TITLE: Coin sequence paradox from Martin Gardner's book
QUESTION [7 upvotes]: "An event less frequent in the long run is likely to happen before a more frequent event!"
How can I show that THTH is more likely to turn up before HTHH with a probability of
9/14, even though waiting time of THTH is 20 and HTHH, 18!
I would be very thankful if you could show me the way of
calculating the probability of turning up earlier, 
and the waiting time. Thank you!

REPLY [2 votes]: While others have already explained the math of how to derive the waiting times and probability that one sequence will arrive first, your fundamental difficulty in understanding is recognizing that distributions can have different shapes, so trends in the mean may not be present when comparing individual cases against each other in the distribution.
Take the following simple problem as an analogy.  
Let System A have a fixed waiting time of 10 (e.g., it's a machine that makes a new widget every 10 minutes).  Meanwhile, let system B have a short waiting time of 8 three-quarters of the time, and a waiting time of 20 the other 25% of the time (75% of the time it takes 8 minutes to make a widget; 25% of the time it has to do 12 minutes of recalibration prior to spending 8 minutes making the next widget).  If you turn on both systems (not knowing if system B needs to recalibrate soon) and ask which system will make a widget first, you'll find system B finishes first 75% of the time, so B appears to be the faster system.  But if you look at their average time to finish, system A has an average wait time of 10, while system B has an average wait time of 
$.75 \times 8 + .25\times20=11$, which indicates system A is the faster system -- even though 75% of the time it finishes second.  
Hence, it isn't paradoxical that something that is faster by average wait time, actually finishes slower more than half the time.