We can consider other patterns in binomial trials; for example, we may wait until we see the pattern SF. What is the average waiting time for this event to occur?

Again we write the sample space and the values of N, the number of trials to obtain SF for the first time. We give the probabilities here for tossing a fair coin, so p = ½ in the table below. The probabilities have not been simplified so that the pattern the exhibit can be seen.
 

N	Points	Probability
2	SF
3	FSF SSF
4	FFSF FSSF SSSF
5	FFFSF FFSSF FSSSF SSSSF
6	FFFFSF FFFSSF FFSSSF FSSSSF SSSSSF

There is a pattern in the sample points: the sample points for n +1 trials can be found by placing F as the first trial and following this by all the points for n trials and finally adding a point that has n S’s followed by a final F.

For example, the points for N = 7 trials are:

F | FFFFSF
F | FFFSSF
F | FFSSSF
F | FSSSSF
F | SSSSSF
------------
SSSSSSF
There are then n -1 points for the event N = n. Each of these has probability , so

First we check that we have allocated a total probability of 1 on the sample space by summing the probabilities:

It follows then that

and subtracting the second series from the first gives us

The right side of this equation is a geometric series with first term and ratio , so

giving S = 1 as we expected.

Now we find the expected waiting time for the event SF. This is

so

and subtracting the second series from the first we have

Now we recognize the quantity in parentheses on the right side of the equation above as the sum of the probabilities on the points in the sample space, or 1.

We have then that

so that

It may strike some that, with a fair coin, the expected waiting time for SS is 6 tosses while the expected waiting time for SF is only 4 tosses. In extended trials, different patterns of success and failure usually have quite different waiting times, even for a fair coin.

Perhaps a good way to convince oneself of this is to do a computer simulation or to toss a fair coin repeatedly.