Sometimes mathematics can be counterintuitive. Take, for example, the following equation:

0.9999… = 1

The ellipsis following the **nines **means that the **nines** continue to infinity. On the surface, this equation seems wrong. I mean, think about it. You can tell that 0.9 is close to 1, and that 0.99 is even closer, but doesn’t it seem like the zero followed by all those nines will just *approach* 1, and not quite make it? It does seem that way. And if it were a *finite* series of nines, then that would be true. But it’s actually *infinite*, which means that there are always, always, always more nines. There are nines upon nines upon nines. A whole universe filled with nines, hidden in those three little dots. And because it’s infinite, it doesn’t just approach 1, it is equal to 1. The strange thing is: 0.999… is actually a different way to write 1.

There are many proofs for this (check out this Wikipedia page to see some of them), but this is the one I like best:

Let x = 0.999... 10x = 9.999... (multiply each side of the equation by 10) 10x = 9 + 0.999... (just split the right side into two) 10x = 9 + x (by the definition of x) 9x = 9 (subtract x from both sides) x = 1 (divide both sides by 9)

Despite this proof (and all the others), many people find this to just *feel* wrong. It’s counter-intuitive. 0.999… should *not* equal 1, many people think. It’s intuitive to me. I guess I’m one of the lucky ones.

But there is another tricky math conundrum that *is* counter-intuitive to me, and which I have never been able to understand. (If my friend Jeff is reading this, he’s shuddering right now, because he knows what’s coming.) Yes, Jeff, it’s the…

### Monty Hall Problem

The Monty Hall problem, stated simply, is this:

You are on a game show (probably hosted by veteran game show host Monty Hall). There are three doors in front of you, Door A, Door B, and Door C. Behind one door is a new car, but you don’t know which one. Behind each of the other two is a goat. You are given the chance to pick one door. Say you pick Door A. Before showing you what’s behind your door, Monty (who knows what’s behind the doors) opens another door (say, Door C). There’s a goat there. So now there are two doors left, A and B. He gives you the choice: “Do you want to keep Door A, or switch to Door B?” What should you do?

The obvious answer, the intuitive answer, the answer that *doesn’t make me angry* is this: You might as well keep Door A, because nothing has changed. There is still a 33% chance the car is behind Door A, and a 33% chance it’s behind Door B.

But NO. That’s not right. The actual answer is that you’re better off switching. Believe it or not, there is now a 67% chance that the car is behind Door B. And it just makes me mad. I’m not going to explain how it works. I’d probably do a lousy job of it. You wouldn’t ask Ayn Rand to explain the beauty of socialism, would you? But unlike Ms. Rand, I am convinced. I do accept that the correct solution to the Monty Hall problem is to switch. Two-thirds of the time, you’d get that car you wanted. I accept that. But I don’t like it. (I also accept that Donald Trump is president — doesn’t mean I have to like it.) If you want to see an explanation, click here. You’ll see that I’m not alone in not liking it — it became quite a controversy, particularly after Marilyn vos Savant (who has the world’s highest IQ, and one must mention that whenever mentioning her name) published her (correct) answer in *Parade* magazine in 1990.

I did come up with an elegant solution of my own a while back. Just decide that you prefer goats to cars, and when Monty opens Door C, run to the door and take that goat. You’ll save thousands on lawn care.

Either way, I’ll stick with the **nines**. They make sense to me.