# Simplifying a seemingly complex probability problem.

We randomly put $100$ numbered balls in $100$ baskets, and if I am to ask what's the probability of the third ball being in a basket between the first and second ball, I know the answer is exactly one third, intuitively. A similar case would be asking what's the probability of one of these balls being bigger than the other. The difference is that for the latter example (with the probability being $1/2$), I can formally prove it using the law of total probability. In the first example I was not able to devise or think of a simple proof, so I was wondering if you could perhaps show me how it can be done using formality rather than common sense.

Thanks a million!

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Assuming only one ball can go in each basket, you can ignore all the other balls and look at the order of balls $1,2,3$. There are $3!=6$ orders of these balls, and ball $3$ is in the middle in $2$ of them.

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I saw that right after typing, removed comment - but thanks for following up! –  gnometorule Feb 20 '13 at 4:54
What's the justification for ignoring the other balls? I was able to get to this situation but didn't know how to justify it –  NBP Feb 20 '13 at 5:12
@NBP The justification is, basically, that in the question you're asking, all of the other 97 balls are meaningless. The question you're asking is "what is the probability that ball 3 is between ball 1 and ball 2?" It doesn't matter if ball 1 is in the first basket, and ball 2 is in the 93rd basket. Imagine you line up all these baskets on a cliff. Shove all the baskets containing balls numbered 4 to 100 off the cliff. You're left with balls 1, 2 and 3. There are only 6 ways these can be arranged, regardless of their original position in the line of 100 baskets. –  Arkamis Feb 20 '13 at 18:42