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I have the following question:

Let A = {1, 2, 3, . . . , 100}. Let x, y, and z be elements in A that are chosen independently and uniformly at random. What is the probability that x = y = z?

Because if x is chosen first, only y and z have to be equal to it, would this mean that the answer is $\frac{1}{100 \cdot 100}$?

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Yes, you are exactly right. This should be a comment however I don't have enough reputation to put a comment.

If number is picked without replacement, how can they be equal?

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That depends on whether $x, y, z$ are necessarily distinct (chosen without replacement). If they are necessarily distinct, then the answer is (EDIT) $0$, obviously.

If they are not necessarily distinct (chosen with replacement), then your answer of $1/100^2 = 0.0001$ is correct.

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  • $\begingroup$ If they are necessarily distinct, the probability that they are equal is zero. $\endgroup$ – Jean-Claude Arbaut Apr 9 '15 at 23:22
  • $\begingroup$ Haha, duh, yes. Consider me suitably chastened. $\endgroup$ – Brian Tung Apr 9 '15 at 23:26

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