3 ms·

I believe this issue is pre-dated even further, back when iPod users didn't like that occasionally a random shuffle would play the same song back-to-back. So th

by bumby 3y ago

I believe this issue is pre-dated even further, back when iPod users didn't like that occasionally a random shuffle would play the same song back-to-back. So the 'random' shuffle was made a bit less random.

saalweachter 3y ago
CD players inspired the same discussion. Probably doesn't go back further than that, unless jukeboxes has a random play option.
antiterra 3y ago
I think the issue was playing the same artist or album back-to-back. So they made ‘smart shuffle’ in 2005. (https://www.wired.com/story/requiem-for-the-ipod-shuffle/ https://www.wired.com/story/requiem-for-the-ipod-shuffle/) A shuffle already implies shuffling like a deck of cards, so you wouldn’t get duplicates unless you had two of the same card, and I that’s how it was described in the manual.
- jimbobthrowawy 3y ago
  Is there any algorithm that would do that without actually storing a list of indices, or track IDs? I can't think of why someone would prefer a truly random song being played from a playlist rather than the "deck of cards" method.
hgomersall 3y ago
It can be biased and still random.
- lostlogin 3y ago
  Can you explain that?
  - drdeca 3y ago
    A non-uniform probability distribution is still a probability distribution, and, as long as it doesn’t assign probability 1 to any particular outcome, is still “random”.
Dylan16807 3y ago
You don't need to make it less random to achieve that. A 100% random shuffle won't have that issue, you just have to make sure it's actually a shuffle and not something else.
- xtagon 3y ago
  Given a set of songs or albums {a, b, c} Suppose shuffling results in ordered set [b, a, c] Suppose user plays through the entire set. Now it's shuffled again before repating. Suppose the new shuffle results in ordered set [c, a, b] The user now hears c play after c
  - Dylan16807 3y ago
    I'm going to ignore the other ways to handle that and just point out that that is multiple shuffles and won't be a problem if the user initiates all shuffles.
    - xtagon 3y ago
      If the user has to initiate all shuffles, it won't play continuously. If you handle it another way, that demonstrates that it doesn't "just" work with a pure shuffle without using those tricks.
      - Dylan16807 3y ago
        1. Without a clearer spec, we should not assume the user wants any repeats at all. 2. Your claim is not even true as written. 3. I'm tempted to assume the user has a reasonable amount of music, so it will last as long as necessary. And you can't force me not to assume that.

by bumby 3y ago

saalweachter 3y ago

CD players inspired the same discussion. Probably doesn't go back further than that, unless jukeboxes has a random play option.

antiterra 3y ago

I think the issue was playing the same artist or album back-to-back. So they made ‘smart shuffle’ in 2005. (https://www.wired.com/story/requiem-for-the-ipod-shuffle/ https://www.wired.com/story/requiem-for-the-ipod-shuffle/) A shuffle already implies shuffling like a deck of cards, so you wouldn’t get duplicates unless you had two of the same card, and I that’s how it was described in the manual.

jimbobthrowawy 3y ago
Is there any algorithm that would do that without actually storing a list of indices, or track IDs? I can't think of why someone would prefer a truly random song being played from a playlist rather than the "deck of cards" method.

hgomersall 3y ago

It can be biased and still random.

lostlogin 3y ago
Can you explain that?
- drdeca 3y ago
  A non-uniform probability distribution is still a probability distribution, and, as long as it doesn’t assign probability 1 to any particular outcome, is still “random”.

Dylan16807 3y ago

You don't need to make it less random to achieve that. A 100% random shuffle won't have that issue, you just have to make sure it's actually a shuffle and not something else.

xtagon 3y ago
Given a set of songs or albums {a, b, c} Suppose shuffling results in ordered set [b, a, c] Suppose user plays through the entire set. Now it's shuffled again before repating. Suppose the new shuffle results in ordered set [c, a, b] The user now hears c play after c
- Dylan16807 3y ago
  I'm going to ignore the other ways to handle that and just point out that that is multiple shuffles and won't be a problem if the user initiates all shuffles.
  - xtagon 3y ago
    If the user has to initiate all shuffles, it won't play continuously. If you handle it another way, that demonstrates that it doesn't "just" work with a pure shuffle without using those tricks.
    - Dylan16807 3y ago
      1. Without a clearer spec, we should not assume the user wants any repeats at all. 2. Your claim is not even true as written. 3. I'm tempted to assume the user has a reasonable amount of music, so it will last as long as necessary. And you can't force me not to assume that.