# 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_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.

• 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)