Some of the cleverest puzzles that people have ever created tend to be the ones with the most counterintuitive solutions: You swear that the answer has to be X, but it turns out to be something very different. And there is a particular reason why the answer is what it is.
A great example of this is the famous Monty Hall problem (no relation to Monty Python). It works like this: Suppose you're on a game show. You have three doors in front of you. Behind one of the doors is a brand-new car, and behind the other two are piles of rotten cheese (the original problem uses goats, but some people would want to have a goat. I assume far fewer people want rotten cheese). You have no way of knowing where the car is and where the cheese is (the cheese is tightly sealed, so you can't detect it by smell). You have to open exactly one of the doors, and you get whatever is behind that door.
However, the host makes things interesting. Once you choose a door, he stops you before you open it. He opens one of the other two doors and reveals a sealed container of rotten cheese, meaning that the car is behind one of the two unopened doors. Then he asks you a question: Should you stay with your original pick, or should you switch to the other unopened door?
The intuitive answer is that it doesn't matter: it's a 50/50 shot. So just go with whichever way you feel more comfortable. But in one of the most bizarre twists in mathematics, that's not the best strategy. Instead, you should switch to the other door, because it has not a 1-in-2, but 2-in-3, chance of having the car behind it.
Let's agree that at the very start, your initial guess had a 1/3 chance of picking the car door. Once another door is opened, if you switch, you will lose. This is a sure-fire outcome, given the original pick. Translation: a 1-in-3 chance of switching resulting in a loss.
Likewise, your initial guess had a 2/3 chance of picking a cheese door. Remember the part where the host opens up another door that has cheese behind it? That condition is the key to the entire setup: Whichever of the other two doors he opens, it MUST be a cheese door. So what happens if you pick one of the cheese doors, which you have a 2/3 chance of doing? The host HAS to open the other cheese door. And guess what's behind the third door? "A new caaaaarrr!" How do you get that new car? By switching your initial guess. If you initially picked a cheese door and then switch your pick, you will win.
There are a few other ways of proving this result, but I think that breaking it down into if-then scenarios makes it easier to understand. Even then, I'll bet that several of you aren't buying it. Nope, there's gotta be some kind of catch that I haven't explained yet. So how could we verify these results? With our second-best tool available: simulation.
You can simulate this scenario with two players and three playing cards. Let one of the cards be the "car," and the other two be the "cheese." The host follows exactly the same rules as the game show:
(1) The dealer shuffles the three cards and places them face-down, but he or she knows where the "car" is.
(2) The other player points to one of the cards but does not turn it over.
(3) The dealer turns over one of the other two cards that is a "cheese."
(4) The dealer asks the player whether he or she wants to stay with the original pick or switch to the other face-down card. The other player answers.
(5) The appropriate card is revealed.
If you do this enough times, you'll notice something: The results will decidedly tilt in favor of "switch wins." Over the long run, you should expect to get about twice as many "switch wins" as "stay wins." But is it entirely possible that any given play results in a "stay win"? Of course it is. A 1-in-3 chance is unlikely but certainly not out of the question. Does beating these odds on any given play disprove these results? Of course not--the sample size is much too small.
However, if an accurate simulation with, say, 100 repetitions, or even better, 1000 repetitions, implied that the odds were close to 50/50, would that say something significant? Sure. Without getting into the statistical analysis, it's a safe bet that if your simulation results after 1000 repetitions come very close to 2/3 chance of success, the actual odds are NOT 50%.
Now what the hell does this have to do with racism? Plenty.
A very common excuse that racism deniers will give is that "each case should be decided individually." For example, take the claim that whether George Zimmerman's acquittal had anything to do with race should be decided on its own merits, not compared to any outside case. Or whether a black man's being arrested had anything to do with his being arrested in the first place should entirely depend on the immediate circumstances, not on the fact that a black person is more likely to be arrested for the exact same crime as a white person is. By this sleight of hand, they try to cover up the overall trend, which would expose the racism to those who might be otherwise blind to it. It would be like trying to judge the probability of this car-and-cheese game based on a single trial. Switching is more likely to win, but could it lose? Of course. One-in-three is a fairly low probability, but by no means is it insurmountable. So someone could play the game one time and mistakenly conclude that the game does indeed have 50/50 odds.
That's why whites should look at the overall picture to confirm the presence of racism in our society. To privileged people, not every given act of racism is obvious racism. But the overall trend is blatantly skewed in one direction. In other words, when whites try to cover up racism by claiming that we should consider allegedly racist acts on a case-by-case basis, they are doing so to consciously or unconsciously avoid scrutiny of the bigger issue, thus maintaining the system of white privilege. Or, in short, they are committing racism.