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Therefore, the circular convolution is equal to the sum of the values of the circularly shifted sequence x((n from k = 0 to k = 3. Because a ( n ) is

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When a pair of dice is tossed, there are 36 possible outcomes (6 outcomes on the first die, and 6 outcomes on the second for each outcome on the first die) Of those 36 outcomes, 1 will produce a sum of 2 (1 + 1), 2 will produce a sum of 3 (1 + 2 or 2 + 1), and 1 will produce a sum of 12 (6 + 6) So there are a total of 4 chances out of 36 of the event 2, 3, or 12 happening That s a probability of 4/36 = 1/9 Similarly, there are 6 ways that a sum of 7 will occur and 2 ways that a sum of 11 will occur, so the probability of the event 7 or 11 is 8/36 = 2/9 The other probabilities on the first level of the tree are computed similarly.

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(recall lhat .u(n) is considered to be a sequence of length five), .r((-n))s is formed by reading the sequence values

TRANSITION DIAGRAMS A transition diagram is a tree or graph (see 15) whose internal nodes represent different states or situations that may occur during a multistage process. As in a decision tree, each leaf represents a different outcome from the process. Each branch is labeled with the conditional probability that the resulting child event will occur, given that the parent event has occurred. EXAMPLE 10.4 The Game of Craps

backward, beginning with n = 0:

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To see how the probabilities are computed for the second level of the tree, consider the case where the point is 4 If the next toss comes up 4, X wins If it comes up 7, Y wins Otherwise, that step is repeated The transition diagram shown in Figure 107 summarizes those three possibilities The probabilities 1/12, 1/6, and 3/4 are computed as shown in the transition diagram in Figure 107: Figure 107 The game of craps P(4) = 3/36 = 1/12 P(7) = 6/36 = 1/3 P(2,3,5,6,8,9,10,11, or 12) = 27/36 = 3/4 So once the point 4 has been established on the first toss, X has a probability of 1/12 of winning on the second toss and a probability of 3/4 of getting to the third toss.

Thus, y(0) is the sum of the first four values of ~ ( ( - n ) ) which gives v ( O ) = 6 . Circularly shifting this sequence to ~, the right by I, we have

The game of craps is a dice game played by two players, X and Y. First X tosses the pair of dice. If the sum of the dice is 7 or 11, X wins the game. If the sum is 2, 3, or 12, Y wins. Otherwise, the sum is designated as the point, to be matched by another toss. So if neither player has won on the first toss, then the dice are tossed repeatedly until either the point comes up or a 7 comes up. If a 7 comes up first, Y wins. Otherwise, X wins when the point comes up. The transition diagram shown in Figure 10.6 models the game of craps.

and summing the first four values gives y ( l ) = 6. Continuing with this process, we find y(2) = 7, y(3) = 9, and y(4) = 8.

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