# Clarifications: Carl has more than a 50% chance of arriving first

I do not have enough "reputation points" to comment on this post: Carl has more than a 50% chance of arriving first but I have some clarification questions.

Walter and Carl both often need to travel from Location A to Location B. Walter walks, and his travel time is Normal with mean w minutes and standard deviation σ minutes (travel time can’t be negative without using a tachyon beam, but assume that w is so much larger than σ that the chance of a negative travel time is negligible). Carl drives his car, and his travel time is Normal with mean c minutes and standard deviation 2σ minutes (the standard deviation is larger for Carl due to variability in traffic conditions). Walter’s travel time is independent of Carl’s. On a certain day, Walter and Carl leave from Location A to Location B at the same time.

(a) Find the probability that Carl arrives first (in terms of Φ and the parameters)

$$\Phi\Bigl(\frac{w-c}{\sqrt5 \sigma}\Bigr)$$

(b) Give a fully simplified criterion (not in terms of Φ ), such that Carl has more than a 50% chance of arriving first if and only if the criterion is satisfied.

"We need to find conditions on $$w,c, \sigma$$ such that $$\Phi\big(\frac{w-c}{\sqrt{5}\sigma}\big) > 0.5$$ Since the standard normal quantile function $$\Phi^{-1}$$ is strictly increasing, applying it to both sides leaves the inequality unchanged, and we obtain $$\frac{w-c}{\sqrt{5}\sigma}>0$$. Since $$\sigma\geq0$$ this occurs if and only if $$w>c$$."

How do you read this critereon? I am reading it as "average time for Walter needs to be greater than average time for Carl, regardless of any other metrics such as standard deviation. Is that correct?

(c) Walter and Carl want to make it to a meeting at Location B that is scheduled to begin w+10 minutes after they depart from Location A. Give a fully simplified criterion (not in terms of Φ) such that Carl is more likely than Walter to make it on time for the meeting if and only if the criterion is satisfied.

we first need to calculate the probability of making it to the meeting on time for both Carl and Walter:

$$p_C \equiv P(\mbox{Carl on time}) = P(C\leq w + 10) = \Phi\left(\frac{w-c+10}{2\sigma}\right)$$ $$p_W \equiv P(\mbox{Walter on time}) = P(W\leq w + 10) = \Phi\left(\frac{10}{\sigma}\right)$$

Now, since $$\Phi$$ is monotonic, $$p_C > p_W$$ if and only if:

$$\frac{w-c+10}{2\sigma} > \frac{10}{\sigma}$$

Simplifying, this is equivalent to $$w-c > 10$$.

How did they get the functions in $$\Phi$$ ? I am lost on the methodology for plugging in; Part (a) made sense, but I do not get the same answer for (c).

• I agree with your interpretation of (b). For (c), how would you express $P(C\leq k)$ for some arbitrary number $k$, in terms of $\Phi$? – Arthur Oct 25 '18 at 19:00