What is either player's chance of winning this game? I came across the following interesting puzzle in a magazine.

Suppose we have $101$ suitcases, numbered $0, \dots, 100$, with a bottle of rum in suitcase $x$, and a bottle-opener in suitcase $y$ such that $|x-y|\leq 50$.
We have $2$ players playing the $T$-suitcase game, where player $A$ is allowed to choose $x$, and $y$ in the above described manner. Player $B$ is allowed to open $T$ suitcases, but they all have to be consecutive numbers. The first suitcase he is allowed to choose any number $0,1,\dots, 100$, and for the other suitcases he is only allowed to open a suitcase left or right from an already opened suitcase. Player $B$ wins if he can find both the rum and the bottle-opener, and player $A$ wins if he doesn't.
What are the chances of winning for player $A$, and $B$ in the cases $T=60, 70, 80$, considering the fact they both try to optimize their winning chances?

I found this extremely interesting, but I didn't get further than a few observations. For example, it is clear that $2T-101$ suitcases in the middle will always be opened, so player $A$ will never put both the rum, and opener in these suitcases.
I've been reading a little about game theory, and I was wondering if maybe some ideas concerning the Nash-Equilibrium can be used, and try to come up with an equilibrium situation where both player $A$, and $B$ will not change their current strategy since it would result in a lower winning chance. The prisoners dilemma is a nice illustration of such principle. To use an idea of a Nash Equilibrium, I guess I need to make a list of possible strategies for $A$, and $B$, but I'm not sure how.
I'm not interested in detailed solutions, but I wonder if someone could provide some enlightening ideas, or insight into the problem. I really like this problem.
 A: Below I will provide a full solution, but in an attempt to honor the spirit of the question, I will first try to provide some ideas and hints about how one can think about this problem.
Ideas
In a vacuum, it is hard to say what $B$ should do. However, if we assume that $B$ knows $A$'s strategy, it is generally much easier to see what $B$ should do. (Of course, this will be trivial unless $A$'s strategy is random.) Thus, we can analyze a strategy of $A$ by assuming $B$ knows that this strategy is being used. This will provide a lower bound on the effectiveness of $A$.
On the other hand, we can also try to analyze a strategy of $B$ without respect for the strategy of $A$. Certain strategies of $B$ will have some probability of guessing correctly no matter what strategy $A$ chooses. (Again, in the non-trivial case, such a strategy will be random.) This can provide an upper bound of the effectiveness of any strategy of $A$.
If the lower bound and upper bound meet, then the problem is solved.
Simple strategies
One simple strategy of $A$ is to choose with equal probability between $(0,50)$ and $(50,100)$. Even if $B$ knows that $A$ is employing this strategy, guessing $50$ provides no information, and $0$ and $100$ cannot both be covered (assuming $T < 101$), so $B$ must resort to guessing. Thus, $A$ can certainly always win with at least probability $\frac{1}{2}$.
One simple strategy of $B$ is to break up the suitcases into overlapping intervals such that any possible pair chosen by $A$ is in at least one such interval. If there are $n$ intervals, then $B$ can certainly always win with at least probability $\frac{1}{n}$.
(However, these strategies are not always optimal, depending on $T$.)
Full solution
Case 1: $100 \ge T\ge 76$
In this case, $B$ can cover all possibilities with only $2$ intervals: $[0,75]$ and $[25,100]$. For example, if $A$ tries to avoid the first interval, $76$ may be chosen, but in this case the smallest second choice is $26$ which is also in the second interval. This means $A$ has an upper bound of $\frac{1}{2}$, and with the same lower bound provided above, the optimal chance of winning for $A$ is indeed $\frac{1}{2}$.
Case 2: $75 \ge T \ge 67$
$A$ has an improved strategy available. Since $B$ can no longer cover all possibilies with $2$ intervals, $A$ should introduce a possibility that is covered by neither $[0,74]$ nor $[26,100]$. In fact, the only choice is $(25,75)$. So, the proposed strategy for $A$ is to choose with equal probability between $(0,25)$, $(25,75)$, and $(75,100)$.
Suppose $B$ knows this strategy is being used. Then $B$ gains no information from guessing any suitcase except $0$, $25$, $75$, or $100$. Guessing $0$ or $100$ only allows $B$ to win if those guesses are correct, so only with probability $\frac{1}{3}$. Consider then if $25$ is guessed. If the guess is wrong, then $B$ is out of luck because $100$ is unreachable. If the guess is right, then both $(0,25)$ and $(25,75)$ are still possibilities, and $B$ must choose between them. Thus, there is still only a $\frac{1}{3}$ chance for $B$ to succeed. The situation is the same for guessing $75$.
Thus, this strategy ensures that $A$ can win with probability at least $\frac{2}{3}$.
$B$ can still cover all possibilities with only $3$ intervals. In the case of $67$, these will be $[0,66]$, $[17,83]$, and $[34,100]$. This means $A$ also has an upper bound of $\frac{2}{3}$, so answer is $\frac{2}{3}$.
Case 3: $66 \ge T \ge 51$
This case is trickier.
In fact, $B$ now has a more complicated strategy still ensures victory with probability at least $\frac{1}{3}$. Consider, $B$ must still cover the possibility that $A$ may choose $0$ or $100$, so $[0,50]$ and $[50,100]$ are still good guesses. So, when will these possibilities fail?
The third possibility, of course, is that one of $x$ and $y$ is in $[0,49]$ and the other is in $[51,100]$. $B$ can guarantee finding both $x$ and $y$ in such a case by exploiting the ability to open cases one at a time. $B$ should first guess $50$, then start going left until finding one of $x$ and $y$. Then $B$ should start going right from $50$ until finding the other.
$A$'s optimal strategy, then, remains the same as before, so the optimal chance of $A$ winning is still $\frac{2}{3}$.
Note that $B$'s strategy here also works for Case 2, so these cases can be combined if desired.
A: Well if I were A I would put the rum in suitcase 0 and the opener in 100. Then there is no contiguous selection of suitcases that would obtain both. I would always win. 
