(This result should be familiar, since if the probability to remain in a state is 1 p, then the average number of steps to leave the state is

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1 How many coin flis on average does it take to get n consecutive heads? 1 The rocess of fliing n consecutive heads can be described by a Markov chain in which the states corresond to the number of consecutive heads in a row, as deicted below. H In this 1H 1 language, H the question 1H becomes 2H how many stes does it take on average to 1 2H get from the state H to the state nh? 1 d) H 1 2H 1 n 1 1 a) 1 H 1H 3H 1 b) c) 1H1 1 3H H nh Assume the coin has robability of coming u heads. Begin with the case deicted in fig. (a), and let A 1 be the average number of flis on average before getting the first head. If the first fli is heads (robability ), then the answer is 1; if, on the other hand, the first fli is tails (robability 1 ), then one fli is wasted and there remain A 1 to go. These two observations give an equation for A 1 : A 1 = (1 )(1 + A 1 ) + 1, (a1) A 1 = 1. (a2) (This result should be familiar, since if the robability to remain in a state is 1, then the average number of stes to leave the state is k=1 k(1 )k 1 = (1/ 2 ) = 1/.) For = 1/2, we find A 1 = 2, so on average two flis are required to get the first head if the coin is fair. Now consider A 2, the average number of flis to get two heads in a row (fig. (b)). Again, if the first fli is wasted on a tails, there s a term (1 )(1 + A 2 ) on the right side. But now if the first fli is heads, there are two ossibilities for what haens next. If the next fli is tails, the first two flis are wasted and we re back where we started. See footnote 2 below. 1 INFO Nov 5

2 But if the next fli is a head, then the goal is accomlished in two flis. This gives the equation A 2 = (1 )(1 + A 2 ) + (1 )(2 + A 2 ) + 2 2, (b1) A 2 = (b2) For = 1/2, we find A 2 = 6, so on average six flis are required to get 2 heads in a row if the coin is fair. Similar reasoning for A 3, the average number of flis to get three heads in a row (fig. (c)) gives A 3 = (1 )(1 + A 3 ) + (1 )(2 + A 3 ) + 2 (1 )(3 + A 3 ) + 3 3, (c1) A 3 = (c2) For = 1/2, we find A 3 = 14, so on average fourteen flis are required to get 3 heads in a row if the coin is fair. In general, the average number of flis to get n heads in a row (fig. (d)), A n, satisfies A n = (1 )(1+A n )+(1 )(2+A n )+ 2 (1 )(3+A n )+...+ n 1 (1 )(n+a n )+ n n. (d1) Regrouing terms on the right hand side and using n 1 = 1 n 1 gives A n = A n (1 )( n 1 ) + (1 )( n n 1 ) + n n = A n (1 n ) + ( n n 1 n n ) + n n = A n n A n + ( n 1 ). This results in A n = n 1 = 1 n n n (1 ) = n 1 1. (d2) For = 1/2, we find A n = 2 n+1 2 flis required to get n heads in a row if the coin is fair, and the number grows exonentially in n. 1 To rove this, let S n = n 1 k= k and note that 1 + S n = S n + n. 2 INFO Nov 5

3 H H 1 H A slight generalization of this roblem is to have a different robability for each successive 1 1H head, 1 2H i.e., to switch to a coin with robability j of getting a head when going for the j th 1 n1 nh 1n n1 n head in a row, as deicted in fig. (e) below: n1 n e) n H f) nh The average number of flis A n to get n heads in a row now satisfies A n = (1 1 )(1 + A n ) + 1 (1 2 )(2 + A n ) (1 3 )(3 + A n ) ( 1 2 n 1 )(1 n )(n + A n ) + ( 1 2 n ) n. (e1) Algebra similar to that leading from (d1) to (d2) now results in A n = n n. (e2) Note 1: All of the above results can derived from a single recursion equation, as suggested by fig. (f). Suose A n 1, the average number of flis required to reach n 1 successive heads is known. Then A n can be determined without knowing the recise details of what haens for the first n 1 flis, as deicted by the ellisis ( ) in fig. (f). It takes an average of A n 1 stes to reach the state (n 1)H. If the next fli is heads (robability n ), then the answer is A n 1 + 1; if, on the other hand, the next fli is tails (robability 1 n ), then A n flis have been wasted and there remain A n to go. These two observations give an equation for A n in terms of A n 1 : A n = (1 n )(A n A n ) + n (A n 1 + 1), (f1) A n = (A n 1 + 1) 1 n. Starting from A =, the above equation gives A 1 = 1/ 1, A 2 = (1/ 1 + 1)/ 2 = (1 + 1 )/ 1 2, A 3 = ( )/ 1 2 3, and by induction gives (e2) for A n. For 1 = 2 =... = n =, these are equivalent to (a2), (b2), (c2), (d2). So eq. (f2), describing fig. (f), embodies the content of all of the revious equations. Note 2: Another direct way to derive all of the above is based on the relation between figures (a) and (f). The robability to loo around k times, i.e., to get to the state (n 1)H and go back to H exactly k times before ultimately getting to nh is (1 n ) k n, and the average number of stes for this rocess is (A n 1 + 1)k + A n = (k + 1)(A n 1 + 1). (f2) 3 INFO Nov 5

