关于 goroutine 调度问题

学习 goroutine 的时候看过这个例子，不太明白为什么会这样？请看代码：

// This sample program demonstrates how to create goroutines and
// how the scheduler behaves.
package main

import (
    "fmt"
    "runtime"
    "sync"
)

// main is the entry point for all Go programs.
func main() {
    // Allocate 1 logical processor for the scheduler to use.
    runtime.GOMAXPROCS(1)

    // wg is used to wait for the program to finish.
    // Add a count of two, one for each goroutine.
    var wg sync.WaitGroup
    wg.Add(2)

    fmt.Println("Start Goroutines")

    // Declare an anonymous function and create a goroutine.
    go func() {
        // Schedule the call to Done to tell main we are done.
        defer wg.Done()

        // Display the alphabet three times
        for count := 0; count < 3; count++ {
            for char := 'a'; char < 'a'+26; char++ {
                fmt.Printf("%c ", char)
            }
        }
    }()

    // Declare an anonymous function and create a goroutine.
    go func() {
        // Schedule the call to Done to tell main we are done.
        defer wg.Done()

        // Display the alphabet three times
        for count := 0; count < 3; count++ {
            for char := 'A'; char < 'A'+26; char++ {
                fmt.Printf("%c ", char)
            }
        }
    }()

    // Wait for the goroutines to finish.
    fmt.Println("Waiting To Finish")
    wg.Wait()

    fmt.Println("\nTerminating Program")
}

这段代码为什么始终先输出第二个 goroutine 的内容，即大写字母，再输出第一个 goroutine 的内容小写字母？

goroutine 不是无序的吗，为什么这个例子总是顺序一一，并且第二个 goroutine 先打印？