# Urn probability function

Suppose I have an urn with an infinite number of balls which can be either red or white. I do not know what the proportion of each colour is, but I do know it's a fixed proportion. After drawing $N$ balls, I have observed $r$ red ones and $w$ white ones.

I believe the probability that I will observe a red ball on the next draw is given by Laplace's Law of Succession, $\frac{r+1}{N+2}$. However, how sure should I be of that? That is, before I drew any balls, I believed any proportion other than $0$ or $1$ was the true one. After I drew those $N$ balls, what should be my estimate pdf over the possible values for the proportion of red balls in the urn?

• The probability you will observe a red ball on the next draw is always the same: it equals the proportion of red balls. Are you perhaps asking how to update an estimate of that probability? If so, you are either performing a Bayesian analysis (whose answer depends on your prior probability distribution for the proportion in the urn) or a non-Bayesian analysis (which, depending on its form, can yield a "fuzzy" probability or a "p-box" or a host of other non-probabilistic representations of your uncertainty). In any case, this question is unanswerable without more information. – whuber Nov 8 '13 at 17:19
• Yes, sorry, that's what I meant, it's an estimate of the probability. – Pedro Carvalho Nov 8 '13 at 17:28

Okay, so, my first idea was using Jaynes' $A_p$ distribution, in which case we can define $A$ = next draw will be red and $N_r$ = out of $N$ draws, $r$ were red. With an ignorant prior distribution $(A_p|X) = 1$, I get that

$$(A_p|N_rX) = (A_p|X)\frac{P(N_r|A_p)}{P(N_r|X)}$$

We know that

$$P(N_r|A_p) = \binom{N}{r}p^r(1-p)^{N-r}$$

And we can find

$$P(N_r|X) = \int^1_0(N_rA_p|X)dp = \int^1_0P(N_r|A_p)(A_p|X)dp = \int^1_0\binom{N}{r}p^r(1-p)^{N-r}dp$$

from which

$$P(N_r|X) = \frac 1 {N+1}, 0 \leq r \leq N$$

And then

$$(A_p|N_r) = (N+1)\binom{N}{r}p^r(1-p)^{N-r}$$

And that looks like my distribution. Is that correct?