Thread: Logic Puzzle, can any one solve this?

Suppose someone tells you that the number of marbles in a jar is 100 with a margin of error of 5, and that 5 has a margin of error 2.

What is the true margin of error of the number of marbles in the jar expressed as a single number?

Hopefully this question should be hard enough to be a puzzle.

Seven, no?
The margin of error could be as much as 7 and as little as 3, so we should account for a margin of error at greatest of seven.
I'm not sure, maybe I don't have the best idea of what is meant by margin of error, but it makes sense to me.

I'm not sure if you can stack margin of errors like that.

Originally Posted by Sneaky

Suppose someone tells you that the number of marbles in a jar is 100 with a margin of error of 5, and that 5 has a margin of error 2.

What is the true margin of error of the number of marbles in the jar expressed as a single number?

Hopefully this question should be hard enough to be a puzzle.

Well, the margin of error on 100 could be 7 or 6 or 5 or 4 or 3

If all margins of error are equally likely, would it not make sense to take the average margin of error?

(7 + 6 + 5 + 4 + 3)/5 = 5

My guess is margin of error stays 5, but I don't like it.

If the largest margin of error could be 7, then shouldn't we consider that...just to be safe?

As eulcer observed?

I think I figured it out.
If the margin of error is 5 with a margin of error of 2, then, the number of marbles can be

Code:

93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 // error of 7
   94 95 96 97 98 99 100 101 102 103 104 105 106         // error of 6
      95 96 97 98 99 100 101 102 103 104 105                // error of 5
         96 97 98 99 100 101 102 103 104                        // error of 4
            97 98 99 100 101 102 103                               // error of 3

Does anyone see where to go from here, how do I express the margin of error as a single number? Seems like you can have a margin of error but some parts of it are more likely of appearing.

So there's 55 numbers.
93: 1/55
94: 2/55
95: 3/55
96: 4/55
97: 5/55
98: 5/55
99: 5/55
100: 5/55
101: 5/55
102: 5/55
103: 5/55
104: 4/55
105: 3/55
106: 2/55
107: 1/55

Originally Posted by Sneaky

I think I figured it out.
If the margin of error is 5 with a margin of error of 2, then, the number of marbles can be

Code:

93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 // error of 7
   94 95 96 97 98 99 100 101 102 103 104 105 106         // error of 6
      95 96 97 98 99 100 101 102 103 104 105                // error of 5
         96 97 98 99 100 101 102 103 104                        // error of 4
            97 98 99 100 101 102 103                               // error of 3

Does anyone see where to go from here, how do I express the margin of error as a single number? Seems like you can have a margin of error but some parts of it are more likely of appearing.

So there's 55 numbers.
93: 1/55
94: 2/55
95: 3/55
96: 4/55
97: 5/55
98: 5/55
99: 5/55
100: 5/55
101: 5/55
102: 5/55
103: 5/55
104: 4/55
105: 3/55
106: 2/55
107: 1/55

You should first decide what the term "margin of error" means, it is ambiguous. From what you post it seems you think is is the semi-range.

You seem to assume that you have uniform distributions for your first two margin of error values, but you will not have a uniform distribution for the final result.

CB