I apologize for being slow writing the conclusion of the explanation for why my Dearly Beloved and I would expect one more ride following our plan to keep re-riding Disaster Transport as long as a fairly flipped coin came up tails. It’s been a busy week, and actually, I’d got stuck trying to think of a way to explain the sum I needed to take using only formulas that a normal person might find, or believe. I think I have it.

What I wanted to know was the expected value of the number of rides, which is equal to the expected number of tails in a row that a flipped coin would give. The probability of getting zero tails in a row (a head right away) is one-half. The probability of one tail (followed by a heads) is one-quarter. Two tails followed by a heads is one-eighth, and so on. The expected value of an outcome is found by multiplying each possible outcome by its probability, and then adding all these products up. This sum, written this way, is , which is not one of the most immediately familiar infinite series forms but which — with a little manipulation, done in the last entry in this number — there is a formula for. Where I was dissatisfied was that I didn’t know where the formula was from, or how to prove it was right, so couldn’t feel quite satisfied with using that. (Actually, I have a fair idea how I might prove it was right, but we’ll let that go.)

The best approach, I conclude, is that which Bug was doing originally. The original problem is a pain to deal with, so, let’s look for something equivalent that’s easier to do. (And, actually, I could do the original problem using this same routine, although it’s a *lot* of work — never hard, just tedious, but it is tedious that we really, really want to get away from.) That’s one of the standard mathematical tricks that you often don’t get to do in homework or on exams, since they mean to test knowledge of a particular technique. But when you’re doing real mathematics, for fun or curiosity, you can look for easier yet equivalent problems.

So, here’s the equivalent and easier problem: think of the infinite number of imaginable rides that might come up, if you re-ride every time a flipped coin comes up tails, and if none of the real-word problems were to stop you. What’s the chance of getting ride number 1? One-half. What’s the chance of getting ride number 2? That’s one-quarter, since you can get it only if the first and second coin flips were tails. If we look at how many rides we expect from the first two possible ones, that probability is one times one-half plus one times one-quarter, or three-quarters. The chance of getting that third ride is one-eight. The expected value of the rides in the first three attempts would be one times one-half plus one times one-quarter plus one times one-eighth, or seven-eighths.

Where this is going is another infinite series: , which is a very familiar form of infinite series known as the geometric series. There are very familiar formulas for working out this sum, and proving they’re true can be done with only a bit of cleverness and be reasonably convincing. But if we’re interested only in this one specific problem, we can do a remarkably convincing wordless proof:

Please imagine the zig-zagging of lines to carry on ad infinitum; each new diagonal or horizontal is meant to split the remaining area in half, and ought to be reasonably convincing that the sum of all these powers of a half adds up to one. (I apologize for the poor registry of the lines, but I spent *far* too long wrestling with Manga Studio to get things looking that tolerable. I’ll clean it up for the book publication.)

And so this is why we could expect that once we began going back around for new rides on Disaster Transport, using a coin flip after each ride to determine whether we go around again, we would expect just the one further ride.

I’d hope everyone next wonders how many rides we did take. Well, on our first ride — after which we were to flip a coin for re-rides — we had a great ride. Some of the other riders were also talking about how this was going definitely to be their last time around, and they’d come to the park to get in a last ride before this odd little landmark was closed, and talked with the attendants about what the ride meant to them. Maybe it was how we were ready to feel a ride was special, maybe it was the added touch of knowing, briefly, some of the other people we were sharing this final ride with, but the roller coaster felt more fun that time, and we decided that this was the best final ride we could get. So we didn’t flip the coin at all, but we were most satisfied.

To gather all these entries together, and leave me keenly aware how much I need to come up with a better naming scheme for open-ended series like this: