Finding a Nash equilibrium

Question

I'm doing the exercises at the end of the paper A Brief Introduction to the Basics of Game Theory by Matthew O. Jackson. I would be grateful if somebody could provide me with solutions to it. I'm not sure about question 2:

Two hotels are considering a location along a newly constructed highway through the desert. The highway is 500 miles long with an exit every 50 miles (including both ends). The hotels may choose to to locate at any exit. These will be the only hotels for any traveler using the highway. Each traveler has their own most preferred location along the highway (at some exit) for a hotel, and will choose to go the hotel closest to that location. Travelers most preferred locations are distributed evenly, so that each exit has the same number of travelers who prefer that exit. If both hotels are the same distance from a traveler’s most preferred location, then that traveler flips a coin to determine which hotel to stay at. A hotel would each like to maximize the number of travelers who stay at it. If Hotel 1 locates at the 100 mile exit, where should Hotel 2 locate? Given Hotel 2’s location that you just found, where would Hotel 1 prefer to locate? Which pairs of locations form Nash equilibria?

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My answers are:

1. hotel 2 should locate at 50 mile exit
2. hotel 1 would prefer to locate 50 mile exit
3. 50, 50 is a Nash equilibrium

However, my gut feeling is telling me I might be wrong. Could you help me with my answer?

Answer 1

Hint: The exits are at $\{0, 50, 100, \ldots, 500\}$. If Hotel 1 is at the 100 mile exit, then Hotel 2 should take the 150 mile exit. Then they get all travellers that prefer something in $[150, 500]$, i.e. $\frac{9}{11}$ of the travellers. Now, if Hotel 2 is at 150 miles, where would Hotel 1 want to be using the same line of thought?

For the last part, find configurations where neither hotel could gain anything from moving.

It is like when two people try to guess a number between $1$ and $100$, say. If the first one guesses $10$, the second one will be annoying and guess $11$, so that they will win for all results in $[11,100]$, i.e. $\frac{9}{10}$ chance of winning.