Problem: What is the expected number of sequences of $3$ heads in $50$ tosses of a coin?

I am a bit confused about this problem in my course. So far we defined expectation value as:

$$E[X]= \sum_{x=0 }^n x \cdot P[X=x] $$

Which tells us the expected value, or mean of a certain experiment. However, now they go and just pick a certain value without explaining how this is done. Normally instead of 3, it would say "$X$". How is this done?

I know that there are $2^{50}$ possible sequences as every entry can be either heads or tails.

  • $\begingroup$ Some clarification might be needed. Suppose your string was $HHHH$. Is that two sequences of three heads? $\endgroup$ – lulu Dec 11 '18 at 15:54
  • $\begingroup$ I think they count that as 1. So far we have only been counting $3$ and then we stop. $\endgroup$ – Wesley Strik Dec 11 '18 at 15:56
  • $\begingroup$ Well, you need to clarify that point. Either way, though, you can do it with indicator variables. For $i\in \{1,\cdots, 48\}$ let $X_i$ be the indicator variable which is $1$ if a "good" block (however defined) starts at the $i^{th}$ slot. Easy to compute the expected value of $X_i$ and then use Linearity. $\endgroup$ – lulu Dec 11 '18 at 15:57
  • $\begingroup$ This sounds way too advanced for a first lecture into introductory probability theory. We only know linearity and coin tosses. $\endgroup$ – Wesley Strik Dec 11 '18 at 16:00
  • 1
    $\begingroup$ I don't see an easier way of doing the problem. And indicator variables are pretty easy to use. I suggest looking that up. $\endgroup$ – lulu Dec 11 '18 at 16:05

This is a straight forward exercise in the use of indicator variables.

Note that a string of the form $HHH$ can start anywhere from the first slot to the $48^{th}$. For $i\in \{1,\cdots, 48\}$ let $X_i$ be the indicator variable for the $i^{th}$ slot. Thus, $X_i=1$ if a good sequence begins on the $i^{th}$ slot and $X_i=0$ otherwise.

By Linearity $$E=E\big [\sum_{i=1}^{48}X_i\big ] =\sum_{i=1}^{48}E[X_i]$$

Now, the $X_i$ don't all have the same expectation, $X_1$ and $X_{48}$ are different than all the others (which all equal each other).

To handle $X_1$ note that the only good sequence that starts in the first slot is $HHHT$. Thus the probability of starting with a good sequence is $\frac 1{16}$, so $E[X_1]=\frac 1{16}$. A similar computation shows that $E[X_{48}]=\frac 1{16}$ as well.

For $1<i<48$ we get a good sequence starting in slot $i$ by $THHHT$, where the first $H$ is in the $i^{th}$ slot. Thus the probability that a good string starts in slot $i$ is $\frac 1{32}$

Combining all this we see that $$\boxed {E=2\times \frac 1{16}+46\times \frac 1{32}=\frac {25}{16}=1.5625}$$

  • $\begingroup$ Why is $TTT$ a good string? As I read the problem we need $THHHT$ if we are in the middle. We can delete one $T$ at the ends. $\endgroup$ – Ross Millikan Dec 11 '18 at 16:23
  • $\begingroup$ @RossMillikan My computation takes all that into account, or at least it intends to. My point was that a $3-$ sequence, either $HHH$ or $TTT$ can start anywhere from slot $1$ to slot $48$. $\endgroup$ – lulu Dec 11 '18 at 17:30
  • $\begingroup$ @RossMillikan Oh, sorry. for some reason I read the problem as looking for three of a kind in a row. Now I see that only Heads is accepted. I'll edit accordingly. $\endgroup$ – lulu Dec 11 '18 at 17:32
  • $\begingroup$ @RossMillikan reviewing the edit history, the problem originally did accept $HHH$ or $TTT$, the OP changed it at some point after I had posted my solution. Anyway, I have edited my solution to correspond to the question in its present form. $\endgroup$ – lulu Dec 11 '18 at 17:35

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