# Probability of uniform distribution

If I have an array of $n$ numbers in the range [0,1) and have the following events for $y$:
for $i=1$ to $n$ :
generate a new number $y$ in [0,1) uniformly and independent of previous
if $y\ge0.5$ then
toss a fair coin if the coin turns up "HEADS" then
$y=y-0.5$

I need to find out what is the probability that y is in the range [0,1/n)
My solution: $1/2+1/2\cdot1/2=3/4$ - the first 1/2 is the probability when y is not greater than 0.5. The second $1/2\cdot1/2$ is the probability when $y>0.5$ AND because of the fair coin - we have 1/2 change of being HEADS.

Is it the correct way ? Do I need to add n into the equation? Please correct me if I am wrong. Thanks!

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Yes, you need your answer to depend on $n$ because obviously the chance of $y<1/2$ is much better than the chance of $y<1/1000000$. What you have computed is the right answer for $n=2$. You'll probably need to treat the case $n=1$ as a special case. –  Henning Makholm Nov 4 '11 at 19:34
The relationship between $y$ and the array of $n$ numbers isn't clear to me. Does $y$ represent one of the $n$ numbers, or is it an additional number? If additional, what is its relationship to the $n$ numbers? If it's one of them, does that mean the same coin toss procedure is applied to all $n$ numbers, and are you interested in all or (at least) one of them lying in the given range? And what's the significance of $n$ turning up again as an interval bound? Please clarify. –  joriki Nov 4 '11 at 19:53
Your edit hasn't resolved any of my questions. You're now generating $n$ numbers $y$ instead of just one, but their relationship to the numbers in the array is still unclear. –  joriki Nov 4 '11 at 22:30

Assume $n$ is fixed beforehand. Then for each $y$ probability that it will be in range [$0,1/n$) is
$$\frac{3}{4}*(\frac{1/n}{1/2})=\frac{3}{4}*\frac{2}{n}=\frac{3}{2n}$$
because with probability $\frac{3}{4}$ $y$ is a random number from [$0,1/2$) and there with probability $\frac{1/n}{1/2}$ it belongs to the range [$0,1/n$).
1. Probability that all $y \in$ [$0,1/n$) is equal to $$(\frac{3}{2n})^n$$ 2. Probability that exists $y \in$ [$0,1/n$) is equal to:$$1-(1-\frac{3}{2n})^n$$ which is $1-$ probability that all $y \in$ [$1/n,1$)