This graph visually represents the states and the transitions of an NFA, highlighting the non-deterministic nature where from state q_G, it can transition to multiple states based on different inputs. Note that accepting states are visualized with double circles while other states are indicated with a single circle. This is common in NFAs where multiple transitions can occur from the same or different states with the same input symbol. The graph illustrates the nondeterministic nature of the automaton.    ■

V = (v₁, v₂, v₃),
Σ = {0, 1},
s = v₁,
F = (v₁, v₂),
δ(v₁,0) = v₂     δ(v₁,1) = v₁,
δ(v₂,0) = v₃     δ(v₂,1) = v₁,
δ(v₃,0) = v₃     δ(v₃,1) = v₃.

Note that in the diagram, states v₁ and v₂ are the accepting states. If this DFA is presented with the input string 011010011, it will go through the following states: v₁, v₂ v₁, v₁, v₂, v₁ v₂, v₃, v₃, v₃.

Hence, the string is not admissible because it is not in the language accepted by this automaton. The language it accepts, as a careful look at the diagram reveals, is the set of all binary strings that contain no successive zeroes.

Now we employ Mathematica to define and plot the corresponding atomation. First, we define the states:

Customary define vertex shape function for double circles

To solve this problem, we need a mathematical model---a sample space that consists of all possible outcomes. We list the outcomes in rows according to the number of tosses needed to get three sequential consecutive head, after that the process is terminated. For instance, HTHHH is such outcome for five flips, where H stands for head and T---for tail. Since there is no (fixed) number that assures that the number of trials does exceed it, the sample space is infinite---it contains all finite strings over the alphabet { T, H } ending with three head. There are infinitely many outcomes, but there are no outcomes that represent infinitely many trials.

A finite state automaton (FSA) or machine is an excellent tool to handle such kind of problems. We consider the transition diagram for the FSA accepting all strings with three consecutive head. Let us denote by p the probability to get head, and q = 1- p to get tail.

Note that the state v₃ is the accepting state and the game is over since we obtain the required number of head. Let d_i (i = 1, 2, 3) be the expected duration of the game from any state i, and d = d₀ be the expected value of X, which corresponds to the duration of the game from the start vertex v₀. We denote by p_ik the probability of getting from state i to state k in one step. Then for every i = 0, 1, 2, 3, we get the equation:

Wolfram code summarizes the above calculations. Start by clearing existing definitions

We summaries expected waiting time for flipping a fair coin:

"HHH" → 14,

"HHT" → 10,

"HTH"" → 20,

"HTT"" → 8,

"THH"" → 18,

"THT"" → 12,

"TTH"" → 6,

"TTT"" → 14.

These values are indeed the expected waiting times for each pattern to occur, calculated by using the transitions between states in the corresponding finite state machine. You can use these values to understand how likely or unlikely certain sequences are to appear in a series of coin flips.

   ■

We flip a fair coin repeatedly and consider the number of times it takes for particular triplets (TTH, or HHH etc.) to appear. If we flip it just three times then each of the eight triplets is equally likely to show up. But if we persist, and ask questions about time to first appearance of each pattern, or about the likelihood of seeing one pattern before another, then inequalities emerge. Not always, by symmetry we expect either of three head or three tail to be produced before the other, but it turns out that any of other triplets is likely to appear before these two. Every (i, j)th entry in the following 8×8-matrix contains the probability that triplet i occurs in a sequence of coin flips before triplet j. The exercise 3 asks you to verify these probabilities.

Similarly to average waiting time, we get

Let us consider one entry, say "5/12" of pattern HHH vs THT. To determine the probability of seeing HHH before THT, we plot a graph of transition probabilities for corresponding DFA.

It should be noted that all states except final states (HHH and THT) should have two outgoing arrows corresponding two outcomes due to flipping a coin---either HEAD or Tail.

