Among my entertainments is listening to the Greatest Generation podcast, hosted by Benjamin Ahr Harrison and Adam Pranica. They recently finished reviewing all the Star Trek: The Next Generation episodes, and have started Deep Space Nine. To add some fun and risk to episode podcasts the hosts proposed to record some episodes while drinking heavily. I am not a fun of recreational over-drinking, but I understand their feelings. There’s an episode where Quark has a sex-change operation because he gave his mother a heart attack right before a politically charged meeting with a leading Ferengi soda executive. Nobody should face that mess sober.

At the end of the episode reviewing “Babel”, Harrison proposed: there’s 15 episodes left in the season. Use a random number generator to pick a number from 1 to 15; if it’s one, they do the next episode (“Captive Pursuit”) drunk. And it was; what are the odds? One in fifteen. I just *said*.

The question: how many episodes would they be doing drunk? As they discussed in the next episode, this would imply they’d always get smashed for the last episode of the season. This is a straightforward expectation-value problem. The expectation value of a thing is the sum of all the possible outcomes times the chance of each outcome. Here, the possible outcome is adding 1 to the number of drunk episodes. The chance of any particular episode being a drunk episode is 1 divided by ‘N’, if ‘N’ is the number of episodes remaining. So the next-to-the-last episode has 1 chance in 2 of being drunk. The second-from-the-last has 1 chance in 3 of being drunk. And so on.

This expectation value isn’t hard to calculate. If we start counting from the last episode of the season, then it’s easy. Add up , ending when we get up to one divided by the number of episodes in the season. 25 or 26, for most seasons of Deep Space Nine. 15, from when they counted here. This is the start of the harmonic series.

The harmonic series gets taught in sequences and series in calculus because it does some neat stuff if you let it go on forever. For example, every term in this sequence gets smaller and smaller. (The “sequence” is the terms that go into the sum: , and so on. The “series” is the sum of a sequence, a single number. I agree it seems weird to call a “series” that sum, but it’s the word we’re stuck with. If it helps, consider: when we talk about “a TV series” we usually mean the whole body of work, not individual episodes.) You can pick any number, however tiny you like. I can then respond with the last term in the sequence bigger than your number. Infinitely many terms in the sequence will be smaller than your pick. And yet: you can pick any number you like, however big. And I can take a finite number of terms in this sequence to make a sum *bigger* than whatever number you liked. The sum will eventually be bigger than 10, bigger than 100, bigger than a googolplex. These two facts are easy to prove, but they seem like they ought to be contradictory. You can see why infinite series are fun and produce much screaming on the part of students.

No Star Trek show has a season has infinitely many episodes, though, however long the second season of Enterprise seemed to drag out. So we don’t have to worry about infinitely many drunk episodes.

Since there were 15 episodes up for drunkenness in the first season of Deep Space Nine the calculation’s easy. I still did it on the computer. For the first season we could expect drunk episodes. This is a number a little bigger than 3.318. So, more likely three drunk episodes, four being likely. For the 25-episode seasons (seasons four and seven, if I’m reading this right), we could expect or just over 3.816 drunk episodes. Likely four, maybe three. For the 26-episode seasons (seasons two, five, and six), we could expect drunk episodes. That’s just over 3.854.

The number of drunk episodes to expect keeps growing. The harmonic series grows without bounds. But it keeps growing slower, compared to the number of terms you add together. You need a 31-episode season to be able to expect at four drunk episodes. To expect five drunk episodes you’d need an 83-episode season. If the guys at Worst Episode Ever, reviewing The Simpsons, did all 625-so-far episodes by this rule we could only expect seven drunk episodes.

Still, three, maybe four, drunk episodes of the 15 remaining first season is a fair number. They shouldn’t likely be evenly spaced. The chance of a drunk episode rises the closer they get to the end of the season. Expected length between drunk episodes is interesting but I don’t want to deal with that. I’ll just say that it probably isn’t the five episodes the quickest, easiest suggested by taking 15 divided by 3.

And it’s moot anyway. The hosts discussed it just before starting “Captive Pursuit”. Pranica pointed out, for example, the smashed-last-episode problem. What they decided they meant was there would be a 1-in-15 chance of recording each episode this season drunk. For the 25- or 26-episode seasons, each episode would get its 1-in-25 or 1-in-26 chance.

