Originally Posted by MaxOctane

p^n + C(n,1)*p^(n-1)*q + C(n,2)*p^(n-2)*q^2 + ... + C(n,r)*p^(r)*q^n-r) = probability that an event will happen at least r times in n trials. Using this equation, you get a probability of 1/30284.

Originally Posted by MaxOctane

...because I thought it meant "random". Even when I push it, it says "Random disc play" or something like that, but my experience with RDM is as such...

I'm driving to work in the morning, and I decide to put my MP3 CD in the player. It has 135 MP3's on it and they are all in the ROOT directory! I hit RDM and listen to about 4 songs before I get to work. At lunch I go to the store and buy a present for my lovely wife whose birthday is on Monday. On the way there, I hear about 3 songs. Construction makes it longer to get there, so I only hear 2 songs on the way back.

So, I dust off the ol discrete math and combinatorics/permutations college book and go to town.

On the way to work, I heard 4 unique songs. On the way to the store, I heard one song which I heard on the way to work. On the way back from the store, I heard the same song again which I heard to work and to the store AND the second song I heard was a song I heard when going to the store!

This means I heard 9 songs total, and 6 unique songs. One song I heard 3 times, and another I heard 2 times.

First let's calculate the probability of listening to 9 unique songs in a row.

135/135*134/135*133/135... 127/135 = .76, so I have a 76% probability of hearing 9 unique songs in a row. I'd take those odds to Vegas any time.

Now, what is the probability that I listen to 9 songs, and hear the same one 3 times and the other 6 are unique out of a total set of 135 unique songs.

135/135*1/135*1/135*134/135*133/135...*129/135 = ~1/21,347 or .00468451% probability. Not quite sure I would take those odds with me to Vegas.

But, I heard the same song 3 times, and another song 2 times. What is the probability of hearing 1 song 3 times, another song 2 times, and the other 4 are unique songs.

135/135*1/135*1/135*134/135*1/134*133/135*132/135*131/135*130/135
which equals ~1/2,753,754 or .0000363141% probability.

What is the probability of listening to 9 songs and hearing AT LEAST 1 song 3 times?

p^n + C(n,1)*p^(n-1)*q + C(n,2)*p^(n-2)*q^2 + ... + C(n,r)*p^(r)*q^n-r) = probability that an event will happen at least r times in n trials. Using this equation, you get a probability of 1/30284.84

Now, It's been awhile since I've done this crap, so I could be wrong. If you decide to correct me, please do so tactfully. Regardless, my point is that the random seed that the GM Radio uses SUCKS!!!! I swear in a Cd full of 135 MP3's I'll hear the same song 8 or 9 times in just a few days. The chances of this being coincidence in practically nil as previously proven.

This makes me want to go to the dealer and ask to have the radio replaced, just to see the look on their face. Maybe if I record a dataset of several hundred songs and plot them on a scatter chart to support my argument. I wonder if they'll do it.

Originally Posted by kwickag

You'd be amazed how often the same number comes up when you use randomizing code in a computer.

Originally Posted by JWLaRue

My assumption is that the algorithm used to randomize the play sequence is one that uses a 'seed' number to figure out what to play. The seed number could be anything from the serial number of the unit to the number of tracks on the CD. This gives only a some-what random sequence.

This would mean that the "random" in these players is only "random" during each play sequence. Each time you start playing the disc again is a "new" random sequence....it does not carry over from each time the CD player is used. In fact, I would expect to see the behavior that you have observed.

-Jeff