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.

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    $\begingroup$ Note that statistical regularity implies that the ratio between the number of heads and the number of tails approaches 1. It does not imply that the difference between the two counts approaches 0. $\endgroup$ Feb 17 '18 at 19:27

$B$ 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.

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    $\begingroup$ Nice solution.thanks. $\endgroup$ Feb 17 '18 at 14:46
  • $\begingroup$ You are welcome. $\endgroup$
    – drhab
    Feb 17 '18 at 15:21
  • $\begingroup$ Awesome solution! $\endgroup$ Dec 11 '18 at 2:33

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.

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    $\begingroup$ Wonderful solution👌the question is also nice ..thanks a lot. $\endgroup$ Feb 17 '18 at 14:42
  • $\begingroup$ I think it's worth pointing out that all the irrelevant information contained in the list of what happened in the first $3$ trials and the first $2/3$ of the last trial might be the examiner's wish to find the students who still believe coins that have come up heads more often have to catch up with tails. That is of course implicit in your bullet points. $\endgroup$ Feb 18 '18 at 1:17

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.


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!


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