# Examples of infinite sample spaces. Math 425 Introduction to Probability Lecture 12. Example of coin tosses. Axiom 3 Strong form

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1 Infinite Discrete Sample Spaces s of infinite sample spaces Math 425 Introduction to Probability Lecture 2 Kenneth Harris Department of Mathematics University of Michigan February 4, 2009 We have been considering finite sample spaces, but some experiments are best modeled by infinite sample spaces. Experiment A: Toss a fair coin until the first head appears, then stop. Experiment C: Toss a fair coin until a run of ten heads in a row appears, then stop. Experiment B: Throw a pair of dice until either a 5 or 7 appears, then stop. No finite sample space is adequate in these cases, since each experiment has (potentially) infinitely many outcomes. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, 2009 / Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Infinite Discrete Sample Spaces of coin tosses Infinite Discrete Sample Spaces Axiom 3 Strong form A. Toss a fair coin until the first head appears, then stop. The sample space for this experiment is We recall the strong form of the Addition Rule (Axiom 3): S { H, TH,, TTH, TTTH,..., T TTT } For each k, a sequence of k tails followed by a single head: T k H. an infinite sequence consisting only tails: T. We can compute the probability of most" of these outcomes: P (T k H) 2 k P (T )??? Axiom (3 Addition rule (strong form)) For any sequence of mutually exclusive events E, E 2,... (so, E i E j whenever i j), P ( E k ) k P (E k ). k Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, /

2 Infinite Discrete Sample Spaces of Coin Tosses S { H, TH,, TTH, TTTH,..., T TTT } Let H be the event that a heads is tossed. Then H c {T } P (T ) P (H c ) P (H) Let H k {T k H} (k 0): the event that heads follows k tails. So, the events H 0, H, H 2,..., are mutually exclusive and H H k Infinite Discrete Sample Spaces continued Then, P (H k ) P (T k H) 2 k. Use the strong form of the Addition Rule: P ( ) H k P (H k ) 2 k. We compute the infinite series shortly. So, P ( H ) P ( and thus, H k ) P ( H c) P ( T ) 0. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Infinite series The Geometric Series Geometric series The Geometric Series Definition Let a, a 2,... be an infinite sequence of real numbers. If the partial sums n s n a k. have a finite limit: k lim s n s, n then we say the infinite series k a k converges with sum s. Otherwise, it diverges. If the series k a k converges as well, then k a k converges absolutely. Theorem The geometric series converges absolutely: ax k a x. From the previous section: 2 k a 2 and x 2 from the Theorem. for any real numbers a, and x <. 2 () k Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, /

3 The Geometric Series The Geometric Series Proof Proof Proof. The nth partial sum is s n Multiply by x (shifting powers by ): n ax k. n+ x s n ax k. k Subtract the previous two results (most terms cancel) Solve for s n, s n x s n a ax k+. s n a( x n+ ) x By definition Since x <, so ax k a( x n+ ) lim s n lim. n n x lim x k 0, k ax k a( x n+ ) lim a n x x. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, 2009 / Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / The monkey and the bard Bernoulli Trials A monkey is sitting at a typewriter and randomly hits keys in a never-ending sequence. What is the probability the monkey eventually types the complete works of Shakespeare? A real monkey will eventually tire of pecking at a typewriter and go searching for a banana. Definition By a sequence of Bernoulli trials, we mean a sequence of trials (repetitions of an experiment) satisfying the following Only two possible mutually exclusive outcomes on each trial. One arbitrarily called success and the other failure. 2 The probability of success on each trial is the same for each trial. 3 The trials are independent. However, this problem is really about a probability model, not a real monkey. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, /

4 s of Bernoulli Trials Sample space s. The following are examples of Bernoulli trials: Flip a coin (heads, tails), Each computer chip in a production line tested (chip passes test, fails test), Rolling a pair of dice for snake-eyes" (double ones, any other value), A patient is prescribed a drug treatment (cured, not cured). A monkey types the complete works of Shakespeare (success, failure). A typical experiment involving a sequence of Bernoulli trials is as follows.. Suppose you are willing to wait indefinitely for the first success in a sequence of Bernoulli trials. What is the sample space of such an experiment? Let s be success, f be failure. Then the sample space S consists of (s) (f, f,..., f, s) only success is last trial (f, f, f...) infinite sequence of failures. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Waiting for success Problem. Suppose the probability of success is p and failure is p for a sequence of Bernoulli trials. What is the probability of each outcome? Solution. Let E k be the event that the first success is on the k + st trial. The finite outcomes, success on the k + st trial, are E k {(f, f,..., f, s)} Since the trials are independent k failures f, first success s P (E k ) p( p) k. Waiting for success Let E be the event that there is eventually some success. So, E E k. The event of no success is E c {(f, f, f,...)}. Since the events E 0, E,... are mutually exclusive, So, P (E) P ( ) E k P (E k ) p( p) k p ( p) P (E c ) P (E) 0. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, /

