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Five street vendors sell oranges. Four of them ask \$1 each, and one asks \$3.

The first approach to a price average would be ( 1 + 1 + 1 + 1 + 3 ) / 5 = 1.4.

The second approach, only considering unique prices, would be ( 1 + 3 ) / 2 = 2.

What are the correct terms for these two different averages, and in what situations is one a better approach than the other?

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I do not think there is a one unique 'correct way'. Different ways to calculate the mean tell you different things. The first one tells you about the mean price over the sample of shops/vendors. The second one tells you about the mean price level.

There are other possibilities. Say each of the cheap vendors sells $\frac{1}{4}$ of all sold oranges. Then $(1\frac{1}{4}+1\frac{1}{4}+1\frac{1}{4}+1\frac{1}{4})$ is the mean price at which oranges are transacted.

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The first one is the only one that makes sense. It's just the average. The second one doesn't have a name I know, and doesn't deserve one.

You could rewrite the first one as $$ \frac{4 \times 1 + 1 \times 3}{4 + 1} $$ and call it a weighted average.

This is the average vendor price. If the vendors had different numbers of oranges you could compute the weighted average price per orange.

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    How do you _know_ that there is _no_ venue of research _whatsoever_ where the second kind of average is relevant? Because if it _is_ useful, however marginal and obscure, then it probably has a name, and definitely deserves one. – 2017-02-03
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    The first is the "average price paid, for all oranges sold", assuming all vendors each get an equal share of customers. I don't think that rules out eventual use of the value in the second approach? – 2017-02-03
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    @Arthur Yes, there may possibly be a context in which the second kind of average makes sense. Even if so I think it's unusual, so I would not imagine a widely known name. just one for that context. – 2017-02-03