Has anyone tried to work the example on pages 713/714 (second
edition)? I cannot got equation 20.1 to match the graph on page 714
For example, consider P(h5|d) = alpha * P(d|h5)P(h5). In this case P
(d = lime|h5) is always 1. For the ten trials,
P(d1 ...d10|h5) = 1**10 = 1. (20.3)
P(hi) = .1
So P(h5|d) = (.1 * alpha) from equation 20.1. Assume alpha = 1/P(d =
lime) = 1/.5 = 2. So after one candy is unwrapped, P(h5|d) = .2 as
shown in figure 20.1(a).
On the next trial (i.e. second candy is also lime) P(d|h5) = 1 (as
above from eqn 20.3). Then P(h5|d) = (1) * (.1) * alpha. The value on
figure 20.1 is shown as approximately .3 - implying alpha = 3
Subsequent calculations for alpha appear to 4, 5 etc.
I can't seem to see how alpha is calculated ..... Any opinions are