Well, in a full motion simulator it's not impossible to surmount--anyone who has ridden on the Mission Space ride (especially the orange version) at Disney World will confirm that (it's an awesome ride, by the way). But unless you have a few million dollars to squander, yes, the feeling of acceleration would fall into that category of "non-visual sensory inputs" that you just can't get from a game. Great point.

LOL, I fail to see how the availability of expensive peripherals that provide improvements negate my comments, which were about "typical" setups. The average gamer has neither a high-tech, $300 wheel nor a $150 helmet (which still doesn't provide any form of peripheral vision).

Have Slashdot stories always been this ridiculous and I just haven't noticed before?

Even using the most realistic settings, most video racing games fail to physically equate to real-life driving—for instance:

• Typical racing wheels allow 1/2 to 3/4 turn of the wheel, lock-to-lock; a typical passenger car has well over 3 turns lock-to-lock, and even a high-performance sports car has over 2.
• Peripheral vision, which is very important in real-life driving, is completely lost in video games; many racing games do not even allow you to look in a different direction than you are going (e.g. turn your head to look left while continuing to drive straight), and those that do allow it require some button-press that is not the same as just turning your head.
• Most non-visual sensory inputs to driving (slight variations in vibrations, etc.) are lost in video games.

So there's no doubt that there's little, if any, physical correlation of video game driving skills to real-life driving skills, even in the most realistic games. However, more abstract skills learned in video gaming, such as situational awareness and reaction times, certainly do apply to real-life driving, especially in high-pressure accident-avoidance situations, where the split-second reaction times honed by racing games are clearly advantageous. Clearly, that was outside the scope of this "study", but the conclusion stated by the title of this post is entirely erroneous.

Did you even bother reading the whole comment? "Agencies often cite more than one exemption when withholding part or all of the material sought in an open-records request."

HA! This has to be the best comment I've seen on Slashdot, ever! You hit the nail on the head!

I took all of my notes throughout university (including engineering courses) using OpenOffice.org. The equation editor in OpenOffice is easy-to-learn, fast (as in, no mouse use required and the keystrokes are all sane), and the completed equations look great. (By default, there isn't a keyboard shortcut for inserting a new equation, so you'll need to manually assign one—I used Ctrl-Shift-F, if I remember correctly.

Your example would almost work as is; it would be entered as:

f_x (x) = int from -infinity to infinity f (x, y) dy

Or, if you prefer your parentheses to stretch (in case you have fractions inside, or what have you):

f_x left ( x right ) = int from -infinity to infinity f left ( x, y right ) dy

Either way, it comes out looking very nice. The one thing that takes some getting used to is that you need to make liberal use of whitespace (e.g. between f and the opening parenthesis of the function), otherwise things will occasionally come out looking a little strange. The best part is, when you don't know what you need to type for a particular symbol, you can select it from the menu and OO will insert the plaintext code, which makes it very easy to learn the code for new items.

From the linked article: "The books published by The Espresso Machine will have a recommended sales price of $8 per copy, although the final decision will be left to each retailer."

Possibly, but this at least eliminates the need for shipping the printed books (which weigh a LOT) all over the world. Plus, for me at least, I can only spend maybe five minutes at a time reading from a computer screenâ"and I'm even a Gen Y computer programmer. There's something about reading from a computer screen that is just a lot harder than reading from a paper.

Okay, you're right in one sense, because I wasn't completely accurate in my previous comment (I wrote it hurriedly during the last few minutes of my lunch break at work)--the probability does in fact increase somewhat with the sample size. However, your description doesn't describe the situation accurately at all; the correct analogy would be that, each of 100 stars has a 1 in 100 chance of having a planet spiraling into it; in this case, viewing all 100 stars does not yield a probability of 1 of seeing a planet spiraling into it. Let me try another simplified description. First, consider a single, six-sided die. One of the six sides has a one on it; if you examine a single side of the die (roll it), you have a 1 in 6 chance of seeing the one; if you examine three of the six sides, you have a 1 in 2 chance of seeing the one; if you examine all six sides, you have a 1 in 1 chance (probability of 1) of seeing the one. This is analogous to your "1 out of 100 stars" situation above, but it is not analogous to real life. Now consider six normal, six-sided dice. If you roll all six of them, what is the probability that at least one of them will come up with a one? You will probably immediately realize the probability is not 1, but calculating it is a bit of a math problem--it's been a while since my college statistics class, but if I remember correctly, the correct way of finding it is to calculate the probability that none of the dice will be a one, and then subtract that from 1, thus:

• For each die, the probability of it not being a one is 5/6;
• Thus, the probability of none of the six dice being a one is (5/6)^6, or about 0.335;
• Thus, the probability of at least one of the six dice being a one is 1 - 0.335, or about 0.665, which is significantly less than 1.

Going back to your 1 in 100 probability, if there are 100 stars and each has a 1 in 100 chance of having a planet spiraling into it, then the probability of any of the 100 stars having a planet spiraling into it is 0.634. Examining only 50 of the 100 leaves a probability of 0.5 * 0.634 or only 0.317. Now, we're making some huge assumptions about the probabilities of this event occurring; but just for the sake of discussion, let's just say that, for any given star, there is a 1 in 10^15 (one in a trillion) chance that, at the present time, it has a planet spiraling into it. (Given the relatively small number of stars we know of that have any planets at all, I suspect that number is a significant overestimate, but I'll use it.) Using your estimate of 100 billion stars in the universe, that makes the probability that any star exists, anywhere in the universe with a planet spiraling into it about 0.0000999, or 1 in 10,000, which is pretty small. Now I'll assume that we have examined 1 billion of those stars closely enough that we would be able to detect this occurrence (which I would guess is a gross overestimate); that makes the probability of any star that we have examined being in the midst of the occurrence about 0.000000999, or 1 in a million. So, being what I would say is quite generous with all of the numbers, we have a 1 in a million chance of seeing this.