Homework probability problem

JellyFish

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Jan 12, 2009
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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.
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 Bn\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 B1=0 & B2=1\displaystyle B_1=0~\&~B_2=1.
If n3\displaystyle n\ge 3 the Bn=Bn1+Bn2\displaystyle B_n=B_{n-1}+B_{n-2}.

The probability that any one of the strings counted in Bn\displaystyle B_n is 12n\displaystyle \frac{1}{2^n}.

Thus P(X=n)=Bn2n\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, why is it "the probability that any one of the strings counted in Bn\displaystyle B_n is 12n\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 2n\displaystyle 2^n strings.
The probability of getting any particular member from Bn\displaystyle B_n is 12n\displaystyle \frac{1}{2^n}.

Example: if n=7\displaystyle n=7 is a the probability of getting 1001011\displaystyle 1001011 is 127\displaystyle \frac{1}{2^7}.
 
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