I'm reading through Shape by Jordan Ellenberg and came across this claim, modeling the movement of a mosquito as a binomial distribution.
The mosquito is fixed to a straight line. Each day, it can choose to whether to fly a kilometer to the northeast or a kilometer to the southwest. It is unbiased, so each path is equally likely.
The claim is that
The chance that a mosquito on its two-hundredth day of life is at least 40km from home is just under 3 in 1000.
A footnote adds that the exact computation is
"What is the probability that a binomial random variable with $p = 0.5$ and $n=200$ takes value at least $120$?"
To cover a distance of, say 40 km northeast, the mosquito would need to have moved 120 km northeast and 80 km southwest. Representing the movement northeast as a "success", this could be the binomial distribution described above. $$ X \sim Bin(0.5,200) $$ And that we are looking to calculate $$ P(X \geq 120) $$
However, it strikes me that the chance that the mosquito is at least 40km from home would be twice of this computed value, since it could be a net 40km northeast or a net 40km southwest. Am I misunderstanding the claim above or modeling the probability incorrectly?
An unambiguous way to model the problem would be to consider the IID random variables $$X_1, X_2, \ldots, X_{200} \in \{-1,1\}$$ where $$\Pr[X_i = 1] = \Pr[X_i = -1] = \frac{1}{2}.$$ Then the distance of the mosquito from the origin after $200$ days will simply be $$D = \left|\sum_{i=1}^{200} X_i\right|.$$ To transform this into a binomial model, we note that $$\frac{X_i + 1}{2} \sim \operatorname{Bernoulli}(p = 1/2).$$ Therefore $$D = \left|\sum_{i=1}^{200} \left(2 \cdot \frac{X_i + 1}{2} - 1 \right)\right| = 2 \left| \sum_{i=1}^{200} \frac{X_i + 1}{2} - 100\right| = 2|Y - 100|,$$ where $$Y \sim \operatorname{Binomial}(n = 200, p = 1/2).$$ It follows that $$\begin{align} \Pr[D \ge 40] &= \Pr[2|Y - 100| \ge 40] \\ &= \Pr[|Y - 100| \ge 20] \\ &= 1 - \Pr[|Y - 100| < 20] \\ &= 1 - \Pr[-20 < Y - 100 < 20] \\ &= 1 - \Pr[80 < Y < 120] \\ &= 1 - \sum_{y = 81}^{119} \binom{200}{y} 2^{-200} \\ &= \small{\frac{570980841170325361571226368804554522130683181662812200441}{100433627766186892221372 630771322662657637687111424552206336}} \\ &\approx 0.00568516. \end{align}$$ This is twice the claimed probability.