If you have 1024 people, each flipping a coin 10 times, is one person guarenteed to get 10 tails in a row?

Question

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?

Answer 1

"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 $0.6323$

Answer 2

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.