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

(figure 20.1(a)).

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

greatly appreciated.

Paul Comitz