HMMT, 2014

Solution:

Answer: $0.2937156494680644 \ldots$ Note that any up-right path must pass through exactly one point of the form $(n,-n)$ (i.e. a point on the upper-left to lower-right diagonal), and the number of such paths is $\binom{800}{400-n}^{2}$ because there are $\binom{800}{400-n}$ up-right paths from $(-400,-400)$ to $(n,-n)$ and another $\binom{800}{400-n}$ from $(n,-n)$ to $(400,400)$. An up-right path contains a point $(x, y)$ with $|x|,|y| \leq 10$ if and only if $-10 \leq n \leq 10$, so the probability that this happens is

$$\frac{\sum_{n=-10}^{10}\binom{800}{400-n}^{2}}{\sum_{n=-400}^{400}\binom{800}{400-n}^{2}}=\frac{\sum_{n=-10}^{10}\binom{800}{400-n}^{2}}{\binom{1600}{800}}$$

To estimate this, recall that if we normalize $\binom{800}{n}$ to be a probability density function, then it will be approximately normal with mean 400 and variance $800 \cdot \frac{1}{4}=200$. If this is squared, then it is proportional to a normal distribution with half the variance and the same mean, because the probability density function of a normal distribution is proportional to $e^{-\frac{(x-\mu)^{2}}{2 \sigma^{2}}}$, where $\mu$ is the mean and $\sigma^{2}$ is the variance.
Therefore, the $\binom{800}{n}^{2}$ probability density function is roughly proportional to a normal distribution with mean 400 and variance 100, or standard deviation 10. So $\sum_{n=-10}^{10}\binom{800}{400-n}^{2}$ represents roughly one standard deviation. Recall that approximately 68 percent of a normal distribution lies within one standard deviation of the mean (look up the 68-95-99.7 rule to read more), so a good guess would be around .32. This guess can be improved by noting that we're actually summing 21 values instead of 20, so you'd have approximately $.68 \cdot \frac{21}{20} \approx .71$ of the normal distribution, giving an answer of .29.