Question about logic riddle (patch of lily pads that doubles every day) I wasn't sure if the Math SE is the right SE to post this question, but after reading this question at the Meta SE, I think this SE is the right place.
In a video on YouTube about logic riddles, one of the riddles is:
In a lake, there is a patch lily pads that doubles in size every day. It takes the patch 48 days to cover the entire lake. How long does it take for the patch to cover half of the lake?
The answer is 47 days, because if the lake is half full at day 47 and the patch doubles every day, then on day 48 the lake must be entire full. More formally the part of the lake covered with the patch $c$ is given by $$c(t) = 2\times c(t-1)$$
There are many variations of variations of this riddle on the Internet, but the essence is the same.
However I think that day 45 is more likely. Let me explain why.
Firstly the function of $c$ is really given by $$c(t) = \min\{2\times c(t-1), 1\}$$ since the lake can't be covered for more than $100\%$.
Say the initial part of the lake covered is $1/2^k = (1/2)^k$ with $k \in \mathbb{R}$, so $c(0) = (1/2)^k$. In this case the lake would be covered in $k$ days, since $$\frac{1}{2^k} \times 2^k = 1$$
But for any other $k \in \mathbb{R}^+ - \mathbb{N}$ (so for every postive real non intiger value of $k$), which is far more likely since $\mathbb{N} \subset \mathbb{R}^+$, you would need $k + 1$ days to cover the entire lake.
A concrete example: say $0.4\text{ }(=40\%)$ of the lake is covered at day the beginning of day $x - 2$, so $c(x-2) = 0.4$. This would mean $c(x-1) = 2\times c(x-2) = 0.8$ and $$c(x) = \min\{c(x-1),1\} = \min\{1.6,1\} = 1$$ thus the lake is entirely covered at day $x$.
So the lake is entirely covered at the beginning of day $x$, even since day $x-1$ begins with $0.8 = 80%$ of coverage. If we take $x = 48$, the lake is covered at day 48 like in the original riddle, but the lake isn't half covered at day $47 = x - 1$. So it must be covered half in the course of day $x - 2$.
Am I missing something or is the lake indeed almost always half covered two days instead of one day before it is fully covered?
Disclaimer: This is not meant as critic on the maker of the video. Her videos are awesome! 
Edit: Yes, I'm probably overthinking this, act as if I'm not. :)

Suppose we think of the growth as a continuous process. If $c_0$ is the initial coverage of the lake at the start of day 1, then the coverage of the lake at the end of day $t$ is $c(t)=c_02^t$. Now $c(t)=1$ when $c_02^t=1$ i.e at $$t_1=-\frac{\ln c_0}{\ln 2}$$
while $c(t)=1/2$ when $c_02^t=1/2$, i.e. at $$t_{1/2}=-\frac{\ln c_0}{\ln 2}-1$$
So, as was obvious without any calculation, $t_{1/2}=t_1-1$. If, for example $t_1$ turns out to be $47.879$, then $t_{1/2}=46.879$. That is if the lake is covered during the 48th day, then it is half covered during the 47th.