4 Summing over k gives A n = k= (A n 1 + 1)(k + 1)(1 n ) k n. Using (1 ) + 3(1 ) = 1 2, it follows that A n = (A n 1 + 1) n (k + 1)(1 n ) k 1 = (A n 1 + 1) n k= 2 n = (A n 1 + 1) 1 n, reroducing the generating equation (f2). (The 1/ n is now recognized as the same familiar 1/ mentioned after eq. (a2).) As usual, there s more structure in these robability distributions than just the average number of stes. Let n (M) denote the robability of reaching n consecutive heads only after exactly M flis of a single coin, each fli with robability of heads. Then the average satisfies A n = M= M n(m). For n = 1, the robability to reach the first head in M flis is the robability of M 1 tails and one head, hence 1 (M) = M. The average number of flis until the first head is k= (k + 1)(1 )k = 1/. The robability distribution 1 (M) is shown for a fair coin ( = 1/2) in the first figure on the next age. Additional figures show the robability distributions for n = 2, 3, 4, 5, 1. In general, the robability vanishes, n (M) =, for M < n since it s imossible to have n consecutive heads with fewer than n total flis. The first non-zero robability is n (M = n) = n, corresonding to all heads for the first n flis. For the next n values of M, from M = n + 1 through M = 2n flis, the robability is constant, n (M) = n+1, since it is fully characterized by just the last n + 1 flis (i.e., a tail followed by n heads, and anything can haen in the first M (n + 1) flis). For larger values of M, n (M) becomes the robability of not having more than n 1 consecutive heads in the first M (n + 1) flis, then followed by a final tail and n heads in the last n+1 flis. For examle, for M = 2n+1 flis, that robability is just all the ways not to have n consecutive flis in the first n flis, then the tail and n heads, so n (M = 2n + 1) = (1 n ) n+1. The robabilities n (M) are thus related to the robability of having no more than n consecutive heads in M (n + 1) flis, in turn equal to 1 minus the robability of having at least n consecutive heads in M (n + 1) flis. In general, the robability of having at least n consecutive heads in N flis of a fair coin (or equivalently the robability of at least n consecutive successes in N Bernoulli trials) is difficult to write down in closed form. To rovide some intution for how those numbers behave, consider the examle of N = 1. 2 To rove this, let S = k= qk = 1 1 q, and note that k= kqk 1 = q S = 1 (1 q) 2. 4 INFO Nov 5

5 .5 n=1.25 n= n=3.6 n= n=5.9 n= Figures: The robabilities n (M) of first fliing n consecutive heads after exactly M flis of a fair ( = 1/2) coin, for n = 1, 2, 3, 4, 5, 1. M is lotted along the horizontal axis. The red line shows the value of the average number of rolls required, eq. (d2): A n = 2 n+1 2 (res., 2,6,14,3,246). The regions indicated in black reresent the first 5% of the robability for each of the grahs. 5 INFO Nov 5

6 What is the robability (n) of n heads in a row somewhere in a sequence of 1 coin flis? For the case of a fair coin ( = 1/2), the results of a numerical simulation (2 million students each fliing 1 times, simulated by a random number generator) are roughly given by: n (n) z(n) Also shown is z(n), the average number of times that a string of n heads occurs (where by definition, for examle, six consecutive heads counts as exactly two occurrences of three consecutive heads). So if a class of 5 students were asked to generate random strings of 1 heads and tails, roughly forty of them (81%) should have at least one occurrence of five consecutive heads, roughly 27 (55%) at least six consecutive heads, and roughly 16 (32%) at least seven consecutive heads (and similarly for consecutive tails). Since these are relatively rare events, the number of times m that a string of n consecutive heads occurs will be roughly Poisson distributed, and determined by the mean number of occurrences, P n (m) = e z(n)( z(n) ) m /m!. For examle, since the mean number of times that five heads in a row occur in a hundred flis is z(5) = 1.56, the robability that five heads in a row occur m times is P n (m) = e 1.56 (1.56) m /m!, and the robabilities of 5 occurrences resectively are.21,.33,.26,.13,.5,.2. Similarly, for n = 7, with a mean z(7) =.37, the robabilities for m =, 1, 2, 3 are.69,.26,.5,.1. (For larger values of n, the average number of occurrences z(n) aoaches the robability (n), so the likelihood is one occurrence if at all, and the robabilities P n (m) are dominated by m =, 1.) 6 INFO Nov 5