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Suppose we have a kind of lottery as follow:

$1.$ You have a $\frac{1}{2}$ possibility of getting a prize on the first try.
$2.$ You have a $\frac{1}{4}$ possibility of getting a prize on the second try.
$\quad\vdots$
$n.$ The probability is $\frac{1}{2^{n}}$ on the $n$th try.
$\quad\vdots$

What is the probability of getting at least one prize?

I know that it is

$$p = 1- \prod _{n=1}^{\infty}\frac{2^n-1}{2^n},$$

But how to calculate it? I don't know. I want to find the answer.

Thanks.

Martin Sleziak
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happa
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1 Answers1

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We have- $$P=\prod_{n=1}^{\infty} \Biggr(\frac{2^n-1}{2^n}\Biggr)=\lim_{N\to \infty}\prod_{k=1}^{N} \Biggr(1-\frac{1}{2^k}\Biggr)=\lim_{N\to \infty} (1/2)^N\prod_{k=1}^{N} \frac{\Biggr(1-\frac{1}{2^k}\Biggr)}{(1-\frac{1}{2})^N}$$ $$=\lim_{N\to\infty}(1/2)^N \times\lim_{N\to\infty}\prod_k\Biggr(\sum_{r=0}^{k-1}(\frac{1}{2})^r \Biggr)$$

Now,let $$L=\lim_{N\to\infty}\prod_k\Biggr(\sum_{r=0}^{k-1}(\frac{1}{2})^r \Biggr)=\lim\prod_k (1+u_k)$$ where $u_k=\Biggr(\sum_{r=0}^{k-1}(\frac{1}{2})^r -1\Biggr)$. Here $\{u_k\} $ is convergent as $k\to \infty$. So $L$ is a finite limit i.e. $L\in \Bbb R$. Therefore, $$P=\lim_N (½)^N ×L=0.$$

This is my approach, but I'm confused that in the following P, each term is $>0$. So how the product can be zero! Please anyone spot any mistake ,please let me know. Thank you.

MAHI
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