An amusement park ride randomly plays one of six songs during the duration of the ride. Suppose you've already heard two of the six songs. What is the probability that if you ride the ride four more times, you will hear the four remaining songs?

Initially, my first thought is just that the probability you hear one of the songs you want to hear the next time you ride is $\frac{4}{6}$, then IF that is a success, your next probability is $\frac{3}{6}$ and so on, so the total probability is $\frac{4}{6}\cdot \frac{3}{6} \cdot \frac{2}{6} \cdot \frac{1}{6}=\frac{1}{54}$. However, it seems like this is probably more nuanced than that - perhaps I need to factor in conditional probability in some way (i.e. What is the probability that you hear song A, given that you have heard song B, etc.). I'd appreciate any explanations. I believe the trouble is, we don't have any further information about how each individual song is distributed, so we're operating under the assumption that they're all evenly distributed, which is quite a large assumption as well.

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    $\begingroup$ $\frac{1}{54}$ is correct, for exactly the reason you describe. The order of the songs is irrelevent. $\endgroup$ Oct 22, 2023 at 20:01
  • $\begingroup$ By common usage, when something is described in a mathematical setting as being "random" with no further explanation, it means the probability is uniformly distributed. If they later claim otherwise, it means they lied to you in the first place by not using the common interpretation, not that you made an unwarranted assumption. Of course, outside of a mathematical setting, it often means they are ignorant of the possibilities, or at least the need to be specific in their choice of wording. $\endgroup$ Oct 23, 2023 at 16:38


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