Thank you, Hurkyl, for taking the time to explain it in detail. That makes sense now.
EDIT:
BTW, the above identity allows one to prove Laplace's celebrated (and abused) Rule of Succession without having to take limits.
An urn contains
n balls. The number
k of red balls is determined randomly so that the possibilities k = 0, 1, ..., n, are equally likely. The remaining
n-k balls are white. You do not know the fraction of red balls. You draw
m balls,
r of which are red. What is the probability that the
(m+1)st is red? Conditional on
k, the probability that
r of the first
m, as well as the
(m+1)st balls are red is
\frac{\binom{k}{r}\binom{n-k}{m-r}}{\binom{n}{m}}\frac{k-r}{n-m} = \frac{r+1}{m+1}\frac{\binom{k}{r+1}\binom{n-k}{(m+1)-(r+1)}}{\binom{n}{m+1}}
Summing over
k gives
\frac{r+1}{m+1}\frac{\binom{n+1}{m+2}}{\binom{n}{m+1}}
which is proportional to the unconditional probability that
r of the first
m as well as the
(m+1)st balls are red. The constant of proportionality is P(k) = 1/(n+1). Similarly, the unconditional probability that
r of the first
m balls are red is proportional to
\frac{\binom{n+1}{m+1}}{\binom{n}{m}} by the same constant P(k) = 1/(n+1).
Finally, divide the first by the second and apply Bayes' theorem to get
P{\text{(m+1)st ball is red} | \text{r of the first m were red}} = \frac{(r+1)\binom{n}{m}\binom{n+1}{m+2}}{(m+1)\binom{n+1}{m+1}\binom{n}{m+1}} = \frac{r+1}{m+2}
So, of course, if you flip a coin once and it comes up heads, that means it comes up heads on the next flip with probability 2/3!
