A coin is flipped three times. What is the probability that the first two flips were heads?

My rationale would be, the sample space is $2^{3}$ and can be described as:

$$\{hhh , hht , hth , htt , thh , tht , tth , ttt \}$$

Then, we get

$$ \{hhh, hht \} $$ which is $\frac{2}{8}$.

My issue with this method is that it is not mathematical. What if I was asked 300 coin flips, and I cannot count my sample space. How would I rationalize through this problem?


  • 2
    $\begingroup$ It is very mathematical. Getting an answer by checking all the cases is a valid form of proof, and it's used in actual proofs from time to time, like in the proof of the four colour theorem, that every Rubik's cube is solvable in 20 moves or fewer, or that no sudoku with 16 or fewer numbers written in is uniquely solvable. So while it can get exhausting, don't think it isn't mathematical. $\endgroup$ – Arthur Sep 27 '17 at 19:51
  • $\begingroup$ That being said, wanting a more elegant solution is quite understandable. $\endgroup$ – Arthur Sep 27 '17 at 19:52
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    $\begingroup$ Nothing wrong with what you wrote, but it isn't necessary. The question only concerns the first two tosses. The two tosses are independent and each is $H$ with probability $\frac 12$ so the answer is $\frac 12\times \frac 12=\frac 14$. This would be true even if you went on to toss the coin another trillion times. $\endgroup$ – lulu Sep 27 '17 at 19:53
  • $\begingroup$ You may want to give an explicit problem you're having trouble with. The natural generalization of the one you've mentioned is sort of a poor test case. The total number of coin flips is irrelevant to the probability the first two are heads. $\endgroup$ – spaceisdarkgreen Sep 27 '17 at 19:54
  • $\begingroup$ Essentially, my question is: how do I show my derivation using numbers and some sort of formula? $\endgroup$ – 1011011010010100011 Sep 27 '17 at 20:00

By flipping the coin $n$ times, then the probability of having the first $k$ flips as heads is the same as doing for $k$ flips.The probability is $\frac{1}{2}$ for each flip.

$\therefore$ Required probability $= (\frac{1}{2})^k = \frac{1}{2^k}$


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