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In a pool game, there are 15 numbered balls, of which one is black and the other 14 consist of 7 colored pairs. For instance, balls 1 and 9 could be yellow, 2 and 10 blue, and so on.

At the beginning of a game, the balls are arranged in an equilateral triangle. Assuming the arrangement is performed at random and uniformly, what is the probability that two of the balls placed in the triangle vertices will have the same color?

The answer to the problem is 1/5, and the calculation is quite simple. However, my first (dull) attempt to solve it was to condition the arrangements on the position of the black ball, like the following:

Consider events S = {arrangements with two vertices with the same color} and B = {arrangements with the black ball in one of the corners}. We have:

$$ P(B) = 3/15 \\ P(S|B) = \frac{3 \times 7 \times 2 \times 12!}{15!} = \frac{1}{65} $$

Explained: Ways to pick position of the black ball = 3, to pick a color = 7, to pick an ordering among the two balls with the same color = 2, to pick the other 12 balls = 12!

For the event that the black ball is not in one of the corners, we have:

$$ P(B^c) = 12/15 \\ P(S|B^c) = \frac{3 \times 7 \times 2 \times 12 \times 12!}{15!} = \frac{12}{65} $$

Which is similar to the calculation of P(S|B), except that we have 12 possibilities to pick a third vertex (cannot be black).

By applying the formula of conditional probabilities, we should get:

$$ P(S \cap B) = P(S|B) \times P(B) = \frac{3}{975} \\ P(S \cap B^c) = P(S|B^c) \times P(B^c) = \frac{144}{975} \\ P(S) = P(S \cap B) + P(S \cap B^c) = \frac{147}{975} = \frac{49}{325} $$

Which is clearly not 1/5. Why didn't this reasoning provide the correct answer?

Thank you

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up vote 1 down vote accepted

Your calculation of $P(S|B)$ is off. If you are given that a black ball is in one of the vertices, then the total number of outcomes in the reduced sample space is not $15!$; it's $3\cdot 14!$.

Your calculation of $P(S|B^C)$ suffers from a similar error.

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Thank you! I picked this answer because it points out that I did not reduce the sample space when considering a conditioned event, which is fundamental. But thanks Brian for also pointing out that I actually calculated the intersection! – Caetano Sauer Mar 8 '12 at 17:16

The calculation for what you’ve called $P(S\mid B)$ actually gives $P(S\cap B)$, and the calculation for what you’ve called $P(S\mid B^c)$ actually gives $P(S\cap B^c)$. This is especially clear in the first one: $3\cdot 7\cdot 2\cdot 12!$ counts the arrangements that have the black ball in a corner and balls of the same color in the other two corners.

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