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Let $\{X_i\}$ be $n$ iid dunif(0, u) (discrete uniform) random variables with u>n. How do I compute the probability that $\{X_{i+1}\}$ > $\{X_i\}$ for all i?

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The title and body contain different questions. The question in the title would correspond to the question in the body with $\gt$ replaced by $\ge$. Please clarify which of these is your actual question. – joriki Aug 30 '12 at 1:17
youre right. i meant strictly greater than – Wuschelbeutel Kartoffelhuhn Aug 30 '12 at 1:20
up vote 4 down vote accepted

Because of a comment by the OP, I will interpret the uniform as having possible values $1,2,3,\dots,u$, a total of $u$ values. It is easy to alter the expression below to deal with another interpretation.

Record the result of the $n$ experiments as a sequence $(x_1,x_2,\dots,x_n)$. Then all $u^n$ sequences are equally likely.

The number of ways to choose a strictly increasing sequence $(x_1,x_2,\dots,x_n)$ is $\binom{u}{n}$. for there are $\binom{u}{n}$ ways to select the set of values, and for every such set, only one sequence made up of elements of the set qualifies as increasing.

For the probability, divide by $u^n$. The result is $$\frac{\binom{u}{n}}{u^n}.$$

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In the books I've read, the second argument of dunif() is always part of the sample space. Thanks a lot for the answer. – Wuschelbeutel Kartoffelhuhn Aug 30 '12 at 1:18
The answer I got from simulations corresponds to the one I get from your formula. Thanks again. – Wuschelbeutel Kartoffelhuhn Aug 30 '12 at 1:25

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