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Notice,
that in the above figure we have only stated constraints on the
probability assignment. An actual probability assignment has
not been made. If the penny is a "fair" coin and the
procedure for tossing the coins is "fair," then it
is reasonable to assume that the possible events are equiprobable.
Each of the two coins can come up heads or tails. Thus, if these
events are equiprobable then each has a probability of 1/2 or
.5. And, the outcome for one coin has no influence on the outcome
for another coin - the events are independent. Then the
probability of the joint events can be obtained by multiplying
their probabilities. Thus, the likelihood that they both come
up heads is .5 x .5 or .25; that the first comes up heads and
the second tails is also .25; that the second comes up heads
and the first tails is .25; and that they both come up tails
if .25. Note that the probability of at least one of the coins
coming up heads is .75 since we must add the cases that can yield
this outcome.
- Note
that in this simple case where the outcomes of the coin tosses
were independent, we needn't worry about belief revision.
Independence implies that the previous history of the
outcomes has no influence on the current likelihood of either
coming up heads or tails. However, what if we had rather special
coins? Perhaps one of the coins, we aren't sure which will come
up heads with a somewhat higher probability whenever the other
coin has come up tails on 4 or more of the most recent 6 trials.
In this case, the outcomes are not independent and we will constantly
have to revise our estimates based on prior outcomes of the coin
tosses.
- Now
that we have examined a very simple world involving only two
coins we are ready to consider the general case. Note that in
the general case we can not assume that the events are
equiprobable and independent. Consequently, belief revision will
clearly be a computationally expensive task. And, another difficult
problem is to determine exactly how to assign probabilities to
a particular event.
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- For
example, pretend that we have 30 pennies that are tossed on every
trial. And assume that these pennies are very special pennies
that can influence the outcome of some of their fellow pennies.
For example, penny 1 and penny 9 may be particularly
in sympathy with each other and always take on exactly the same
value. And, half of the time penny 6 will take on the
opposite value of penny 1, and so on. How can we determine,
for example, the probability of penny 6 coming up heads?
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- The
next figure below discusses this general problem of assigning
probabilities to a set of propositions in a coherent fashion
for this general case.
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