We're tossing a coin until two heads or two tails in a row occur. The game ended with a tail. What's the probability that it started with a head?
Let's say we denote the game as a sequence of heads and tails, e.g. $(T_1, H_2, T_3, H_5, H_6)$ is a game that started with a tail and ended with a head.
In this notation, I need to find $P(H_1 | T_{n-1}T_{n})$.
$$P(H_1 | T_{n-1}T_{n}) = \dfrac{P(H_1 \cap T_{n-1}T_{n})}{P(T_{n-1}T_{n})}$$
For a given $n$, there is exactly one sequence starting with a head and ending with two tails: $(H_1, T_2, H_3, …, H_{n-2}, T_{n-1}, T_n)$ – this is the event mentioned in the numerator. Now, there are two options for the event in the denominator: either the game is $(H_1, T_2, H_3, …, H_{n-2}, T_{n-1}, T_n)$, or $(T_1, H_2, T_3, …, H_{n-2}, T_{n-1}, T_n)$ – they differ in length by 1, though.
How do I calculate their probabilities? I was thinking of calculating discrete probabilities of sequences of length $n$, but since there are two options for the last event, I'm not sure how it'll work.
Best Answer
If I may, there is an easier approach to that problem.
We know the game ended with tails, so we have one of the following states:
$(T, T), (H, T, T), (T, H, T, T), (H, T, H, T, T), (T, H, T, H, T, T), \cdots $
You get the pattern.
Now notice that if you have a sequence of $n $ flips, the probability you got that sequence was $\frac{1}{2^n} $ right? Because the outcome of one flip was not influenced by the other.
Now we can start by infering this: the first sequence does not start with heads and has probability $\frac14$. The sequence afterwards starts with heads and has probability $\frac12\frac14$ i.e. half the probability of occurring when compared to the previous one. Doing this for all pairs of sequences, we see that each tail-starting sequence has double the probability of happening when compared to a heads-starting sequence and this can only happen if the probability of the game starting with tails is $66\% $ and with heads is $33\% $.
Another way of doing this is by explicitly summing all the probabilities of all sequences that start with heads. That sum is
$$\sum_{i = 1}^{\infty} \frac{1}{2^{2i + 1}} = \frac16$$
This is $P(\text{starts with heads|ends with tails}) $. Now all we have to do is divide by the probability it ended with double tails, since that is already given, to get $P(\text{starts with heads})$. The probability it ended with double tails is given by summing the probabilities of all these sequences (show it equals $\frac12$).
Now $\frac16/\frac12 = \frac13$ which is the result we obtained intuitively.