For two stopping states, HHH (set i = 1) and THT (set i = 2), we introduce leading numbers: \[ \varepsilon_r (1,1) = 1 \quad (r = 1, 2, 3), \qquad \varepsilon_1 (2,2) = 1, \quad \varepsilon_3 (2,2) = 1. \] All other leading numbers are zeroes (RJB explains). The corresponding coefficients are \[ e_{1,1} = 1 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4}, \qquad e_{2,2} = 1 + \frac{1}{4} = \frac{5}{4} . \] This leads to system of two equations: \[ \frac{7}{4}\, x_1 = \frac{1}{8} , \qquad \frac{5}{4}\, x_2 = \frac{1}{8} . \] You don't need a computer to solve these two equations: \[ x_1 = \frac{1}{14} , \quad x_2 = \frac{1}{10} \qquad\Longrightarrow \qquad x_1 + x_2 = \frac{6}{35} . \]

Luckily, the captured man had a laptop with a software algebra system, and the prisoner was able to make use of it. The computer scientist spent a few minutes on the problem and asked the pirate to use letters on all faces (which would make the die more attractive). He suggested and the captain agreed to use letters L, I, V, E, D, and A on it, and let the deciding words to be LIVE or DEAD. What is the expected number of rolls for each of three words?

Solution.     First, we draw a DFA that simulates appearance of the five letters. Note that solid lines indicate movements with probability ⅙ and dotted lines show transitions with different probabilities (mostly ½ by default, but some with ⅔). There are two accepting states.

Then we calculate the expected number of rolls to reach one of the final states. Denoting this number by d, we derive the following equation \[ d=1+ \frac{2}{3}\,d + \frac{1}{6}\, d_l +\frac{1}{6}\,d_k , \] where d₁ and d_k are expected numbers to reach the final states from states ``L'' and ``K,'' respectively. These expected values are also expressed via expected values to get to the final states from other states: \[ d_l = 1 +\frac{1}{6} \,d_l + \frac{1}{6} \,d_k + \frac{1}{6} \,d_{li} + \frac{1}{2} \,d , \qquad d_k = 1 +\frac{1}{6} \,d_l + \frac{1}{6} \,d_{k} + \frac{1}{6} \,d_{ki} + \frac{1}{2} \,d . \] Similarly, we obtain \begin{align*} d_{li} &= 1 +\frac{1}{6} \,d_l + \frac{1}{6} \,d_k + \frac{1}{6} \,d_{liv} + \frac{1}{2} \,d , \quad &d_{ki} &= 1 +\frac{1}{6} \,d_{kil} + \frac{1}{6} \,d_k + \frac{2}{3} \,d , \\ d_{liv} &= 1 +\frac{1}{6} \,d_l + \frac{1}{6} \,d_k + \frac{1}{2} \,d , \quad &d_{kil} &= 1 +\frac{1}{6} \,d_{li} + \frac{1}{6} \,d_k + \frac{1}{2} \,d . \\ \end{align*} Solving this system of equations, we obtain the expected number of rolls to finish the game: \[ d=\frac{279936}{431} \approx 649.50348 \quad\mbox{or about 650 rolls.} \] Now we consider the case when there is only one stopping pattern---either KILL or LIVE. Since they are similar, we draw only one DFA:

Solving the following system of equations \begin{align*} d&= 1+\frac{5}{6}\, d +\frac{1}{6}\, d_{l} , \quad &d_{l} &= 1 +\frac{1}{6}\, d_{l} +\frac{1}{6}\, d_{li} + \frac{2}{3}\, d , \\ d_{li} &= 1+ \frac{1}{6}\, d_{liv} + \frac{1}{6}\, d_{l} + \frac{2}{3}\, d , \quad &d_{liv} &= 1 + \frac{1}{6}\, d_{l} + \frac{2}{3}\, d , \end{align*} for d, the expected value to reach the final state LIVE, we obtain \[ d=1296 . \] The probability for the machine to be at a particular state (there are nine of them) depends on the number of rolls. However, their numerical values are not our immediate concern. Instead, we are looking for the case when the number of trials increases without bounds trying to evaluate the `"ultimate'' probability for infinite sequence of letters (if they are not terminated by two absorbing states). We denote by p_k, p_l, p_ki, p_li, p_kil, and p_liv the limiting ``probabilities'' to reach states ``K,'' ``L,'' ``KI,'' ``LI,'' ``KIL,'' ``LIV,'' respectively.

