Will it be HEADS ?

[  maths  probability  ] Last Updated: 02 Jun 2021

Question

A box contains two coins: a regular coin and one fake two-headed coin (\(P(H)=1\)). I choose a coin at random and toss it \(n\) times. If the first \(n\) coin tosses result in heads, what is the probability that the \((n+1)^{th}\) coin toss will also result in heads?

Answer

Let us chart down what we know:

\[A \Rightarrow \text{Event that first 'n' tosses are heads}\] \[C_1 \Rightarrow \text{Event that regular coin is chosen}\] \[\text{Probability of heads given regular coin, } P(H|C_1) = \frac{1}{2}\] \[C_2 \Rightarrow \text{Event that two-headed coin is chosen}\] \[\text{Probability of heads given two-headed coin, } P(H|C_2) = 1\]

To find,

\[B \Rightarrow \text{Event that } (n + 1)^{th} \text{ toss is heads}\]

Now, let’s calculate the probability of event \(A\) happening for each of the coins:

\[P(A|C_1) = \frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\text{...n times} = \frac{1}{2^n}\] \[P(A|C_2) = 1*1*1*\text{...n times} = 1\]

\(P(A) = P(A \cap C_1) + P(A \cap C_2)\) \(P(A) = P(A|C_1)P(C_1) + P(A|C_2)P(C_2)\) \(P(A) = \Big[\frac{1}{2^n}*\frac{1}{2}\Big] + \Big[1*\frac{1}{2} \Big] = \frac{1}{2}*\Big[\frac{1}{2^n} + 1\Big]\)

Now, before we calculate \(P(B)\), we first need to know the probability of the coin based on \(n\) observations of heads. Also, one needs to note that \((n+1)^{th}\) toss is independent of other tosses, so we have:

\[P(B) = P(C_1|A)P(H|C_1) + P(C_2|A)P(H|C_2)\]

where,

\[P(C_1|A) \rightarrow \text{Probability that regular coin is chosen given the event A}\] \[P(C_2|A) \rightarrow \text{Probability that two-headed coin is chosen given the event A}\]

Now, let’s calculate, \(P(C_1|A) \text{ & } P(C_2|A)\)

\[P(C_1|A) = \frac{P(A \cap C_1)}{P(A)}\] \[P(C_1|A) = \frac{\frac{1}{2^n}*\frac{1}{2}}{\frac{1}{2}*\Big[\frac{1}{2^n} + 1\Big]} = \frac{1}{2^n + 1}\] \[P(C_2|A) = \frac{P(A \cap C_2)}{P(A)}\] \[P(C_2|A) = \frac{\Big[1*\frac{1}{2} \Big]}{\frac{1}{2}*\Big[\frac{1}{2^n} + 1\Big]} = \frac{2^n}{2^n + 1}\]

\(P(C_2|A)\) shows us that, higher the \(n\), closer will be probability that two-headed coin is chosen to \(1\).

Now plugging the above values will give,

\[P(B) = \frac{1}{2^n + 1}*\frac{1}{2} + \frac{2^n}{2^n + 1}*1\] \[\boxed{P(B) = \frac{2^{n+1} + 1}{2^{n+1} + 2}}\]

So, the probability \(P(B)\) tends to \(1\) as \(n\) tends to \(\infty\).

References

1. Probability course chapter 1 problems (Problem 36)