The law of large numbers allows us to obtain certainty from uncertainty and order from chaos. If an experiment is run repeatedly, any experiment, the average outcome is arbitrarily close to the expected outcome.
Imagine someone will pay off your mortgage, completely, if ten tosses of a fair coin yield between 4 and 6 heads. However, if the number of heads is not close to 5, i.e. less then 4 or more than 6, you will be banished to Siberia. You'd probably decline, politely. A big pay-off with probability 2/3 does not compensate for a "horrible fate" with probability 1/3. (My apologies to those who live happily in Siberia.)
Now change the game slightly. The coin is tossed a million times, and your mortgage is paid in full if there are between 400,000 and 600,000 heads. This time you take the bet. There is virtually no chance of being sent to Siberia. The number of heads will be awfully close to 500,000. This is the law of large numbers.
Let the random variable x have mean m and standard deviation s. Call this x1. Let x2 be the average of two successive trials of x. Roll the die twice, for instance, and take the average. In general, xn is the average of n independent trials, using the variable x. The law of large numbers says that the standard deviation of xn approaches 0. In other words, xn clusters arbitrarily close to its mean. In our coin toss experiment, the fraction of "heads" approaches 1/2.
In the last section we showed that mean and variance both commute with addition. Thus the mean of xn is nm/n, or m. The variance of xn is ns2/n2, or s2/n. The standard deviation is divided by sqrt(n).
As you recall, the odds of being a distance c from the mean is bounded by the variance over c2. Apply the law of large numbers and divide the variance by n. Thus the probability of straying far from the mean is inversely proportional to n.
Let x be the flip of a coin, where heads = 1 and tails = -1. The mean is 0 and the variance is 1. Flip the coin a million times and take the average. The mean is still 0, the variance becomes 0.000001, and the standard deviation is 0.001. Is the average likely to exceed ±0.2? This is the ratio that corresponds to 400,000 or 600,000 heads. Set c = 0.2, and divide the variance by c2. The result is 0.000025. And the odds of "losing" may be even less than 1 in 40,000; this is an upper bound. So, I think I'd take the bet. The law of large numbers is on my side.