Three coins are flipped. If at least one of them comes up heads, then what is the probability that they all come up heads?

The answer to this is 1/7, as it is the #(ways to have all heads)/#(ways to have at least one head).

Why are these two answers different? Why is it that the "at least one" is a weaker condition, when at least simply means that one of the three coins is heads, which is the same as choosing a specific head, like the first coin? I would greatly appreciate some intuition here. Thanks!

Why is the at least one case less likely? If anything, shouldn't it be more likely because it doesn't restrict a specific coin to come up heads?

• Suppose you had a thousand tosses. Saying that "at least one $H$" appears is practically meaningless (since it would be astonishing to throw $1000$ Tails in a row). Specifically, that assumption only rules out one possible sequence out of $2^{1000}$. But saying "the first toss was $H$" rules out half the total number of sequences.
– lulu
Jun 13, 2020 at 17:56
• @lulu : You should post that as an answer. Jun 13, 2020 at 19:18

General result: Let $$A, B, C$$ be events in a probability space, all with positive probability. If $$A\subseteq B\subseteq C$$, then:

$$P(A \mid B) \ge P(A\mid C)$$.

This follows because $$A$$ 'takes up more of ' $$B$$ than it does of $$C$$.

If you apply the above general result to these events in the space of flipping three coins:

$$A$$ is the event of all three coming up heads;

$$B$$ is the event of the first coin coming up heads; and

$$C$$ is the event of at least one coin coming up heads;

you get that $$A$$ given $$B$$ is more likely than $$A$$ given $$C$$, as desired.

• OHHHH THANK YOU Jun 13, 2020 at 21:03

$$\begin{array}{c} & HHH \\ & HHt \\ & HtH \\ & tHH \\ & Htt \\ & tHt \\ & ttH \end{array}$$ The seven outcomes in which at least one "head" appears are listed above. In just one instance, all three are "heads."