Probability

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_1|A) : 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_1|A) : P(B_2 |A) = (P(A|B_1) P(B_1)) : (P(A|B_2) P (B_2))
• = ( P(A|B_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

  post-odd = likelihood ratio of the event A * pre-odd

• Another solution is to compute directly P(B_1|A) by using Bayes and total probability theorems,

  P(B_1|A) = P(A|B_1) P(B_1) / P(A) = P(A|B_1) P(B_1) / (P(A|B_1) P(B_1) + P(A|B_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)