For some purposes, the [weak law of large numbers][1] is superior to the strong law. This is one of those purposes. The weak law is quantitative, unlike the strong law.

Let $X$ be the arithmetic mean of a random sample of size $n$ chosen from a given distribution with mean $\mu$ and variance $\sigma^2$. Then for any $\epsilon>0$ $$ P(|X-\mu| \ge \epsilon) \le \frac{\sigma^2}{ n \epsilon^2} $$

Example: tossing a fair coin (mean $\mu=1/2$, variance $\sigma=1/4$), I want to get within $\epsilon = 0.1$ of the mean $1/2$. If I toss it $n=1000$ times, then I will be within that error with probability at least $$ 1-\frac{\sigma^2}{n\epsilon^2} = \frac{159}{160} \approx 0.99 $$ [1]: http://mathworld.wolfram.com/WeakLawofLargeNumbers.html

For some purposes, the [weak law of large numbers][1] is superior to the strong law. This is one of those purposes. The weak law is quantitative, unlike the strong law.

Let $X$ be the arithmetic mean of a random sample of size $n$ chosen from a given distribution with mean $\mu$ and variance $\sigma^2$. Then for any $\epsilon>0$ $$ P(|X-\mu| \ge \epsilon) \le \frac{\sigma^2}{ n \epsilon^2} $$

Example: tossing a fair coin (mean $\mu=1/2$, variance $\sigma=1/4$), I want to get within $\epsilon = 0.1$ of the mean $1/2$. If I toss it $n=1000$ times, then I will be within that error with probability at least $$ 1-\frac{\sigma^2}{n\epsilon^2} = \frac{159}{160} \approx 0.99 $$

Now $X = H/n$ where $H$ is the number of times my coin came up on heads. So this means (fair coin, tossing 1000 times) $$ P(400 < H < 600) > 0.99 $$ [1]: http://mathworld.wolfram.com/WeakLawofLargeNumbers.html

For some purposes, the weak law of large numbers is superior to the strong law. This is one of those purposes. The weak law is quantitative, unlike the strong law.

Let $X$ be a random sample of size $n$ from a given distribution with mean $\mu$ and variance $\sigma^2$. Then for any $\epsilon>0$ $$ P(|X-\mu| \ge \epsilon) \le \frac{\sigma^2}{ n \epsilon^2} $$

For some purposes, the [weak law of large numbers][1] is superior to the strong law. This is one of those purposes. The weak law is quantitative, unlike the strong law.

Let $X$ be the arithmetic mean of a random sample of size $n$ chosen from a given distribution with mean $\mu$ and variance $\sigma^2$. Then for any $\epsilon>0$ $$ P(|X-\mu| \ge \epsilon) \le \frac{\sigma^2}{ n \epsilon^2} $$

Example: tossing a fair coin (mean $\mu=1/2$, variance $\sigma=1/4$), I want to get within $\epsilon = 0.1$ of the mean $1/2$. If I toss it $n=1000$ times, then I will be within that error with probability at least $$ 1-\frac{\sigma^2}{n\epsilon^2} = \frac{159}{160} \approx 0.99 $$ [1]: http://mathworld.wolfram.com/WeakLawofLargeNumbers.html