How do I prove using conditional probabilities that two people have the same probability of picking a blue ball, despite who picks first?

Question

Imagine there are two people: Alice and Bob. Alice and Bob both pick a ball out of a box of nine balls, without replacement. There are 6 blue balls and 3 red balls. If one picks a blue ball, that person wins money (win). If one picks a red ball, that person has to pay money (lose).

How do I prove, using conditional probabilities, that Alice and Bob both have the same probability of winning, regardless of who picks a ball first?

My approach:

Event A = Alice wins, event B = Bob picks a red ball.

$$P(B|A) = \frac{P(A|B)P(B)}{P(A|B)P(B) + P(A|B^c)P(B^c)}$$

This comes down to:

$$P(B|A)=\frac{\frac{6}{8} * \frac{3}{9}}{\frac{6}{8}*\frac{3}{9}+\frac{5}{8}*\frac{6}{9}} = \frac{3}{8}$$

The same can be said for the opposite case, so event A = Bob wins, event B = Alice picks a red ball. So their probabilities of winning (choosing a blue ball) are the same, regardless of who picks the first ball.

Answer 1

No, your reasoning only illustrates that the problem is symetric with respect to Alice and Bob.

You need to show that the one that picks the first ball has the same probability of winning than the one that picks the second ball.

For the one that picks the first ball, the probability of winning is: $\frac{6}{9} = \frac{2}{3}$.

For the one that picks the second ball, the probability of winning is $ \frac{2}{3}\cdot\frac{5}{8} + \frac{1}{3}\cdot\frac{6}{8} = \frac{10+6}{24} = \frac{16}{24} = \frac{2}{3}$

The first term in the first sum above corresponds to the case where a blue ball was picked first and the second term corresponds to the case where a red ball was picked first. Indeed there is 2/3 probability that the first pick is blue, and in that case there are 5 blue balls among the 8 remaining balls. There is a 1/3 probability that the first pick is red, and in that case there are 6 blue among the 8 remaining balls.

So in both cases, the probability of winning is $\frac{2}{3}$.

Answer 2

Let $A_r$ denote the event that Alice picks a red ball.

Let $A_b$ denote the event that Alice picks a blue ball.

Let $B_r$ denote the event that Bob picks a red ball.

Let $B_b$ denote the event that Bob picks a blue ball.

It is understood, that Alice picks first, without replacement, so that Bob is then selecting from $(8)$ balls, rather than $(9)$.

I am assuming that if Alice and Bob both pick a ball of the same color (i.e. both red or both blue), then no one wins.

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Edit
See my comments following the answer of citronrose.

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Alice's probability of winning is $p(A_b) \times p(B_r|A_b).$

This equals $\frac{6}{9} \times \frac{3}{8} = \frac{18}{72}.$

The $\frac{3}{8}$ factor is explained by presuming that after Alice picks a blue ball, there are $8$ balls left, $3$ of which are red.

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Bob's probability of winning is $p(A_r) \times p(B_b|A_r)$.

This equals $\frac{3}{9} \times \frac{6}{8} = \frac{18}{72}.$

The $\frac{6}{8}$ factor is explained by presuming that after Alice picks a red ball, there are $8$ balls left, $6$ of which are blue.