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Bayes Billiard Balls 10/28

First argument:
Suppose we throw 1 red billiard ball on the table and measure how far it goes on the scale from 0 to 1, call this value x, then throw n balls, what is the distribution of the number of balls to the left of the red ball?

\begin{displaymath}P(left =k\vert x)=\binom{n}{k}x^k(1-x)^{n-k}\end{displaymath}

Now what is x distribution? It is Uniform(0,1), and the overall distribution of the number of successes is the sum for all possible x's:

\begin{displaymath}P(left =k)=\int_0^1 \binom{n}{k}x^k(1-x)^{n-k} dx=\binom{n}{k}\int_0^1 x^k(1-x)^{n-k} dx\end{displaymath}

Second way:
Suppose I throw all the balls down first, and choose which is to be the red one, then the probability that the red one has k to the left of it is: $\frac{1}{n+1}$. So we have:

\begin{displaymath}P(left =k)=\int_0^1 \binom{n}{k}x^k(1-x)^{n-k} dx=\binom{n}{k}\int_0^1 x^k(1-x)^{n-k} dx=\frac{1}{n+1}\end{displaymath}

Which tells us that:

\begin{displaymath}\int_0^1 x^k(1-x)^{n-k} dx=\frac{k!(n-k)!}{(n+1)!}

This constant actually has a special name it is B(k+1,n-k+1) and it comes in for the following important density function, whose support is [0,1].

Susan Holmes