why does count++ (instead of count = count + 1) change the way the map is returned in Golang.

As explained you have to initalize the global Source used by rand.Intn() and other functions of the rand package.

Besides using rand.Seed() option, you also can create the source using the methods rand.NewSource() and rand.New().

source := rand.NewSource(time.Now().UnixNano())
r := rand.New(source)
randomNumber := r.Intn(10)

fmt.Println("A random number: ", randomNumber)