5 Back to the Monkey The complete works of Shakespeare consists of L symbols (spaces, letters, punctuation, etc.). A monkey beating on the keys at random has exceeding small, but nonzero probability p of typing a sequence of L symbols which is exactly the complete works of Shakespeare. Let each trial consist of L symbols banged-out by the Monkey. Success in a trial occurs when the monkey has typed the complete works of Shakespeare in the trial. The probability of eventually succeeding is one, since the only other outcome, the infinite sequence of failures, (f, f, f,...), has probability zero. is played with a pair of dice. The player (or shooter") rolls once: If 7 or show, she wins, if 2, 3 or 2 shows she loses, and if any other number shows (the gambler s point"), she must keep rolling the dice until she gets a 7 before her point appears (she loses), or her point appears before the 7 (she wins). What is the probability of winning at craps? Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Sample space The sample space S consists of the following sequences: (7), (), (2), (3), (2) (d, a 2, a 3,..., a n, d) where d 7 and a i d, 7 (d, a 2, a 3,..., a n, 7) where d 7 and a i d, 7 (d, a 2, a 3,...) where d 7 and a i d, 7 Let E d be the event that the first throw is d and the gambler eventually wins. Let E d,n be the event that the first roll is d and the gambler wins on the nth throw. (So, d 4, 5, 6, 8, 9, 0). Computing the odds Let d 4. The probability that the gambler rolls 4 and wins on the nth throw, the outcome (4, a 2, a 3,..., a n, 4): P (E 4,n ) ( 3 36 )2 ( )n 2 The probability that the gambler wins some time (when she threw a 4 on the first toss): P (E 4 ) P ( E 4,n ) ( 3 36 )2 ( )k. This is a geometric series, n2 P (E 4 ) ( 3 36 )2 ( 27 ( 3 36 ) )k 38. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, /

6 Computing the odds Computing the odds The probability of rolling a 4 or 0 are 3 36, so P (E 4 ) P (E 0 ) ( 3 36 )2 ( )k 38 The probability of rolling a 5 or 9 are 4 36, so P (E 5 ) P (E 9 ) ( 4 36 )2 ( )k 2 45 The probability of winning at black jack is The probability of throwing a 7 or on the first throw is The probability of rolling a 6 or 8 are 5 36, so P (E 6 ) P (E 8 ) ( 5 36 )2 ( )k Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / continued The ideas of independence and conditioning are remarkably effective at working together to provide neat solutions to a wide range of problems. A coin shows a head with probability p, or a tail with probability p. It is flipped repeatedly until the first head appears. What is the probability that a heads appears on an even number of tosses? Let E be the event of heads eventually appearing on an even toss. By the partition rule, conditioning on the outcome of the first toss (H or T ): since is odd, P(E H) 0. P (E) P(E H) P (H) + P(E T ) P (T ) 0 p + P(E T ) ( p), What about P(E T )? The key is that we now need an odd number of tosses to succeed. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, /

7 continued continued Let O be the event that the first head is on odd throw, and N be that no head is tossed. But, so Thus, P (E c ) P (O) + P (N) P (O) + 0 P (O). P (O) P(E T ), P(E T ) P (E). P (E) P(E T ) ( p) ( P (E)) ( p) P (E) ( P (E)) ( p) Let q p, and solve for P (E): P (E) ( P (E)) q P (E) q + q p + ( p) p 2 p. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / continued P (E) p 2 p. Consider a fair coin (p 2 ). The probability of tossing heads on an even throw is P (E) 3 The probability of tossing heads on an odd throw is Does this seem right? P (O) Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Method 2 Method 2. Let E n (n 0) be the event that heads first appears on the 2nth toss. Let q p. P (E n ) pq 2n pq ( q 2) n Since the events E, E 2,... are mutually exclusive, P (E) P (E n ) n0 pq ( q 2) n n0 pq p( p) q2 ( p) 2 p 2 p Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, /

8 A problem of Huygens : Huygen s problem Two players, A and B, take turns at throwing dice; each needs some score to win. If one player does not throw the required score, the play continues with the next person throwing. At each of their attempts A wins with probability α and B wins with probability β. (a) What is the probability that A wins if A throws first? (b) What is the probability that A wins if A throws second? Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / A problem of Huygens Solution to Huygen s problem Let p be the probability A wins when A has the first throw, and p 2 be the probability A wins when B has the first throw. By conditioning on the outcome of the first throw, when A is first, p α + ( α)p 2. When B is first, conditioning on the first throw gives (If B throws their score, A loses.) Solving this pair gives p 2 ( β)p. p p 2 α α + β αβ ( β)α α + β αβ Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / A problem of Huygens A problem of Huygens Suppose A and B are tossing a fair coin, and each needs a head. A throws first. (Here, α β 2 ). The probability that A wins is p The problem is equivalent to tossing a heads on an odd throw. The probability that B wins is p 2 2 p 3 Suppose A and B are throwing a pair of dice. A needs a 5 and B a 7. (Here, α 2 and β 6 ). The probability that A wins if A throws first is p The probability that A wins if A throws second is p p The problem is equivalent to tossing a heads on an even throw. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, / Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 2 February 4, /

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