gulp-watch works only on initial run - gulp

Question

What I do?
Run gulp (SCSS files are being processed, I get a CSS file)
I change any SCSS file again
Expected:
CSS file from 1. is updated with the changes from 2.
What happens?
CSS file from 1. isn't changed
Command line output:
$ gulp
[09:24:28] Using gulpfile c:\Users\User\_dev\github\project\gulpfile.js
[09:24:28] Starting 'sass'...
[09:24:28] Finished 'sass' after 98 ms
[09:24:28] Starting 'default'...
[09:24:28] Finished 'default' after 7.31 μs
[09:24:35] sass-watch saw _base.scss was changed
[09:25:39] sass-watch saw _base.scss was changed
gulpfile.js:
gulp.task('sass', function() {
watch({ glob: 'css/**/*.{scss,sass}', name: 'sass-watch'})
.pipe(plumber())
.pipe(sass())
.pipe(gulp.dest('css'))
});
gulp.task('default', ['sass']);
Notes:
Issue on GitHub (gulp)
Issue on GitHub (gulp-watch)
gulpfile.js on GitHub Gist)
OS: Win7
node: 0.10.29
npm: 1.4.14
The way the source files are piped in is not important. The result stays the same when using gulp.src()

Answer 1

I dont think your sass task is correctly written.
Try something like this:
var gulp = require('gulp');
var sass = require('gulp-sass')
gulp.task('sass', function () {
gulp.src('PATH-TO-SASS-FILES/*.scss')
.pipe(sass())
.pipe(gulp.dest('./css'));
});