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So once the point 4 has been established on the first toss, X has a probability of (3/4)(1/12) of winning on the third toss and a probability of (3/4)(3/4) of getting to the fourth toss Similarly, once the point 4 has been established on the first toss, X has a probability of (3/4)(1/12) + (3/4)(3/4)(1/12) of winning on the fourth toss, and so on Summing these partial probabilities, we find that once the point 4 has been established on the first toss, the probability that X wins on any toss thereafter is 1 3 1 3 21 3 31 3 41 3 51 - P 4 = ----- + -- ----- + -- ----- + -- ----- + -- ----- + -- ----- + 12 4 12 4 12 4 12 4 12 4 12 = ----------3 1 -4.

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into the expression for y(n) as follows:

CHAP. 10]

Let y(n) be Ihe linear convolution of the two finite-length sequences, h ( n )and r ( n ) ,of length N ,

1 12 = ---------1 4 1 = -3 1----12

Derive the following relationship between y(n) and yN(n):

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A stochastic process is a process that can be analyzed by a transition diagram, that is, it can be decomposed into sequences of events whose conditional probabilities can be computed. The game of craps is actually an infinite stochastic process since there is no limit to the number of events that could occur. As with the analysis in Example 10.4, most infinite stochastic processes can be reformulated into an equivalent finite stochastic process that is amenable to (finite) computers. Note that, unlike other tree models, decision trees and transition trees are usually drawn from left to right to suggest the time-dependent movement from one node to the next. ORDERED TREES Here is the recursive definition of an ordered tree: An ordered tree is either the empty set or a pair T = (r, S), where r is a node and S is a sequence of disjoint ordered trees, none of which contains r. The node r is called the root of the tree T, and the elements of the sequence S are its subtrees. The sequence S of course may be empty, in which case T is a singleton. The restriction that none of the subtrees contains the root applies recursively: x cannot be in any subtree, or in any subtree of any subtree, and so on. Note that this definition is the same as that for unordered trees except for the facts that the subtrees are in a sequence instead of a set and an ordered tree may be empty. Consequently, if two unordered trees have the same subsets, then they are equal; but as ordered trees, they won t be equal unless their equal subtrees are in the same order. Also subtrees of an ordered set may be empty. EXAMPLE 10.5 Unequal Ordered Trees

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There are several ways toderive this relationship. One is to examine what happens when the DTFT of y(n) is sampled. Alternatively, this result may be derived from a systems point of view as follows. First, note that yN(n) is equal to one period of the linear. convolution of the finite-length sequence h(n) with the periodic sequencei(n):

This calculation applies the formula for geometric series. (See page 323.) If the probability is 1/3 that X wins once the point 4 has been established on the first toss, the probability that Y wins at that point must be 2/3. The other probabilities at the second level are computed similarly. Now we can calculate the probability that X wins the game from the main transition diagram: 1 5 5 1 1 2 1 P = -- + ----- P 4 + -- P 5 + ----- P 6 + ----- P8 + -- P9 + ----- P 10 9 36 36 9 12 9 12 2 1 1 1 2 5 5 5 5 1 2 1 1 - - - - - - = -- + ----- -- + -- -- + ----- ----- + ----- ----- + -- -- + ----- -9 12 3 9 5 36 11 36 11 9 5 12 3 244 = -------495 = 0.4929 So the probability that X wins is 49.29 percent, and the probability that Y wins is 50.71 percent.

then the periodic sequence .f(n) is formed by linearly convolving x ( n ) with pN(n):

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