Game backlog calculator: whether the pile ever ends, and the rate that decides
Your data
The last field is the one people leave out, and it is the one that decides. Sales, gifts and anything a subscription drops in your lap all count, whether or not you meant to acquire them.
Results
| Games you finish each month | — |
| Buying rate that cancels that out exactly | — |
| Net change in the pile each month | — |
| Months until the pile is gone | — |
| Which is this many years | — |
| Hours a week just to stand still | — |
The second row is the whole page in one number. There is a buying rate above which the pile never empties, it does not depend on how big the pile is, and it is exactly the rate at which you finish things. One game past it and the answer stops being a date.
What the buying rate does to the wait
| New ones each month | Months until it is gone | Years |
|---|
Read that table downwards and watch the wait stop behaving. The steps in the first column are even, and the ones in the second are not: each equal push towards the break-even rate costs more than the one before it, until the last step costs everything at once. It is a hyperbola, and the bold row is where it stops having a finite answer.
The consequence is that two people with the same pile can have completely different futures. Give both of them the same forty games and the same hours in the evening, let one bring home one a month and the other two, and the first is done in a few years while the second is never done at all. Nothing about the pile explains that. The rates do all of it.
It also explains why buying less feels so much more effective than playing more, even though both move the same subtraction. Playing more is bounded by the hours in your evening, and buying less is not bounded by anything, so near the break-even rate a single purchase avoided moves the answer further than an extra hour a week.
The last row of the results turns that around and gives the number that is actually usable. It is not how long the pile takes, it is how many hours a week your current buying rate already commits you to before the pile shrinks by a single game.
An average hour count for a pile of very different games is a rough thing, and the page cannot know that the long one you keep postponing is four times the average. What it can tell you is which side of the threshold you are on, and that answer is not sensitive to the average being a little off.
Why does the size of the pile matter so little?
Because the pile is drained by a difference between two rates, and the size only decides how long that difference has to run. If the difference is comfortable the pile is short work whatever its size, and if the difference is nearly nothing no size is small enough.
Two people with the same forty games and the same evenings can therefore have completely different futures, one finishing in a few years and the other never finishing, purely from how often something new arrives.
Is there really a buying rate where it becomes impossible?
Yes, and it is not a vague warning: it is exactly the rate at which you finish games. Above it the pile grows every month, so there is no date to compute, and below it there always is one.
That threshold does not depend on the size of the pile at all, which is why a purge of the library changes the date but never changes which side of the line you are standing on.
Why does the wait grow so fast near that line?
Because you are dividing by the difference, and the difference is heading for zero. Equal steps towards the threshold do not cost equal amounts of time, they cost more and more, and the last step costs all of it at once.
The table on the page walks the buying rate up in equal steps for exactly that reason. The first steps look mild, the later ones do not, and the row at the threshold has no answer to print.
Is it better to buy less or to play more?
They move the same subtraction, so on paper neither is special. In practice they are not symmetric: the hours you can play are bounded by your evenings, and the games you do not buy are bounded by nothing.
Near the threshold that asymmetry is stark. One purchase avoided can move the answer further than an extra hour a week, because it is acting on the side of the subtraction that still has room.
My games are wildly different lengths. Is one average any use?
For the date, only roughly, and the page will not pretend otherwise. A pile with one enormous game in it behaves differently from a pile of equal ones with the same total.
For the question that matters it is quite robust, because which side of the threshold you are on is decided by a comparison rather than by a precise value. Being ten or twenty per hundred wrong about the average rarely moves you across the line, and if it does, you were standing on it.
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