A fair coin is flipped $3$ times. Consider a random variable $X$ which is the number of runs. Number of runs is the number of changes of letter $H$ and $T$. For example, $HHH$ has one run, $TTH$ has two runs and $THT$ has three runs. Find the probability distribution of the random variable $X$.

My work: I don't understand the phrasing of this question. In examples in my textbook and online $X$ is defined as the number of heads or tails. But I can't follow where the example in this question is going. I would think that $TTH$ and $THT$ would both have 2 runs since $HHH$ only has one. I don't know what zero runs would be either. Can anyone give me guidance on what exactly this question means? I'm pretty sure I can solve it once I understand what the number of runs means.

The outcomes would be:

$HHH$ $X=1$




$TTH$ $X=2$


$THT$ $X=3$


I don't know what number of $X$ would correspond with each.

  • $\begingroup$ Why should you think that there must be an outcome described by the event zero runs? In my interpretation, $\text{Onerun}=\{HHH,TTT\}, \text{Tworuns}=\{HHT,HTT,TTH,THH\}, \text{Threeruns}=\{HTH,THT\}$. Something like $TTHHHHTTHTHHHT$ would have $7$ runs. $\underbrace{TT}_1 \underbrace{HHHH}_2 \underbrace{TT}_3 \underbrace{H}_4 \underbrace{T}_5 \underbrace{HHH}_6 \underbrace{T}_7$ $\endgroup$ – JMoravitz Feb 28 '16 at 1:13
  • $\begingroup$ Thank you that makes sense to me! $\endgroup$ – Koalafications Feb 28 '16 at 1:20

I think you are just missing what the book means by a "run". E.g.


has 4 runs: a run of heads, followed by a run of tails, followed by a run of heads, followed by a run of tails. All the H's or T's that are next to each other get lumped together to form a single "run", regardless of how many there are. Thus HHH and TTT have 1 run, HTT, HHT, THH, TTH have 2 runs, and HTH and THT have 3 runs.

Thus $P(X=1) = 2/8, P(X=2) = 4/8, P(X=3) = 2/8$ is the distribution of $X$.

  • $\begingroup$ Yep that was it! Haha I was looking at it like some sort of pattern but it was really that simple. Guess I like to over complicate things. Thank you though! $\endgroup$ – Koalafications Feb 28 '16 at 1:20

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