Table 75 Sample point TH TTH HTH TTTH HTTH HHTH TTTTH HTTTH HHTTH HHHTH

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Expected Waiting Time for TH

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This observation makes the sample space fairly easy to write out and it also makes the calculation of probabilities fairly simple Note that the probability of any sample point has the factor qp for the sequence TH at the nth toss If the sample point begins with no heads, then its probability has a factor of qn 2 If the sample point begins with H, then its probability has a factor of pqn 3 If the sample point begins with HH, then its probability has a factor of p2 qn 4 This pattern continues until we come to the sample point with n 2 H s followed by TH The probability of this point has a factor of pn 2 Consider the case for n = 5 shown in the sample space above The probabilities of the points add to qp(q3 + pq2 + p2 q + p3 ) This can be recognized as qp(q4 p4 )/(q p) This pattern continues, and by letting X denote the total number of tosses necessary, we nd that P(X = n) = qp(qn 1 pn 1 ) q p for n = 2, 3, 4,

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The formula would not work for q = p = 1/2 In that case the sample points are equally likely, each having probability (1/2)n and, as we have seen, there are n 1 of them It follows in the case where q = p = 1/2 that P(X = n) = n 1 2n for n = 2, 3, 4,

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EXPECTED WAITING TIME FOR TH

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Using the formula above,

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E[N] =

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qp(qn 1 pn 1 ) q p

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To calculate this sum, consider

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n qn 1 = 2q + 3q2 + 4q3 +

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7

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Waiting Time Problems

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Now S + 1 = 1 + 2q + 3q2 + 4q3 + and we have seen that the left-hand side of this equation is 1/p2 So S = 1/p2 1 E[N] = 1 qp 1 q p p2 1 1 q2 = 1 qp

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The formula above applies only if p = q In the case p = q,we rst consider / P(X = n + 1) n = P(X = n) 2(n 1) So

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2(n 1)P(X = n + 1) =

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n=2 n=2

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nP(X = n)

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We can write this as

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[(n + 1) 2]P(X = n + 1) =

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nP(X = n)

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and from this it follows that 2E[N] 4 = E[N] so E[N] = 4 This may be a surprise We found that the average waiting time for HH with a fair coin is six (By symmetry, the average waiting time for TT is six tosses and the average waiting time for HT is four tosses) There is apparently no intuitive reason for this to be so, but these results can easily be veri ed by simulation to provide some evidence that they are correct We will simply state the average waiting times for each of the patterns of length 3 with a fair coin in Table 76

Table 76 Pattern HHH THH HTH TTH THT HTT HHT TTT Average waiting time 14 8 10 8 10 8 8 14

We continue this chapter with an intriguing game

AN UNFAIR GAME WITH A FAIR COIN

Here is a game with a fair coin In fact, it is an unfair game with a fair coin!

Three Tosses

Consider the patterns HH, TH, HT , and TT for two tosses of a fair coin If you choose one of these patterns, I will choose another and then we toss a fair coin until one of these patterns occurs The winner is the person who chose the rstoccurring pattern For example, if you choose HH, I will choose TH If we see the sequence HTTTTH, then I win since the pattern TH occurred before the pattern HH My probability of beating you is 3/4! Here is why If the rst two tosses are HH, you win If the rst two tosses are TH, I win If the rst two tosses are HT then this can be followed by any number of T s, but eventually H will occur and I win If the rst two tosses are TT then this can be followed by any number of T s but eventually H will occur and I will win If you are to win, you must toss HH on the rst two tosses; this is the only way you can win and it has probability 1/4 If a T is tossed at any time, I will win, so my probability of winning is 3/4 The fact that the patterns HH, TH, HT , and TT are equally likely for a fair coin may be observed by a game player who may think that any choice is equally good is irrelevant; the choice of pattern is crucial, as is the fact that he is allowed to make the rst choice If you choose TT , then the only way you can win is by tossing two tails on the rst two tosses I will choose HT and I will win 3/4 of the time A sensible choice for you is either TH or HT, and then my probability of beating you is only 1/2 If we consider patterns with three tosses, as shown in the table above, my minimum probability of beating you is 2/3! (And if you do not choose well, I can increase that probability to 3/4 or 7/8!) This can be tried by simulation It is puzzling to note that no matter what pattern you choose, I will probably beat you This means that if we play the game twice, and I win in the rst game, then you can choose the pattern I chose on the rst game, and I can still probably beat you Probabilities then are not transitive, so if pattern A beats pattern B and pattern B beats pattern C, then it does not follow that pattern A will necessarily beat pattern C