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I am curious about how other people think (model) probability questions?

Here is a sample 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?

  1. What is your first thought when you see this problem?
  2. Do you see images in your head? If so, what kind of images show up in your head? Are there images of hats in a bag for example?
  3. How do you model the problem?
  4. Why do you choose your model?
  5. What's your answer?
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    Ok, this did convinced me enough to answer your question (so you won't leave empty-handed), but I still think it's not a good fit for math.SE. For your convenience: 1. Permutation. 2. Yes, bipartite graphs, no. 3. $|\Omega|=|S_n|$. 4. It is exactly the abstract formulation of the problem. 5. This is a well-known problem, with $\lim_{n\to\infty} P(A) = e^{-1}$ as stated by Somabha Mukherjee below.2012-05-19

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Imagine you have a refridgerator, a giraffe , a monkey, a dog, a cat, a mouse, a woman and an elephant together with enough food to feed them for a week. You also have access to a car, a ship and a tricycle.

Picture this: Put the giraffe on the elephant, the monkey on the giraffe, the dog on the monkey, the cat on the dog, the mouse on the cat.

The giraffe is afraid of the elephant, the elephant is afraid of the mouse, the mouse is afraid of the cat, the cat is afraid of the dog , the dog is afraid of the man and the man is afraid of the wife.

So if you are the giraffe, what do you do now ?

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    I really appreciate the tongue-in-cheekiness of this answer. I hope that's what was intended.2012-05-23
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This is just a simple problem on derangement, which says that in how many ways can $n$ letters be put in $m$ envelopes, such that no letter goes into its correct envelope. And by simple inclusion-exclusion or using a recursion relation involving derangements of $n,n-1$ and $n-2$ objects, the answer comes out to be: $n!(\frac{1}{1!} - \frac{1}{2!} + \frac {1}{3!} - ... +- \frac {1}{n!})$. So, the required probability is actually $e^{-1}$ as $n \rightarrow \infty$.