How do I figure this out?

[dstebbins]

I'm taking a summer semester right now, with a REALLY tough math class. Here's an example of the kind of stuff I've got.

 

Your watching a movie on a Youtube playlist with x videos. On the first video, you click the random button and click "play next" once to take you away from the first video, and click "play next" a second time to take you to another video that could be any one of the x number of videos, including the first one. After that, you watch x videos in a row without touching the keyboard or mouse. What are the odds that the entire movie will be played in perfect chronological order?

 

Needless to say, my instructor is a big computer nerd.

 

This is the first week of the class (block 2 of Summer Semester), and he wanted to give us a taste, so if we didn't know beforehand what we were getting ourselves into, it's best if we drop the class if it's not needed for our degrees (which, for me, it's not). Is this problem easier to solve than I think it is, or should I drop? HELP!

[Bignose]

I am a little confused. Is this a homework problem? Are you supposed to know it before you went in? Or is it something that you are going to learn how to do? Because if this is a class on probability, then these are the kind of problems you will learn how to answer. Whether it is "hard" or not is going to depend on how well you learn the material (which will be a combination of your study habits and will to learn and some natural ability, too).

 

One problem isn't a good indicator of how hard a class it will be. What does the syllabus look like? How about the text? How about the impressions other people who have taken the class have?

[dstebbins]

Well, I just took this course as an elective, and I don't have to answer this particular problem; it's just designed to give me an idea of what I'm getting myself into if I don't drop the class.

 

My question is, am I panicking over nothing? How is this problem worked, and I can decide for myself if I'm ready for this class.

[Bignose]

It is a problem based on conditional probabilities.

 

First, you calculate the odds that the 1st video in the playlist is the earliest chronologically (assuming some distribution of the random variables, probably uniform). Then, with the condition that the 1st video is the earliest, what is the probability that the 2nd video in the playlist is 2nd earliest. Then with the condition of both the 1st and the 2nd video are in the right order, what is the probability that the 3rd is also in order, etc. etc.

 

If you know how to calculate those probabilities -- this is probably one of the many topics the class will cover -- it isn't hard. A lot of times the trickiest part of a probability question is to make sure you are answering the question as asked. The recent raffle thread in the other math sub-forum is a good example. I didn't read the question exactly right and wasn't answering it exactly right.

[dstebbins]

It is a problem based on conditional probabilities.

 

First, you calculate the odds that the 1st video in the playlist is the earliest chronologically (assuming some distribution of the random variables, probably uniform). Then, with the condition that the 1st video is the earliest, what is the probability that the 2nd video in the playlist is 2nd earliest. Then with the condition of both the 1st and the 2nd video are in the right order, what is the probability that the 3rd is also in order, etc. etc.

 

If you know how to calculate those probabilities -- this is probably one of the many topics the class will cover -- it isn't hard. A lot of times the trickiest part of a probability question is to make sure you are answering the question as asked. The recent raffle thread in the other math sub-forum is a good example. I didn't read the question exactly right and wasn't answering it exactly right.

 

Well, needless to say, the probability that the first one in the playlist is the earliest is one in x. So, I guess, if that were the case (the one in x chance happened), then y would be returned 1. If it didn't happen, then y would return 0.

 

I can figure that, no problem, but that would only work if every trial was independent of every other trial. I can also figure out (with relative ease) the odds of inter-dependent trials like card-counting at the casinos, but this is sort of a mixture between the two.

 

How would I figure the probability that all the y's would return 1, and not one y would return 0?

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Wait, something just came to me.

 

Let's pretend that there are 20 videos in the playlist. If the first one (chosen completely at random) is video 1, then that's a one in twenty chance of y = 1.

 

If y = 1 the first time, then the chances of y = 1 the second time would be one twentieth of the original chance, or one in four hundred.

 

Therefore, the chance that we'd have twenty y = 1 would be one in 20^20, or 1.048576e+026 (copied and pasted from my Graph Calc program).

 

Is that the jist of it?

[Bignose]

No, you aren't using the conditions right.

 

Going with 20 again. To start, assuming a uniform distribution, it is indeed a 1 in 20 chance that the 1st video is the oldest.

 

But, now you need to use that condition that the 1st video is already the oldest. That leaves 19 more to be distributed uniformly again. So, the chance that the second video is the second oldest is now 1 in 19.

 

And so on.

 

You end up with 1 in 20! chance, not 20^20.

[dstebbins]

No, you aren't using the conditions right.

 

Going with 20 again. To start, assuming a uniform distribution, it is indeed a 1 in 20 chance that the 1st video is the oldest.

 

But, now you need to use that condition that the 1st video is already the oldest. That leaves 19 more to be distributed uniformly again. So, the chance that the second video is the second oldest is now 1 in 19.

 

And so on.

 

You end up with 1 in 20! chance, not 20^20.

 

I don't think you understand the problem.

 

See, when you click "randomize" on a Youtube playlist, whenever it's done with one video, it goes to a different random one, but after that one, it could go to any video in the playlist other than the one it just finished. You could, theoretically, end up watching the same two videos over and over again, if randomness decides so.

 

Let me try a different situation. Suppose you have a sack full of twenty numbered balls, with the numbers 1 through 20 on them. You pull out one ball and record its number. If you pull out the #1 ball, you check the box that says "success?" and draw another ball, but immediately after you draw the second ball (and this is very important), you put the ball from Stage 1 back into the sack.

 

Stages 3 through 20 are the same as stage 2 (draw another ball and discard the previous ball immediately afterwards), so all the stages except 1 have a 1 in 19 chance.

 

So, in that case, would my chances of it being in perfect numerical order be (1/20)*(1/19)^19?

[Bignose]

Then that sounds right. It wasn't very clear what the situation was -- I was thinking of a fixed list that gets rearranged, but doesn't re-sample with replacement. All of these conditions (and how to use the mathematical terminology to explain it better) should be covered by such a class.

[dstebbins]

Then that sounds right. It wasn't very clear what the situation was -- I was thinking of a fixed list that gets rearranged, but doesn't re-sample with replacement. All of these conditions (and how to use the mathematical terminology to explain it better) should be covered by such a class.

 

You've obviously never watched a randomized Youtube playlist. I often do that with playlists from Whose Line is it Anyway, watching games like Hoedown and Scenes from a Hat. Often, I think to myself "Wait, didn't I just watch this one?"