7
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Also, what about in general, for some value p, which has the value 2 in the given formula?


MOTIVATION:

I was wondering the probability of never getting tails if one forever flipped a coin whose probability of landing tails decreased (in this case, geometrically) each flip.


The probability of tails on flip i is $\frac{1}{2^i}$, and the probability of heads on flip i is $1-\frac{1}{2^i}$. So, first flip the coin is 50-50, next it is 75-25, etc. And the probability of never landing tails is equal to the probability of always landing heads, which is the infinite product of the heads probabilities, yielding $\prod_{i=1}^\infty 1-\frac{1}{2^i}$. => ($\frac{1}{2} * \frac{3}{4} * \frac{7}{8} ...$)

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    $\begingroup$ I don't think there's any analytic solution. The answer is approx 0.28878809 $\endgroup$ – Shalop Mar 22 '15 at 5:09
  • $\begingroup$ Why don't you think there's any analytic solution? Also, might there be a very good approximation (besides the decimal approximation you provided, which I thank you for)? Like, a function that is very close to the partial products? $\endgroup$ – tscizzle Mar 22 '15 at 5:12
  • $\begingroup$ I just plugged it into Wolfram Alpha, and it gave the answer in terms of a Pochammer symbol. Just go there, type in "product 1-1/2^i from i=1 to infty" and you can get the answer to a huge number of decimal places. $\endgroup$ – Shalop Mar 22 '15 at 5:14
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    $\begingroup$ Just to add, this number is also the measure of a (generalized) Cantor set, where we take the interval $[0,1]$ and then remove the middle half, and then remove the middle fourths from the two remaining intervals, and then remove the middle eights from the four remaining intervals, and so on... $\endgroup$ – Shalop Mar 22 '15 at 5:22
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    $\begingroup$ No, not true. The infinite product $1-\frac{1}{n^2}$ also converges to a nonzero number. In fact, $\Pi_i^{\infty} (1-a_i)$ converges to a nonzero number iff $\sum_i^{\infty} a_i < \infty$. (assuming all $a_i \in [0,1]$). Try to prove it (hint: use logarithms). $\endgroup$ – Shalop Mar 22 '15 at 5:26
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You can use the $q$-Pochhammer symbol to represent the infinite product as

$$ \prod_{i=1}^{\infty}\left( 1-\frac{1}{2^i} \right) = \left( \frac{1}{2}, \frac{1}{2} \right)_{\infty} . $$

Note:

$$ \left( a, q \right)_{\infty}= \prod_{i=1}^{\infty}\left( 1-a q^k \right) $$

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$$\sum_{k\geq 1}\log\left(1-\frac{1}{2^k}\right) = -\sum_{k\geq 1}\sum_{n\geq 1}\frac{1}{n 2^{kn}}=-\sum_{m\geq 1}\frac{1}{2^m}\sum_{d\mid m}\frac{1}{d}=-\sum_{m\geq 1}\frac{\sigma_1(m)}{m 2^m}$$ hence: $$\prod_{i\geq 1}\left(1-\frac{1}{2^i}\right)=\exp\left(-\sum_{m\geq 1}\frac{\sigma_1(m)}{m 2^m}\right)\geq\exp\left(-\sum_{m\geq 1}\frac{m+1}{2^{m+1}}\right)=e^{-3/2}$$ but the bound: $$\sigma_1(m)=\sum_{d\mid m}d \leq \sum_{k=1}^{m}k = \frac{m(m+1)}{2}$$ is obviously very crude.

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