Do I break even in a fair game? Thanks for reading. The following screenshot I took from the book "The Art of Probability" by Richard Hamming.

I'm really confused.
How is it that the probability of heads is $\frac{1}{2}$, meaning that we expect to get heads around half the time after a large number of tosses, yet the author of this book is telling us that we should expect the score to deviate by $\sqrt{n}$ after a large number of tosses?!
I always thought that if I played a fair game, the more times I played that game the more I would expect to "break-even"! But according to this guy, I don't!

Update:
In general, I'm a little confused as to how to interpret the expected value of $X$ vs the standard deviation of $Y$. In my comments (both below and to the top answer thus far) I think I make my confusion pretty clear!
Additionally, let's say I let a new random variable $Z$ correspond to the sum of a bunch of $Y_s$, say $m$ of them. How would I interpret $Z$s expected value and standard deviation?!?!?!
Thank you!
 A: Suppose you start with $\$1000$ and play $100$ games where you bet $\$1$ on each win. That's a fair game, so the expected value you end up with is $\$1000$.
However, that doesn't mean you'll always end up with exactly $\$1000$. What if you lose all $100$ games? It's unlikely, but still possible, and you'd end up with $\$900$. With the same low probability you could win all $100$ games and end up with $\$1100$. In fact, it's rather unlikely that you end up with exactly $\$1000$, the chance for that is only about $8\%$.
Of course you'll still end up close to $\$1000$. But how close? One mathematical way to quantify that is the variance of a distribution. Suppose your score at the end of the game is the square of the difference between your actual balance and the expected balance. Then the variance is simply the expected value of your score. In this case, the variance is exactly $\$^2 100$.
The standard deviation is the square root of the variance, so in this case $\$10$. It's a good approximation (up to a constant factor) of the average absolute deviation of this game, which would be the average absolute number of dollars your balance changes by and is probably what interests you here.
Now the nice thing about variances is that variances of sums of independent variables add up! The variance of a single game is $\$^21$, and all games are independent, so the variance of $n$ games is $\$^2n$. So the standard deviation after $n$ games is $\$\sqrt n$, so the average absolute deviation is about $\$c \sqrt n$ for some constant $c = \sqrt{\frac{2}{\pi}} \approx 0.7979$.
Now let's look at what happened in the end. One player won, and one player lost. The money the winner has is on average exactly $\$1000$ plus the average absolute deviation, since that's the definition of the average absolute deviation. Of course the loser has on average exactly $\$1000$ minus the average absolute deviation.
Note that the winner depends on the outcome of the game, so you can't just go ahead and say "I expect to end up with on average about $\$8$ less money." No, what you can say is that "I expect to end up with on average about $\$8$ less money in case I lose", and you can also say that "I expect to end up with on average about $\$8$ more money in case I win."
To finally answer your question: Most of the time you won't break even, either gaining or losing a few dollars. But on average that gain or loss balances out, so the expected amount of money you gain is zero. Remember that "expected" is a technical term that doesn't quite match up with the regular English word, so for example even though the expected number of heads in a coin flip is $\frac 12$, you shouldn't end up expecting to see half a head.
