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I have a bag with only red and blue balls. I take a ball out and hide it. Now there are x red balls and y blue balls in the bag. The bag was created with the following procedure. For each ball, there is a p chance of it being red, and 1-p chance of it being blue. What is the probability the hidden ball is red, expressed using only x, y, and p.

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Let's look at how we can form the initial configuration: For any ball in the bag, there is a constant probability of p to pick one red ball and q to pick a blue ball; and we know p+q = 1.

When we know there is (x+y+1) balls in the bag, the chance of bag containing (x + 1) red balls and y blue balls is $$ \binom{x+y+1}{x+1}\cdot p ^{x+1} \cdot q^y $$

And the relative probability of taking a red ball from it is $$ P(R) = \binom{x+y+1}{x+1}\cdot p ^{x+1} \cdot q^y \cdot \frac{x+1}{x+y+1} = \binom{x+y}{x}\cdot p ^{x+1} \cdot q^y = \binom{x+y}{x}\cdot p ^{x} \cdot q^y \cdot p $$

Similarly, the chance of the bag containing x red balls and (y+ 1) blue balls is $$ \binom{x+y+1}{y+1} \cdot p ^x \cdot q^{y+1} $$ And the relative probability of taking a blue ball from it is $$ P(B) = \binom{x+y+1}{y+1} \cdot p ^x \cdot q^{y+1} \cdot \frac{y+1}{x+y+1} = \binom{x+y}{x}\cdot p ^{x} \cdot q^{y+1} = \binom{x+y}{x}\cdot p ^{x} \cdot q^y \cdot q$$

Since the hidden ball can only be red or blue, the probability of hidden ball to be red is: $$ \frac{P(R)}{P(R) + P(B)} = \frac{\binom{x+y}{x}\cdot p ^{x} \cdot q^y \cdot p } {\binom{x+y}{x}\cdot p ^{x} \cdot q^y \cdot (p + q)} = \frac{p}{p+q} = p $$

So the probability of hidden ball to be red is p, regardless of what is observed inside the bag.

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