OK, you asked for it.

(snip some incorrect probability calculations)

[*] Ignoring very small second-order terms. For instance, the probability of the next track being from the same album isn't 1/160, it's slightly reduced by the fact you've just played a song from that album, so that not quite 1/160 of the remaining songs are also from that album.

In this case, the "very small second-order terms" add up to a whole lot of error, as the probability you came up with (22.5%) is almost 44% higher than the actual probability (15.6%).

The correct calculation works like this:

We're playing a track. It's in one of our 160 playlists. The only way we can reasonably do this calculation is to assume that all playlists are of identical size (as I mentioned earlier, the calculation becomes really difficult if they vary in size). So, let's assume that each of our 160 playlists has (1449/160), or about 9.06, songs.

The chance that the next track will also come from the same playlist is (((1449/160) - 1)/1448); there are (1449/160) - 1 songs left in your playlist, and 1448 left to be chosen from. Similarly, the chance that the track after that will also come from the same playlist is (((1449/160) - 2)/1447).

The chance of this (3 in a row) happening somewhere in the playlist is therefore about 3.92%, when we multiply these 2 probabilities by the 1446 possible positions in a playlist where a sequence of 4 can start.

Now we factor in the 4th song. The probability of the 4th song not being from the same playlist is ((1449 - (1449/160))/1446); there are all the songs except our one playlist to choose from, and 1446 songs left in total. So the probability of 3 in a row from one playlist, followed by a 4th from another playlist, is a long ugly equation which evaluates to a probability of about 3.91%. Not discarding the extra figures, when we multiply this result by 4 (for the 4 different locations the "not like the others" song can go), we get the aforementioned 15.6%.

This calculation, however, is pretty much worthless for figuring out how often this sort of occurrence will happen in the real world, because nobody's playlists are all the same size, and because it's possible (even likely, given the design of the Empeg software) for items to be in multiple playlists. Bottom line, for any practical distribution of songs on an Empeg, I'm pretty sure the only way to calculate this probability exactly is to use brute force. If that's the case (I haven't proven it, of course), given the fact that in this example of 1449 tracks there are more than 1.18x10^3953 permutations to brute-force through, you wouldn't live anywhere near long enough to get the answer. Neither would our universe, probably.

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Daniel M. Zimmerman, Caltech Computer Science
Mk.2 #060000058, 36GB
Mk.1 #00101, 10GB