Peeking At Cedar Point


A glimpse of the Transport Refreshments stand, in September 2012, hidden from view by the construction fence.

Back a couple months I wrote way too much about the problem of how many rides to expect on Cedar Point’s Disaster Transport, if we chose whether to re-ride it based on a random event. It struck me there’s another problem created by the amusement park’s removal of the indoor bobsled roller coaster. This one is based on Transport Refreshments, the block of food and drink stands which stood by the removed Disaster Transport and Space Spiral.

Specifically: what’s to become of that area? When my Dearly Beloved and I visited in late September the area was walled off, for construction, but one could rationalize any kind of fate for it. The block might get torn down to provide space for new rides; it might be left as-is, with the name Transport Refreshments left as a mysterious reference that new visitors would have to learn something of park history to understand; or the stands might be re-themed to the GateKeeper roller coaster being built. By now, probably, park-watchers really know, but when we visited, there wasn’t any telling, except by peeking over the fence.

The problem is you can’t see very much, because the fence is in the way. I’m tall and can hold my camera pretty high and so could get glimpses showing that the buildings hadn’t as of late September been torn down, and that they even had the sign in place, but that doesn’t mean much.

It does suggest a cute problem, though, one that’s easy to solve using calculus and maybe is solvable by easier tools. That problem’s, how do you get the best view of the hidden Transport Refreshments? Going up close to the fence means the fence obscures more of your field of view; getting farther away — the ground is roughly level here — reduces the field of view obscured by the fence, but also reduces the Transport Refreshments’ angular diameter. There’s probably a best spot to see what’s beyond, but, where is it?

To turn this into a word problem, let’s pretend things are nice round numbers: that the person doing the viewing has eyes about six feet off the ground, that the fence is eight feet tall, and that — four feet past the fence — the main sign for the Transport Refreshments stands twelve feet tall. I am sure these arbitrarily plucked numbers will produce only good results.

It Would Have Been One More Ride Because


Some props from the Disaster Transport bobsled coaster are still in use for Halloweekend.

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.

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The Help Needed To Get to One


This is a view from a lower vantage point of the Disaster Transport roller coaster car.

So, it’s established that my little series, representing the number of rides we could expect to get if we based re-riding on a fair coin flip, is convergent. So trying to figure out the sum will get a meaningful answer. The question is, how do we calculate it?

My first impulse is to see if someone else solved the problem first, for exactly the reasons you might guess. This is a case where mathematics textbooks can have an advantage over the web, really, since an introduction to calculus book is almost certain to have page after page of Common Series Sums. Figuring out the right combination of keywords to search the web for it can be an act of elaborate guesswork. Mercifully, Wikipedia has a List of Mathematical Series which covers my problem exactly. Almost.

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Why Not Infinitely Many More Rides?


The Disaster Transport building was used as a Toy Factory haunted house during Halloweekends.

Returning to the Disaster Transport ride problem: by flipping a coin after each ride of the roller coaster we’d decide whether to go around again. How many more times could I expect to ride? Using the letter k to represent the number of rides, and p(k) to represent the probability of getting that many rides, it’s a straightforward use of the formula for expectation value — the sum of all the possible outcomes times the probability of that particular outcome — to find the expected number of rides.

Where this gets to be a bit of a bother is that there are, properly speaking, infinitely many possible outcomes. There’s no reason, in theory, that a coin couldn’t come up tails every single time, and only the impatience of the Cedar Point management which would keep us from riding a million times, a billion times, an infinite number of times. Common sense tells us this can’t happen; the chance of getting a billion tails in a row is just impossibly tiny, but, how do we know all these outcomes that are incredibly unlikely don’t add up to something moderately likely? It happens in integral calculus all the time that a huge enough pile of tiny things adds up to a moderate thing, so why not here?

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Just One More Ride?


A corner view of Disaster Transport, with the Space Spiral visible behind it.

Given that we know the chance of getting any arbitrary number — let’s say k, because that’s a good arbitrary number — of rides in a row on Disaster Transport, using the scheme where we re-ride if the flipped coin comes up tails and stop if it comes up heads, the natural follow-up to me is: how many more rides can we expect? It’s more likely that we’d get one more ride than two, two more rides than three, three more rides than four; there’s a tiny chance we might get ten more rides; there’s a real if vanishingly tiny chance we’d get a million more rides, if Cedar Point didn’t throw us out of the park and tear the roller coaster down first.

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How Many Last Rides?


Disaster Transport building and the Space Spiral at Cedar Point, July 2012

So our scheme for getting a last ride in on Disaster Transport without knowing in advance it was our last ride was to flip a coin after each ride, and then re-ride if the coin came up tails. (Maybe it was heads. It doesn’t matter, since we’re supposing the coin is equally likely to come up heads as tails.) The obvious question is, how many times could we expect to ride? Or put another way, how many times in a row could I expect a flipped coin to come up tails, before the first time that it came up heads? The probability tool used here is called the geometric distribution.

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Reading the Comics, July 28, 2012


I intend to be back to regular mathematics-based posts soon. I had a fine idea for a couple posts based on Sunday’s closing of the Diaster Transport roller coaster ride at Cedar Point, actually, although I have to technically write them first. (My bride and I made a trip to the park to get a last ride in before its closing, and that lead to inspiration.) But reviews of math-touching comic strips are always good for my readership, if I’m readin the statistics page here right, so let’s see what’s come up since the last recap, going up to the 14th of July.

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