Probabilty question for maths whizz

[mikeplow]

If a Cd has nine tracks on, and is set to play on shuffle, what are the odds that it would play all nine tracks in order from track one to nine?

With thanks for any answer, and explanation, given.

[bibblebub]

1/9!

[scotman]

I believe it is 1/9 x 1/8 x 1/7 etc

[nightmare]

As Bibblebub states, 1/9! i.e. 1/362880

[ChuckFickens]

I make it 362880 to 1

[nightmare]

strictly speaking, the ODDS are actually 362879 to 1 (against)
The probability is 1/362880

[bibblebub]

1/362880 = 362879 to 1 against

[Ellipsis]

You also asked for an explantion mikeplow ...

At the outset, the odds of playing track 1 are 1 in 9.
If track 1 is played, that leaves eight tracks remaining to be played. The odds of playing track 2 are 1 in 8.
And so on down to track 9, which if all the previous eight tracks have been played, must be played - the odds are 1 in 1.

These odds are all multiplicative so you end up with 1/9 x 1/8 etc, which is the same as 1/(9x8x7x6x5x4x3x2x1) which can be written as 1/9! (the ! meaning "factorial" - see http://en.wikipedia.org/wiki/Factorial).

9! = 362,880, so the odds are 1 in 362,880.

Assumptions:
All tracks are played (you don't quit early, have a power cut, etc.)
A perfect random number selector (no bias to certain tracks)
Shuffle mode will always play all tracks before repeating a track