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In the case of 3 draws (N=3) from Uniform[0,1], the expected second highest value would be 1/2. Although unlikely it could happen that all three numbers were less than 1/2. It is exactly this probability that I am interested in. Case of N=3

If I draw N numbers from an uniform distribution. I would like to know the probability (as a function of N) that all N number are below the expected second highest value. It is probably easier to understand if I split it in two. I use the notation that $Y^{(n)}$ is the n'th highest number drawn.

1) When drawing N numbers from an uniform distribution [0,1], the expected second highest value should be (see any textbook on auction theory or order statistics):

$E[Y^{(2)}]=\frac{N-1}{N+1}$

2) The probability that all N numbers are below some threshold $\bar{p}$, could also be expressed as the probability that the highest drawn number is below $\bar{p}$ (where $F()$ is the CDF for the uniform distribution [0,1]):

$Prob[\text{all N draws}≤\bar{p}] = Prob[Y^{(1)}≤\bar{p}] = Prob[F(y)^N≤\bar{p}] = Prob[y≤F^{-1}(\bar{p}^\frac{1}{N})] = Prob[y≤\bar{p}^\frac{1}{N}]$

However I get that the probability is increasing in N when $\bar{p} \in (0,1)$. Intuitively I would expected to find that the probability (of all N draws being less than $\bar{p}$) was decreasing in N. I must have made a mistake somewhere.

The last step is then to combine the two to find: $Prob[\text{all N draws}≤E[Y^{(2)}]]$

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    $\begingroup$ The probability that all $N$ independent draws are below $\bar{p}$ is $(\bar{p})^N$, not $(\bar{p})^{1/N}$. In your first inequality you use $Y^{(1)}$ while you should be using $Y^{(N)}$. $\endgroup$ – Xi'an Dec 16 '14 at 13:02
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    $\begingroup$ It would help other and future users of Cross Validated if you could update your question by correcting the mistakes in the second equation. Thanks. $\endgroup$ – Xi'an Dec 16 '14 at 15:07
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The probability that all $N$ independent uniform draws are below a value $0\le\bar{p}\le1$ is $\bar{p}^N$. The probability of the event is thus $$ \mathbb{E}[Y^{(2)}]^N = \left\{\frac{N-1}{N+1} \right\}^{N} $$ which is increasing with $N$ and with limit $$ \lim_{N\to\infty} \left\{1 - \frac{2}{N+1} \right\}^{N} = e^{-2}\,. $$ As $\mathbb{E}[Y^{(2)}]$ is increasing with $N$ but is less than $1$, it seems difficult to guess how this probability should move with $N$. Nonetheless, it is not that surprising that the probability that the largest uniform out of $N$ is smaller than the expectation of the second largest uniform out of $N$ goes up with $N$.

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