Probability: What is the best strategy for multiple choice game scenario? Here is the scenario:


*

*Each day you get to chose 1 box out of 3 total boxes.

*1 out of those 3 boxes has a rare item, the other 2 boxes have a common item.

*You get to choose 1 box per day (give the constraints above) for 7 days.

*Note that each day the position of the rare item is randomized.

*Total items in boxes: 21 (7 rare, 14 common).

*You get a total of 7 items (rarity determined by random selection)


What is the best strategy to obtain a rare item from this game?
(if possible, show the "math" for it)
I think it is best to pick the same box every day, but I'm not sure how to explain why I think that.
Thanks for your help.
 A: If you pick the same box every day, your chance of success all seven times is $\frac{1}{3^7}$. You have the same chance if you pick a difference box each day, as the results are independent.
To simplify this, take the base case: 2 days, 2 boxes. Your possible configurations are:
RC, RC
RC, CR
CR, RC
CR, CR

On any given day, the probability of you getting a rare from this configuration is 1/2 for any given box. Therefore, the probability of getting at least one rare in two days is 3/4 regardless of which box you choose.
A: I agree with everything @mason and @Jonathan are saying.
To address your idea about choosing the same box (or same question in a multiple choice).


*

*Suppose that you decide to choose the 1st box every time. Now I think we can agree that the chances of a RARE item is $\frac{1}{3}$.

*Now on day two you stick with your strategy and decide to pick box 1. Now according to your description of the problem, the RARE item is randomized each day, so the probability of there being a rare item in box one is still $\frac{1}{3}$. Now if your chances are worse by picking box two or three, then their probabilites $\underline{must\:be\: less\;than\: \frac{1}{3}}$. For simplicity suppose the chances of a RARE in boxes two and three is $\frac{1}{4}$ But then the probability of getting a rare item if I opened all three boxes would be less than 1 $(\frac{1}{3} + \frac{1}{4} + \frac{1}{4} = \frac{10}{12} \ne 1)$! This would violate some of the laws of probability (as well as the condition that each day there is a RARE item among the three boxes).

*Hence either the probability is 1/3 for each box each day, or some rule in the problem you described has been broken!
A: The main question has been ably dealt with by (at this time) Jonathan Rich and FAS.
OP, in the comments, brought up the question of strategy in a multiple choice test, if one is essentially out of time and has to guess. For concreteness let us assume there are $30$ questions, with $3$ choices on each question, a), b), and c), only $1$ of which is correct.
Imagine that the teacher made sure that a) was always the correct answer, and then used a good randomizing device to scramble the labels of each choice. Then we are back at the "rare item" question of the OP, and conclude that there is no strategy available to the student.
But possibly the teacher wanted all correct answers to appear exactly $10$ times. 
Then if we guess a) each time, we will get exactly $10$ right. 
If we randomize our guesses, using a fair die each time to make the decision, then on any question our probability of being right is $1/3$. It turns out that the mean number of correct answers will be $10$, precisely the same as if we guess a) each time. But the variance will be much greater. If the passing mark is $15$, then ticking a) each time leads to a sure fail. A randomized strategy, by contrast, gives a reasonable shot at passing, at the cost of a significant probability of getting a very low mark. 
People making up tests get nervous about having consecutive correct choices be the same. Perhaps one could exploit that nervousness and devise improved guessing strategies. One would need data.   
