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.

Re: Logic Puzzle, can any one solve this?

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.

Re: Logic Puzzle, can any one solve this?

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

Re: Logic Puzzle, can any one solve this?

Quote:

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?

(Thinking)

Re: Logic Puzzle, can any one solve this?

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

Re: Logic Puzzle, can any one solve this?

Quote:

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