Are you bayesian enough to answer?

[Teresa]

The last 10 songs that a media player has played have been from the first half of the list. Given this, a gambler thinks that it is more probable that the next song will more likely be from the second half of the list.
Show with conditional probability that he is wrong!

Thanks a lot, I have been trying all afternoon without results

[dheeraj]

according to conditional probability the probability of some event A occurring is the same whether or not some other event B has been observed it is only when we present some outside condition C that the probabilities change
for example the probability of die A landing on 1 is 1/6 and the probability of A and B adding up to 7 is 5/36(condition C) therefore if we ask the probability of A and b adding up to 8 it is 5/36 if we ask given A is 1 the probability becomes 1/6 because only an outcome of 6 will give us the desired outcome

in your problem the chances of a song landing in the first half of the playlist are 1/2 and are equal to the chances of a song in the second half
the fact that the last 10 songs have been played is irrelevant as each time a new song is played it is chosen randomly and thus has 1/2 chance of being in the first half of the playlist
therefore the probabilities are equal

[Jason]

Ridiculous claim!

Probability of an arbitrarily selected song being from the second half is assumed to be .5 (as in, a fair media player). Now, the probability of having eleven randomly selected songs from the first half is thus (.5)^11 (obviously). Now the probability of getting just ten songs all be from the first half is (.5)^10 (obviously). Now pay careful attention to the next step:

The probability, GIVEN THAT TEN SONGS HAVE BEEN FROM THE FIRST HALF, of having ELEVEN SONGS FROM THE FIRST HALF, is the probability of having eleven songs from the first half divided by the probability of having ten songs from the first half (conditional probability).

(.5)^11 / (.5)^10 = .5

So the probability is still 50-50!

Good luck. =)