We can arrive to state ``K'' either from the starting state (with probability ⅙), or from the same state ``K'' (with probability ⅙), or from the state ``L'' (with probability ⅙), or from the state ``LI'' (with probability ⅙), or from the state ``LIV'' (with probability ⅙), or from the state ``KI'' (with probability ⅙), or from the state ``LIV'' (with probability ⅙). So we have the equation \[ p_k = \frac{1}{6} + \frac{1}{6} \, p_k + \frac{1}{6} \, p_l + \frac{1}{6} \, p_{ki} + \frac{1}{6} \, p_{kil} + \frac{1}{6} \, p_{li} + \frac{1}{6} \, p_{liv} . \] Similarly, we can reach the state ``L'' from the starting state and states ``K,'' ``LI,'' and ``LIV,'' all with the same probability ⅙. This observation we translate into the equation \[ p_l = \frac{1}{6} + \frac{1}{6} \, p_k + \frac{1}{6} \, p_l + \frac{1}{6} \, p_{li} + \frac{1}{6} \, p_{liv} . \] For other states we get similarly \[ p_{li} = \frac{1}{6} \, p_l + \frac{1}{6} \,p_{kil} , \quad p_{liv} = \frac{1}{6} \, p_{li} , \quad p_{ki} = \frac{1}{6} \, p_{k} , \quad p_{kil} = \frac{1}{6} \, p_{ki} . \] Hence, we need to find p_kil and p_liv from this system of algebraic equations because the probabilities to reach the final states are just ⅙ times these probabilities. When a CAS is available, it is not hard to find these probabilities to obtain \[ p_{kil} = \frac{216}{28339} , \quad p_{liv} =\frac{215}{28339} , \] Since the odds of the states KILL vs LIVE are 216 : 215, we get the following probabilities: \[ \Pr [\mbox{KILL}] =\frac{216}{215+216} = \frac{216}{431} \approx 0.5011600928, \quad \Pr [\mbox{LIVE}] =\frac{215}{431} \approx 0.4988399072 . \] Note that there KILL (or LIVE) is the event that a run of consecutive letters KILL (or LIVE occurs before a run of consecutive letters LIVE (or KILL).

The case LIVE vs DEAD can be treated in a similar way. However, we present its solution in the following example, using another technique.

The prisoner was able to convince the pirate to use his die idea by telling him that it would be much more elegant to have 6 full letters on the die, rather than 5 and a blank. The pirate agreed, and he started rolling. The result of 474 rolls was:

vleideedeilelvdllidlvevvdieeaididvldevelvadeeevieaivavidei
eiiiaaddaddivalaeaeeivvilvvlveavaldaeedeilivaveaveleealddl
veaeividevevlldadelddlivavvialvaleeeedaleailvlvvlvilalilai
aleallaevaidveaidddlladdlalilvideiladldeviivlvddadvlvvvlei
vevivllaveldvlvavvevilaiavidvvvlelleldevdldaleiivldveliiee
iialevaledvvlivvvdivdead

Even with a slight advantage, the computer scientist will still lose in the long run. Being an enterprising man, the pirate offered him one last chance to save his life. He noticed that the prisoner was an excellent computer scientist and he had some ideas for a start-up company. If the prisoner would work for 12 hours a day, 7 days a week, he would be spared and if the company was successful, he would be rewarded with 5% of all profits. The computer scientist was shocked! This was a better offer than where he was working before!    ■

Now we calculate the expected number of rolls, denoted by $d$, to see the run of DEAD by solving the following system of algebraic equations: \begin{align*} d&=1+\frac{5}{6}\,d +\frac{1}{6}\,d_{D} , \quad &d_{D} &= 1+\frac{1}{6}\,d_{D} +\frac{2}{3}\,d + \frac{1}{6}\,d_{DE} , \\ d_{DE} &=1+\frac{1}{6}\,d_{D} +\frac{2}{3}\,d + \frac{1}{6}\,d_{DEA} , \quad &d_{DEA} &=1+\frac{5}{6}\,d . \end{align*}

