Probability
⚠️ 
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Imagine you have a jar of 500 coins. 1 out of 500 is a coin with two heads and all the others have a tail and a head. You take a random coin from the jar and flip it 8 times. You observe heads 8 consecutive time. Are the chances that you took the coin with two heads higher than having drawn a regular coin with a head and a tail?
The main tool is Bayes Theorem.
Define A the event of tossing the chosen coin and having heads 8 times, B_1
and B_2
the events of choosing the special and fair coins respectivly. We compute the odd of choosing the special coin over the fair one given the event A.
P(B_1A) : P(B_2 A)
If this odd is greater than 1, then the answer is yes. Otherwise, no.
By Bayes theorem (some manipulations),
P(B_1A) : P(B_2 A) = (P(AB_1) P(B_1)) : (P(AB_2) P (B_2))
= ( P(AB_1)/ P(A B_2) ) * (P(B_1) / P(B_2) (*)
The second ratio is the odd of choosing the special coin over the fair one. It equals 1/499
.
The first ratio is 1/(1/2)^8 = 256
.
So the odd of choosing the special coin over the fair one given the event A is 256/499
which <1. Hence there is a lower chance that we took the special coin than the fair one.
Extra comments:
 From the solution, if there were 9 consecutive heads, then the odd would be 512/499 and hence the answer would be
yes
. 
The formula (*), in general, has the form
postodd = likelihood ratio of the event A * preodd

Another solution is to compute directly
P(B_1A)
by using Bayes and total probability theorems,P(B_1A) = P(AB_1) P(B_1) / P(A) = P(AB_1) P(B_1) / (P(AB_1) P(B_1) + P(AB_2) P(B_2)) = 1 * (1/500) / (1/500 + (1/2)^8 * (499/500))
and see that it is
<1
 The problem contains several implicit assumptions. For example:
 A with one head and tail is called fair under the assumption that both sides have equal chance to land in each toss.
 We assume that each toss results in a head or a tail face but no other scenarios (like standing)