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I was weak at probability / statistic related stuff and was trying to learn the basics online.

Somehow I came across two similar problems, which I thought both can be solved with the same concept (and should have same answer) but turns out it is not.

The first question is: Expected Value of Flips Until HT Consecutively

As @Aditya's answer shown, the answer to this question is 4.

The second question is as followed:

When flipping a fair coin repeatedly, what is the expected value of the number of trials needed to get two tails in a row?

And the solution I know is

Let $E_n$ be the expected # of flips to have $n$ tails in a row.

Then $$E_n = \frac{1}{2}(E_{n-1} + 1) + \frac{1}{2}(E_{n-1} + 1 + E_n)$$

which implies $$E_n = 2E_{n-1} + 2$$

By the fact that $$E_1 = 2$$ We can conclude that the answer is $$E_2 = 2*2+2 = 6$$

I think I am confused but what exactly is the difference between this two questions?

In other words, why can't I use the same method for the first question for finding $E_{HT}$ to find $E_{TT}$ (vice versa)? Similarly, does that mean that $E_{HT}, E_{TT}, E_{HH}, E_{TH}$ are having different values? (Intuitively I thought they are the same...)

Community
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shole
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  • The expected number of throws needed is dependent not only on the length of the sequence, but also on the structure. As soon as $H$ occurs, we only need $T$ to get $HT$. But , whenever $H$ occurs, we need $TT$ and not just $T$ to get $TT$. Therefore we need more trials in average. – Peter Feb 20 '17 at 13:21

1 Answers1

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The situation in which they are the same is slightly different from the situation you are considering. They would be the same if you did the following: flip two coins. If you got HT (resp. TT), stop and count the number of times you've done the experiment. Otherwise repeat. That's because for (ordered) pairs of flips, HT and TT are equally likely. Notice that in this scenario the total number of trials is on average $4$, so the total number of flips is on average $8$.

But you're not doing that. You're flipping coins one at a time and keeping track of the whole sequence so far. This changes matters. Specifically, if you are waiting for HT, once you have gotten an H, you are only waiting for the next T, because flipping another H just returns you back to waiting for a T. By contrast, if you are waiting for TT, once you have gotten a T, getting an H causes you to have to wait for an entire TT sequence again.

You might think about this in terms of a Markov chain with a transition diagram. In the HT case, we can take our states to be "just had a head", "just had a tails", and "done", in which case the transition matrix looks like:

$$P_{HT}=\begin{bmatrix} 1/2 & 0 & 1/2 \\ 1/2 & 1/2 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

and the expected number of flips is $1$ plus the expected time to hit state $3$ under the initial distribution $\begin{bmatrix} 1/2 & 1/2 & 0 \end{bmatrix}$.

In the TT case we instead have:

$$\begin{bmatrix} 1/2 & 1/2 & 0 \\ 1/2 & 0 & 1/2 \\ 0 & 0 & 1 \end{bmatrix}$$

and again the expected number of flips is $1$ plus the expected time to hit state $3$ under the initial distribution $\begin{bmatrix} 1/2 & 1/2 & 0 \end{bmatrix}$. Try drawing these transition diagrams to get a feel for what's going on.

Ian
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  • Thanks Ian, that's makes more sense to me now. Just one more question (opinion based): Sometimes these problems are so alike but the way to solve them is so different, is there a general way to determine when to use which kind of solution, or is it all about experience? – shole Feb 20 '17 at 13:35
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    @shole There are relationships between solution techniques (in particular, these two problems can be solved exactly the same way other than changing the relevant matrix), but a lot of it is just experience. – Ian Feb 20 '17 at 13:41