Two tails in a row - what's the probability that the game started with a head?

Solution 1:

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.

Solution 2:

Suppose the game ends on the $n$th toss with a tail. If $n$ is even, then the first toss must be tails; if $n$ is odd, then the first toss must be heads. For example, if $n=3$, then the sequence must be $HTT$. If $n=4$, it must be $THTT$. If $n=5$, it must be $HTHTT$. Etc.

\begin{align} P(H_1 | \text{ends with tails}) &= \frac{P(H_1,\text{ends with tails})}{P(\text{ends with tails})}\\ &= \frac{\sum_{n=2}^\infty P(H_1, \text{ends on $n$th toss with tails})}{1/2}\\ &= 2(0 + 2^{-3} + 0 + 2^{-5} + 0 + 2^{-7} + \cdots)\\ &= \frac{1}{4} \cdot \frac{1}{1-(1/4)}\\ &= \frac{1}{3}. \end{align}

Solution 3:

Let $E_H, E_T$ be the events of ending heads or tails, respectively, and $B_H, B_T$ those of beginning such. Let $S,L$ be the events of ending on the second toss or ending later.   Then by the law of total probability:

$$\mathsf P(E_T) = \mathsf P(E_T, B_T, S)+\mathsf P(E_T, B_T, L)+\mathsf P(E_T,B_H)$$

We note by symmetry that the unconditional probability of ending on two heads is the same as the unconditional probability of ending on two tails, which must mean:

$$\mathsf P(E_T) = \tfrac 12$$

To begin with a tail and end with a tail on the second toss, requires tossing two tails, obviously, so

$$\mathsf P(E_T,B_T,S)=\tfrac 14$$

To begin with a tail and end with a tail later than the second toss, requires tossing a tail then ending with a tail after restarting with a head.

$$\mathsf P(E_T, B_T, L)= \tfrac 12\mathsf P(E_T, B_H)$$

Put it all together to find $$\mathsf P(B_H\mid E_T)=\frac{\mathsf P(E_T, B_H)}{\mathsf P(E_T)}$$