# 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.

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In the scenario that you are describing, it is impossible to come up with a strategy that is better than any other strategy. – Jonathan Rich Apr 3 '13 at 16:15
Could you explain why? It seems to me that if you pick the same box each day you would have a better opportunity. Like with the Money Hall problem there is a strategy... en.wikipedia.org/wiki/Monty_Hall_problem – mason81 Apr 3 '13 at 16:20
The Monty Hall problem is a very specific problem, which is why the strategy applies. The way your problem is defined, the strategy 1231231 has exactly the same chance for successes as the strategy 1111111. – Jonathan Rich Apr 3 '13 at 16:23
example: if you pick the same box every day (say box 1), then you get 1 in 3 odds each day, but 7 in 21 over all, whereas if you pick a different box each day, you still get 1 in 3 odds each day, but 7 in 63 overall. I'm sure I'm figuring that wrong, but perhaps you could help me understand why I'm wrong? Is it really just 1 in 3 odds period? – mason81 Apr 3 '13 at 16:28
Another thought that to me is similar is with multiple choice tests. I was once told (not that I believe everything I'm told) that if you must guess on a large number of questions on a multiple choice test (say your an out of time on an ACT section, where you are rewarded for correct answers, but not punished for incorrect answers) that it is best to select the same answer (say C) for each... Maybe that's where I'm getting the idea that picking the same box (in the OP scenario)? – mason81 Apr 3 '13 at 16:31

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.

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Thank you for your help on this. I appreciate it. – mason81 Apr 3 '13 at 16:39

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.

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Very interesting. I appreciate your answer and this information. – mason81 Apr 3 '13 at 17:32
You are welcome. I thought it might be fun to think about multiple choice test strategies, where a lot more is going on than in the game scenario you described. – André Nicolas Apr 3 '13 at 17:44
yeah, no doubt. the human element makes for quite an interesting factor. talking about this almost makes me miss those old scantron answer sheets from high school... almost... :) – mason81 Apr 3 '13 at 19:03
@mason81: I once met a teacher who told me she had been bubbling all day. It took a while for my overheated imagination to understand the prosaic reality. – André Nicolas Apr 3 '13 at 19:07

I agree with everything @mason and @Jonathan are saying.

1. 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 $\frac{1}{3}$.
2. 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).