If you have 1024 people, each flipping a coin 10 times, is one person guarenteed to get 10 tails in a row? I came across a form of this question in a video by the YouTube channel 'MindYourDecisions'. 
The video is linked below:
https://youtu.be/a0rCBKsLoBc
In the video, he shows that you do not need 1024 people but 7071 people to have a guarentee of one person getting 10 tails in a row.
Though he did not explain why 1024 people would not be enough. Because my first thought told me that the chance of getting ten tails in a row is $$\left(\frac12\right)^{10}$$ 
So the chance of one person out  of the 1024 people getting 10 tails in a row is $$1024 \times \left(\frac12\right)^{10} = 1$$
So where is the fallacy in this assumption? Why is it wrong to assume 1024 people even though it is $2^{10}$.
Because say on a smaller scale,we had 4 people, each flips a fair coin two times,then wouldn't we be guarenteed to get 2 tails in a row from one person, or is there a fallacy in this assumption as well?
 A: "Because say on a smaller scale, we had 4 people, each flips a fair coin two times,then wouldn't we be guaranteed to get 2 tails in a row from one person, or is there a fallacy in this assumption as well?"
$\bullet$  So there indeed is a fallacy in this assumption and also in the way you have interpreted the video. 
$\bullet$  Although the video begins with unquantifiable terms like "Statistical Certainty" but then he clearly says that he means: for what number of people the probability of getting $10$ heads in a row is $99.9$% ? Which is different from "Guaranteed" to get $10$ heads in a row.
$\bullet$ The probability of getting $10$ heads in a row is 
 $$P(10  \text{ Heads}) = \frac{1}{2}\times \frac{1}{2} ... 10 \text{ times} = \frac{1}{2^{10}} $$
$\bullet $ Probability of not getting any heads is the complement of the above event
$$P(\text{not getting 10 heads})= 1 - P(10  \text{ Heads})$$ 
$\bullet $
$$P(\text{n people not getting 10 heads }) = P(\text{no heads}) \times P(\text{no heads}) ... \text{n times} = (1 - \frac{1}{2^{10}})^{n}$$
$\bullet$ The complement of no one getting 10 heads is that someone gets $10$ heads Hence,
$$P(\text{ Someone gets 10 heads } ) = 1- (1 - \frac{1}{2^{10}})^{n} $$
So the question is for what value of $n$ is $P(\text{ Someone gets 10 heads } )$ equal to $0.999$ and the answer is $n = 7070.09$, if you put $n = 1024$ then you only get $0.6323$
A: There are problems in the assumption, because each people flipping coin is an independent event so two of them may get the same sequence.
Intuitively, will you get a head if you flip a coin twice? Maybe no, because they are independent events and can both be tail.
