Make 2N random selections from size N list whilst avoiding sequential duplicates

Given a list of N objects, how do I make 2N selections in a random order such that no object is chosen twice in succession? (All N objects must be chosen twice.)

My attempt is shown below, and may be useful as a template for testing.

The premise is to generate randomised indices and then check for repeats. When a repetition is found, swap the first repetition with the value before. Finally, if the lower two elements are the same, swap the 2^nd and 3^rd.

<script type="text/javascript">
    function init() {
        for (rep = 0; rep < 100000; rep++) {
            var indices = [0,0, 1,1, 2,2];
            shuffle(indices);

            for (var i = indices.length - 1; i > 1; i--)
                if (indices[i - 1] == indices[i]) {
                    var tmp = indices[i - 1];
                    indices[i - 1] = indices[i - 2];
                    indices[i - 2] = tmp;
                }

            if (indices[0] == indices[1]) {
                var tmp = indices[1];
                indices[1] = indices[2];
                indices[2] = tmp;
            }

            // test
            for (i = indices.length - 1; i > 1; i--)
                if (indices[i - 1] == indices[i])
                    rep = 1E8;  // fail
        }

        dbg.innerHTML = (rep == 1E8) ? "Oh bother." : "OK";
    }

    function randomInt(max) { return Math.floor(Math.random() * max); }
    function shuffle(o) { for (var j, x, i = o.length; i; j = randomInt(i), x = o[--i], o[i] = o[j], o[j] = x); return o; }
</script>

<div>
    <span id="dbg" />
</div>

The failure of the method to deal with the lowest two elements is common to the alternative method of selecting from a reducing list. One potentially ends up with two identical elements remaining.

(Note, the method presented is not recommended for use since does not have a uniform random distribution.)

(Shuffle code taken from Jeff's answer.)