That changes the calculations. Not in spirit: that’s still the same. Count the number of possible outcomes and the chance of each one being a drunk episode and add that all up. But the work gets simpler. Each episode has a 1-in-15 chance of adding 1 to the total of drunk episodes. So the expected number of drunk episodes is the number of episodes (15) times the chance each is a drunk episode (1 divided by 15). We should expect 1 drunk episode. The same reasoning holds for all the other seasons; we should expect 1 drunk episode per season.

Still, since each episode gets an independent draw, there might be two drunk episodes. Could be three. There’s no reason that all 15 couldn’t be drunk. (Except that at the end of reviewing “Captive Pursuit” they drew for the next episode and it’s not to be a drunk one.) What are the chances there’s no drunk episodes? What are the chances there’s two, or three, or eight drunk episodes?

There’s a rule for this. This kind of problem is a mathematically-famous one. We get our results from the “binomial distribution”. This applies whenever there’s a bunch of attempts at something. And each attempt can either clearly succeed or clearly fail. And the chance of success (or failure) each attempt is always the same. That’s what applies here. If there’s ‘N’ episodes, and the chance is ‘p’ that any one will be drunk, then we get the chance ‘y’ of turning up exactly ‘k’ drunk episodes by the formula:

That looks a bit ugly, yeah. (I don’t like using ‘y’ as the name for a probability. I ran out of good letters and didn’t want to do subscripts.) It’s just tedious to calculate is all. Factorials and everything. Better to let the computer work it out. There *is* a formula that’s easy enough to work with, though. That’s because the chance of a drunk episode is the same each episode. I don’t know a formula to get the chance of exactly zero or one or four drunk episodes with the first, one-in-N chance. Probably the only thing to do is run a lot of simulations and trust that’s approximately right.

But for this rule it’s easy enough. There’s this formula, like I said. I figured out the chance of all the possible drunk episode combinations for the seasons. I mean I had the computer work it out. All I figured out was how to make it give me the results in a format I liked. Here’s what I got.

The chance of these many drunk episodes | In a 15-episode season is |
---|---|

0 | 0.355 |

1 | 0.381 |

2 | 0.190 |

3 | 0.059 |

4 | 0.013 |

5 | 0.002 |

6 | 0.000 |

7 | 0.000 |

8 | 0.000 |

9 | 0.000 |

10 | 0.000 |

11 | 0.000 |

12 | 0.000 |

13 | 0.000 |

14 | 0.000 |

15 | 0.000 |

Sorry it’s so dull, but the chance of a one-in-fifteen event happening 15 times in a row? You’d expect that to be pretty small. It’s got a probability of something like 0.000 000 000 000 000 002 28 of happening. Not *technically* impossible, but yeah, impossible.

How about for the 25- and 26-episode seasons? Here’s the chance of all the outcomes:

The chance of these many drunk episodes | In a 25-episode season is |
---|---|

0 | 0.360 |

1 | 0.375 |

2 | 0.188 |

3 | 0.060 |

4 | 0.014 |

5 | 0.002 |

6 | 0.000 |

7 | 0.000 |

8 or more | 0.000 |

And things are a tiny bit different for a 26-episode season.

The chance of these many drunk episodes | In a 26-episode season is |
---|---|

0 | 0.361 |

1 | 0.375 |

2 | 0.188 |

3 | 0.060 |

4 | 0.014 |

5 | 0.002 |

6 | 0.000 |

7 | 0.000 |

7 | 0.000 |

8 or more | 0.000 |

Yes, there’s a greater chance of no drunk episodes. The difference is *really* slight. It only looks so big because of rounding. A no-drunk 25 episode season has a chance of about 0.3604, while a no-drunk 26 episodes season has a chance of about 0.3607. The difference comes from the chance of *lots* of drunk episodes all being even worse somehow.

And there’s some neat implications through this. There’s a slightly better than one in three chance that each of the second through seventh seasons won’t have any drunk episodes. We could expect two dry seasons, hopefully not the one with Quark’s sex-change episode. We can reasonably expect at least one season with two drunk episodes. There’s a slightly more than 40 percent chance that some season will have three drunk episodes. There’s just under a 10 percent chance some season will have four drunk episodes.

There’s no guarantees, though. Probability has a curious blend. There’s no predicting when any drunk episode will come. But we can make meaningful predictions about groups of episodes. These properties seem like they should be contradictions. And they’re not, and that’s wonderful.

I hope you’re going to do a follow-up post to tell us how many it turns out to be!

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I’m curious! … What the heck, I can at least try to stop in as seasons end. (They’re doing only one episode a week, so it’ll be a couple years until the full results are in, barring surprises or changes.)

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