In regards to:


I flip a fair coin independently 6 times. This gives me a sequence of heads and tails. For each consecutive 'HTH' in the sequence you win $5.

So I define a random variable X to be the amount of dollars that I win. For example here is what $X=5$ is: $THTHTT$, and $X=10$ is: $THTHTH$. What is the best way to calculate the expected value of X?

Accepted Answer:

We use the method of indicator random variables, and solve the problem for $n$ tosses. For $k=3$ to $n$, let random variable $X_k$ be $1$ if there is an HTH sequence ending at the $k$-th toss, and let $X_k=0$ otherwise. Then the amount $W$ that we win is given by $$W=5(X_3+X_4+\cdots +X_n).$$ By the linearity of expectation we have $$E(W)=5E(X_3)+5E(X_4)+\cdots+5E(X_n).$$ We have $E(X_i)=\Pr(X_i=1)=\frac{1}{8}$, so $E(W)=\dfrac{5(n-2)}{8}$.

Only minor modification is needed for a biased coin that has probability $p$ of landing heads.

André Nicolas (https://math.stackexchange.com/users/6312/andr%c3%a9-nicolas), Expected value of coin flip sequence, URL (version: 2015-12-21): https://math.stackexchange.com/q/1584597

This user has stated on his profile that he cannot answer questions about past posts:

I am wondering why $Pr(X_i=1) = 1/8$ ? Since there are $6$ coin flips, there are $2^6$ possible outcomes. Did he calculate every single way there can be a sequence of $HTH$ and divided by $2^6$ to get $1/8$ or is there an easier way? I believe there is an easier, more intuitive way, but I don't know how.

What if I decided I wanted my indicator random variable to be $1$ if $X_i, X_{i+1}, X_{i+2}$ creates the sequence $HTH$ respectively ($k = 1$ to $4$ inclusive)?

  • 2
    $\begingroup$ Since we specify the end toss, the event $X_i$ consists of just 3 coin tosses (anything before these flips is fine), of which there are 8 outcomes and exactly one of them is $HTH$. $\endgroup$ – Countingstuff Apr 22 '18 at 16:12

Basically, it is simply

$\frac18$ of the sequences will be $HTH?????$ so $P(X_3=1)=\frac18$
$\frac18$ of the sequences will be $?HTH????$ so $P(X_4=1)=\frac18$
$\frac18$ of the sequences will be $??HTH???$ so $P(X_5=1)=\frac18$
$\frac18$ of the sequences will be $???HTH??$ so $P(X_6=1)=\frac18$
$\frac18$ of the sequences will be $????HTH?$ so $P(X_7=1)=\frac18$
$\frac18$ of the sequences will be $?????HTH$ so $P(X_8=1)=\frac18$

Your suggested indexing simply lowers the indices by $2$.

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If you start with HTH, then the other three outcomes have 8 possibilities.$${HTH, HTT, HHT, HHH, TTH, TTT, THT, THH} So far we have $3$ cases of $10$ dollars and $5$ cases of $5$ dollars.

That gives you $55$ out of these $8$ trials.
Similarly see what happens if the first $HTH$ starts with the second flip of the coin.

Continue with other cases and get the total money out of all 64 cases and divide the result by 64.

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