Expected Number of Coin Flips

1. Mar 2, 2008

mmehdi

A coin having probability p of landing heads is flipped 5 times and then flipped utill the pattern of the first 5 flips is observed again for the first time, including the possiblity of using some of the first flips. If we want to find the expected number of flips of coins.(The lead is that: The expected number of additional flips after the five flips shall give us the answer)

....I have kind of tried to do it a couple of ways, but a solution which holds for all combination of five flips seems is proving to be too confounding.

2. Mar 2, 2008

Hurkyl

Staff Emeritus
Eek, where did this problem come from? The only method I see for solving it is setting up a giant linear difference equation for some auxiliary probabilities, thus allowing me to use a computer algebra package to solve the system and manipulate it to get the answer you want.

3. Mar 2, 2008

mmehdi

After the first five flips, every single flip shall lead to a new sequence of five flips. So the probability that the sequence shall terminate is whether every flip is such that it regenerates the earlier sequence. So I wonder if it could be turned into a burnoulli trial, or the expectation of an indicator function variable.

I have an intuitive feeling that this should have a rather straight forward close form solution.

4. Mar 2, 2008

mmehdi

Expected Number of flips

In order for the sequence to terminate on the sixth flip. The first five will have to be all heads or all tails. Meaning all the flips should be equal to the first flip. For the sequence to terminate in 7 flips the sequence should be (12121). Third flip equal to first flip, fourth equal to 2nd, fifth equal to 3rd, sixth equal to 4th and seventh equal to 5th.

For the sequence to terminate on the 8th flip the sequence will have to be (12312), 9 th flip (12341), and tenth flip (12345).

5. Mar 2, 2008

mmehdi

The problem came in an old exam, three years ago. One of the way to do it, is E[X]=P[X>1]+P[X>2]+P[X>3].......

Where P[X>1]= P[1-P[X=1]. Where P[X=1] which means termination at one additional flip is p^6(1-p)^6. I am just having trouble in extending it to infinite series