We have to find the expected number of tosses till we observe for the first time, at least two heads and at least two tails.
Just in case the question would have asked for the expected number of tosses till we observe for the first time, at least one head and at least one tail, i.e., We observe both sides of the coin for the first time., then it would have been simple.
We make the first toss, whichever side of the coin results, now we just have to toss for the opposite side. Without loss of generality, let’s just assume that the first toss resulted in a tail. Now we know the expected number of tosses till the first occurrence of the head is 2. How?
$X = \frac{1}{2}\left (1\right ) + \frac{1}{2}\left (X+1\right )$
Let X be the expected number of tosses till the first occurrence of head, So either with the probability of half, we obtain head in the very first toss and we are done, or it results in a tail and so we have to again make X number of tosses only that we have wasted one toss, so need to add it.
Okay, so that means, the expected number of tosses till we have observed both sides of a fair coin is, 3.
Now here is another question, How many expected number of coin tosses if we keep tossing repeatedly until we observed for the first time at least two heads? This is simple and the answer is 4. Because first, you toss till the first occurrence of head, which requires 2 tosses, and once you have a head, you just forget about it and toss for the next occurrence of a head which again requires 2 tosses, Hence the expected number of tosses for at least two heads is 4.
So now to answer the original question. Let E be the expected number of required tosses. So,
$E = \frac{1}{2}\left (2+3\right ) + \frac{1}{2}\left (2+4\right )$
i.e., E = 5.5
For the first two tosses, outcomes can be either HT, TH, HH, or TT. In the first two cases, when we get either HT or TH, Then we have seen both sides of the coin once, Now we just have to toss till we observe both sides of the coin once again. The answer to this is 3 as we have calculated above. Similarly, in the third and fourth cases, we have observed either two heads or two tails, and so then we just need to toss for at least two tails or at least two heads respectively, the answer to which is 4.
Hence the answer.
