There\'s more to this one!
The short answer is yes, if each "trial" is the start of 4 flips of a heavily biased coin for example. In that case it would actually take 783 flips since your first trial would be flips 1,2,3,4 and your 780th trial would be 780,781,782,783.
It isn't 783 flips, it's 780 TRIALS, but only when you define a single TRIAL as an entire SEQUENCE of flips that begins immediately after a failure, and either has 4 consecutive sucesses, or it ends in failure. Not every flip starts a new trial. So if S=success and F=failure, and you got SSFF, you would count that as 2 trials, one for SSF, and one for F. These trials are independent, but the individual flips are not independent with respect to starting a run. Then there are (37/7)^4 = 780.57518 trials, but there will be many more flips because there is more than 1 flip per trial on average. The average number of flips per trial is 1.23175 = 1*(30/37) + 2*(7/37)*(30/37) + 3*(7/37)^2*(30/37) + 4*(7/37)^3*(30/37) + 4*(7/37)^4. This is the EV of the number of flips per trial. Note that 3 in a row and 4 in a row both have 4 flips. The average number of flips in 780.57518 trials is 780.57518 * 1.23175 = <font color="red">961.47 flips</font>. This is called the mean recurrence time.
A formula for mean recurrence time u is:
u = (1 - p^r)/(qp^r)
Where p is the probability of success (7/37)
q is the probability of failure (30/37)
r is the number of times in a row (4)
u = [ 1 - (7/37)^4 ]/ [ (30/37)*(7/37)^4 ] = <font color="red">961.47 flips</font>
An interesting thing about mean recurrence time for success runs like this is that the answer changes for different patterns of success runs of the same length. For example, suppose we wanted to know how often SFFS occurred. Even though this sequence has the same length of 4, and the probability of it occuring on any given try is still (7/37)^4 as before, it turns out that the mean recurrence time is not the same. Another example is that a pattern of all heads or all tails occurs less frequently than other patterns of the same length consisting of both heads and tails. This counterintuitive fact can be understood by noting that these other sequences cannot be broken into trials as we did here. When one sequence fails, another can be starting up in the middle. The trials overlap, so there are effectively more trials for the same number of flips, hence it takes fewer flips for it to occur.
Was this a homework problem?
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