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I'm tormented by this apparently simple question: If you toss a fair coin $7$ times in a row, what is the probability of getting an even number of heads? (please note: this is self-study and not a homework problem) I've seen a lot of questions here on "even number of heads" but they deal with biased coins, whereas I can't wrap my head around the seemingly simple problem stated above.

When I encountered this problem I thought like this: The possible scenarios where we can get an even number of heads are:

  • $2$ H, $5$ T
  • $4$ H, $3$ T
  • $6$ H, $1$ T

All we then need to do is add up the number of ways we can achieve these three outcomes, and divide by the total number of possibilities, which works out to be $\frac{\binom{7}{2} + \binom{7}{3} + \binom{7}{6}}{2^7} = \frac{63}{2^7} = 0.4921875$.

This is tantalizingly close to the answer given: $1/2$, but is not quite $1/2$.

I'd like two things here:

  1. A simple explanation for why the final probability is exactly $1/2$ (I'm still in the early phases of basic probability, so wouldn't like an explanation that dives into length formulas or probability distributions).
  2. Why the reasoning I followed doesn't quite work.
The Phenotype
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ankush981
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2 Answers2

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Zero is an even number, so you are missing a scenario. That answers your second question.

For the first question, note the number of heads must be either even or odd. Thus the probability will be $1/2$ if is there are exactly as many ways to get an even number of heads as there are to get an odd number of heads. But the number of ways to get an odd number of heads is the same as the number of ways to get an even number of tails. Since the coin is fair, the number of ways to get an even number of either outcome is the same, which completes the argument.

user1390
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Your reasoning works, but you missed the case $0$ heads, which is an even number.

So your final answer would become

$$\frac{\binom{7}{0}+\binom{7}{2} + \binom{7}{3} + \binom{7}{6}}{2^7}= \frac{64}{2^7} = \frac{1}{2}.$$

The Phenotype
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