The probability of getting one variation of consecutive outcomes in Bernoulli trials before another Consider tossing a fair coin and writing down the outcomes in a sequence of symbols. "H" stands for head and "T" for tail. Let A , B  and C  be the words HTHH , HHTH  and THHH  respectively. What is the probability of encountering each A, B and C as a subword in your sequence of outcomes before both others?
Edit: My question is relateted to this question. While its answers intuitively explain why some patterns hit sooner than others, none of them include a way of calculating those probabilities. Which is what I'm asking.
Edit 2: I have already recieved an answer. Having read the wikipedia articles, this question still stumps me. I'm starting a bounty on it.
 A: To avoid having to consider the initial steps you can start with the position after four tosses at which point each of the possible states has a probability of $1/16$. 
You can turn your question into a Markov process with three absorbing states out of sixteen, and the non-absorbing states having a 0.5 chance of going to one of two other states, making the question arithmetic.  
Run the Markov process by taking the 133rd or higher power of the transition matrix, and you will find that the three absorbing states have probabilities of about 


*

*HHTH 0.307692308

*HTHH 0.326923077

*THHH 0.365384615

A: A beatiful formula exists for the case of two words.  It was discovered by John Conway and is explained at http://plus.maths.org/content/os/issue55/features/nishiyama/index.  I am not aware of any such formula for three or more words.  In that case, you may have to find the eigenvalues of the transition matrix as suggested by whuber.
A: I think you're going to have to resort to simulation to answer your question. The distribution of the number of tosses until your words occur are complex. For example, even the simple word "HH" has a probability of $1 \over 4$ of occurring in 2 tosses, $1 \over 8$ for 3 tosses, $1 \over 8$ for 4 tosses, and then the probability decreases as the number of tosses increases. [There might be some interest in this simple case for its own right, as the number of combinations that end in a first occurrence of "HH" follow the Fibonacci series as $n$ increases! ] 
Anyway, if we let $X$, $Y$, and $Z$ represent the number of tosses until your words A, B, and C are observed, then you are trying to find  
$P[X < Min(Y,Z)], P[Y < Min(X,Z),$ and $P[Z < Min(X,Y)]$ . 
I am suggesting simulation because no one has been able to supply you a complete answer so far. Note that if you can find two of these probabilities, you know the third.
