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In CS, there's a systematic way to check if your code is buggy or not as you write code. Is there a way to check the correctness of your answer to a probability question without using a textbook?

For example, my friend proposed a solution to a probability question that seemed right.

Question: Suppose that each of N men at a party throws his hat into the center of the room. The hats are first mixed up, and then each man randomly selects a hat. What is the probability that none of the men selects his own hat?

Proposed Solution: if we suppose there are 8 men, then the suggestionw as (7 / 8) * (6 / 7) * (5 /6 )... * (1/2) * 1

which for the n case simplifies to 1/n

The argument sounded reasonable, but the answer was wrong which I found out from the textbook. I had to think about it a bit before I realized he had undercounted. Is there a more systematic way to check answers for probability questions?

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    You are not going to like my answer but here goes : The problem lies at "...The argument sounded reasonable, but ...". IMO, it is difficult to verify Probability answers since a) You usually have only one way of approaching the answer (If there are multiple, problem solved) b) You can't conduct experiments or go backwards (Like, if I were to find the inverse of a matrix, I can multiply the answer with the original matrix and check that I get the identity matrix). I suggest, strengthen your reasoning (a proof based course like Real Analysis might help).2012-05-17
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    I do not know of a general procedure. In combinatorial problems, one can compute carefully some "small" cases, as a check against an error in the "general" analysis. It also helps to have tackled enough problems to have made most of the common errors!2012-05-17
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    It's not clear if you really meant $ (7 / 8) * (6 / 7) * (5 /6 )$ or if you meant to continue the product for 7 terms, but note that in your expression the 7s and 6es cancel from the numerator and denominator, leaving just $\frac58$. If you meant to continue the product down to $…(2/3)(1/2)$ then the cancellation goes even farther and leaves you with just $\frac18$.2012-05-17
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    yea sry i'll edit it. the proposed solution was suggesting 1/n as the final answer2012-05-18

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