An unbiased coin is tossed six times in a row. Which statement describing the last two coin tosses has the highest probability of being correct? 
An unbiased coin is tossed six times in a row and four different such trials are conducted. One trial implies six tosses of the coin. If H stands for head and T stands for tail, the following are the observations from the four trials:
  $$\text{(1) HTHTHT}\quad\text{(2) TTHHHT}\quad\text{(3) HTTHHT}\quad\text{(4) HHHT_ _}$$
  Which statement describing the last two coin tosses of the fourth trial has the highest probability of being correct?
(A) Two $\text T$ will occur.
  (B) One $\text H$ and one $\text T$ will occur.
  (C) Two $\text H$ will occur.
  (D) One $\text H$ will be followed by one $\text T$.

I think option A is correct and the reason is statistical regularity. Am I correct? If not then please help me how to do this problem. Any help would be appreciated. Thanks in advance.
 A: $B$ is correct here. It has probability $\frac12$ in contrast to the other options that all have probability $\frac14$.
A) TT has probability $\frac14$
B) HT or TH has probability $\frac14+\frac14$ (summation of two probabilities of mutually exclusive events)
C) HH has probability $\frac14$
D) HT has probability $\frac14$
Essential for this conclusion is the fact that the coin is unbiased.
A: A is not correct; B is. Statistical regularity – more often called independence – means that


*

*the results of the three previous trials do not affect the fourth trial's outcomes

*the four prior tosses of the coin in the fourth trial do not affect the last two tosses


Therefore, each of $\text{HH, HT, TH, TT}$ has a $\frac14$ chance of occurring. With regards to the options, only option B has a $\frac12$ chance; the others have a $\frac14$ chance.
A: B is correct.  It is the probability of two outcomes out of four equally likely outcomes and equals 1/2.  The others are the probability of one outcome and equal 1/4.
A: The results from the trials are a red herring. The actual chances of 3 H and 3T out of any 6 tosses is statistically correct, but in the real world...
Any toss will result in 50/50 H/T. So B is closest to what 'should' happen. One more H and one more T, but in random order.
As already stated, the other options have half the chance of option B. Even though a pattern seems to be emerging that might make A the correct answer. Wrong!
