Homework probability problem

I am looking for some guidance with this question involving discrete random varaibles.


Let X denote the random variable counting the number of tosses of a fair coin until (and including) the event of two consecutive heads. Let pn = P(X = n). Find p2. For n ? 3, find a recursive formula for pn in terms of pn-1 and pn-2 using the Low of Total Probability based on the fact that in order to obtain two consecutive heads for the first time in n tosses, either heads followed by tails occurred in the first two tosses and consecutive heads occurred for the first time in the subsequent n - 2 tosses, or tails occurred in the first toss and consecutive heads occurred in the subsequent n-1 tosses. Use this recursive formula to find the expected value of X.

Thank you

JellyFish said:

Let X denote the random variable counting the number of tosses of a fair coin until (and including) the event of two consecutive heads. Let pn = P(X = n). Find p2. For n ? 3, find a recursive formula for pn in terms of pn-1 and pn-2 using the Low of Total Probability based on the fact that in order to obtain two consecutive heads for the first time in n tosses, either heads followed by tails occurred in the first two tosses and consecutive heads occurred for the first time in the subsequent n - 2 tosses, or tails occurred in the first toss and consecutive heads occurred in the subsequent n-1 tosses. Use this recursive formula to find the expected value of X.

Click to expand...

I will tell you that I cannot use the above reasoning to solve.
However because this is a rather well known problem, I can show you one solution.

Let \(\displaystyle B_n\) be the number of strings of H’s & T’s, of lengh n, that end in HH but do not have that pair anywhere else in the string.
We can see that \(\displaystyle B_1=0~\&~B_2=1\).
If \(\displaystyle n\ge 3\) the \(\displaystyle B_n=B_{n-1}+B_{n-2}\).

The probability that any one of the strings counted in \(\displaystyle B_n\) is \(\displaystyle \frac{1}{2^n}\).

Thus \(\displaystyle P(X=n)=\frac{B_n}{2^n}\).

Homework probability problem

Are you looking for the probability that you'll do your homework?

pka said:

JellyFish said:

Let X denote the random variable counting the number of tosses of a fair coin until (and including) the event of two consecutive heads. Let pn = P(X = n). Find p2. For n ? 3, find a recursive formula for pn in terms of pn-1 and pn-2 using the Low of Total Probability based on the fact that in order to obtain two consecutive heads for the first time in n tosses, either heads followed by tails occurred in the first two tosses and consecutive heads occurred for the first time in the subsequent n - 2 tosses, or tails occurred in the first toss and consecutive heads occurred in the subsequent n-1 tosses. Use this recursive formula to find the expected value of X.

Click to expand...

I will tell you that I cannot use the above reasoning to solve.
However because this is a rather well known problem, I can show you one solution.

Let \(\displaystyle B_n\) be the number of strings of H’s & T’s, of lengh n, that end in HH but do not have that pair anywhere else in the string.
We can see that \(\displaystyle B_1=0~\&~B_2=1\).
If \(\displaystyle n\ge 3\) the \(\displaystyle B_n=B_{n-1}+B_{n-2}\).

The probability that any one of the strings counted in \(\displaystyle B_n\) is \(\displaystyle \frac{1}{2^n}\).

Thus \(\displaystyle P(X=n)=\frac{B_n}{2^n}\).

Click to expand...

pka, why is it "the probability that any one of the strings counted in \(\displaystyle B_n\) is \(\displaystyle \frac{1}{2^n}\)."?
i've tried hard to understand it, but i failed.
if you can explain how you came out with that, i'll really appreciate it.
thank you.

The collection of bit-strings (0’s & 1’) having length n contains \(\displaystyle 2^n\) strings.
The probability of getting any particular member from \(\displaystyle B_n\) is \(\displaystyle \frac{1}{2^n}\).

Example: if \(\displaystyle n=7\) is a the probability of getting \(\displaystyle 1001011\) is \(\displaystyle \frac{1}{2^7}\).