Wirt writes off three correct answers and pats us on the head,
>
> Don't ever let anyone tell that probability calculations are easy.

He then offers a different method, giving "weights" to the starting states.
I would like to examine the method, for 8 tosses.

1 - HHHHHH       1
2 - THHHHHH      1
3 - xTHHHHHH     1

At this point, if I understand Wirt correctly, the probability of 6 heads
in a row in 8 tosses should be 1*(1/64)+1*(1/64)+1*(1/64) = 3/64.  Let's look
at this another way to check our work.  What is the probability of throwing
6 heads in 6 throws?  1 in 64--we seem all to agree on that.  What about
case 2 above?  What is the probability of throwing THHHHHH in 7 throws?
1 in 128 (using the same method we used on case 1).  What about case 3?
Well, case three is two outcomes--HTHHHHHH and TTHHHHHH--in 8 throws.  That
comes to 2 in 256.  So the probability of reaching one of the above outcomes
should be 1/64 + 1/128 + 2/256 = 2/64, not 3/64.  What's wrong here?
What's wrong is that in his analysis, Wirt appears to have forgotten that
the probability of *starting* in a given state is not equal across the states.

Give it another shot, Wirt :-).  Or borrow Gavin's 26,000,000 coins and give
them a toss or two :-).

Ted

P.S.  There is one point Wirt makes with which I fully agree:

>                       I don't see how this question could have been asked on
> a test. It simply requires too much thought for an hour-long test.

Right!
--
Ted Ashton ([log in to unmask]), Info Sys, Southern Adventist University
          ==========================================================
Numbers are intellectual witnesses that belong only to mankind.
                                       -- Balzac, Honore de (1799 - 1850)