1. Find the expected number of flips of a biased coin to see r (r ≥ 3) consecutive head by deriving a corresponding enumerator.
2. Let M = { V, Σ, δ, v₀, F } be a DFA, where V = { v₀, v₁, v₂, v₃ }, Σ = {0, 1}, F = { v₃ }, and the transition function is given by the following table:
   

   State 0 1 v₀ v₀ v₁ v₁ v₁ v₂ v₂ v₁ v₃ v₃ v₃ v₁

   Which of the following strings are in L(M)? (a)     ε (b)     001101, (c)     1110, (d)     11011.

3. Find the mean waiting time and verify stopping probabilities for each pair of the eight triplets when flipping a fair coin in the table.

   	HHH	HHT	HTT	HTH	THT	THH	TTH	TTT
   HHH	---	½	⅖	⅖	5/12	⅛	3/10	½
   HHT	½	---	⅔	⅔	⅝	¼	½	7/10
   HTT	⅗	⅓	---	½	½	½	¾	⅞
   HTH	⅗	⅓	½	---	½	½	⅜	7/12
   THT	7/12	⅜	½	½	---	½	⅓	⅗
   THH	⅞	¾	½	½	½	---	⅓	⅗
   TTH	7/10	½	¼	⅝	⅔	⅔	---	½
   TTT	½	3/10	⅛	½	⅖	⅖	½	---

4. A coin is tossed successively until either two consecutive head occur in a row or four tail occur in a row. What is the probability that the sequence ends with two consecutive head? What is expected number of tosses for this event?
5. A coin is tossed successively until one of the following patterns is observed: either a run of H's of length m or a run of T's of length n. We know from Example~\ref{M582.3} that the expected waiting time of the run of H's is 2^m+1 - 2, while for the run of T's it is 2ⁿ⁺¹ - 2. Find the expected number of tosses until either of these two runs happens for the first time and show that for a fair coin this number is \[ \left( \frac{1}{2^{m+1} -2} +\frac{1}{2^{n+1} -2} \right)^{-1} = \frac{2^{n+m} - 2^{n} - 2^{m} +1}{2^{n-1} + 2^{m-1} -1} \,. \]
6. Suppose you flip a fair coin without stopping, what is the average waiting time until the first occurrence of the pattern occurs? Write the enumerator for each event.
   1.     HHHHH,
      
   2.     HTHTH,
      
   3.     HHHTH,
      
   4.     HHHHT .

7. What is the probability that in a stochastic sequence of head (H) and tail (T) the pattern HHHHT appears earlier than HHHTH? Assume that tail lands with probability q.
8. (a)    On average, how many times do you need to choose digits from the alphabet Σ = { 0, 1, … , 9 } (taken randomly with replacement) to see 4-long sequence (i, i, i, i), i ∈ Σ?
   (b)    A pseudo-random generator produces r-tuples of integers (i₁, i₂, … , i_r ) from the alphabet Σ = { 0, 1, … , 9 }. What is expected waiting time (measured in the number of calls of the random generator) for the sequence (i, i, … , i)? DFA:
9. The goal of this exercise is to show that some waiting time problems may have counterintuitive solutions. Given two sequences of patterns, HTHT and THTT, a coin is flipped repeatedly until one of the patterns appears as a run; what pattern win? Find the expected number of tosses needed to see a winner; also, determine the expected running time (=number of flips) for each of the patterns. As a check on your answers for a fair coin: the odds are 9 to 5 in favor of getting HTHT first, though its expected number of tosses till first arrival is 20, whereas the pattern THTT needs only 18 flips on the average.

10. Suppose, for instance, that A flips a coin until she observes either HHHTH or HTHTH, while B flips another coin until he observes either HHHHT or HHHTH. Here H stands for head and T---for tail. It is assumed that both coins have the same probability p of head on each trial. For what p the average waiting times for A and for B will be the same?

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