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10 Powerful Ideas

10 Powerful Ideas. Lecture 23. Paradoxes and Payoffs of Probability. A Tale of Two Brothers. Carl and Hal ( cburch@cs.cmu.edu , hburch@cs.cmu.edu ).

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10 Powerful Ideas

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  1. 10 Powerful Ideas Lecture 23 Paradoxes and Payoffs of Probability

  2. A Tale of Two Brothers Carl and Hal (cburch@cs.cmu.edu, hburch@cs.cmu.edu). As competitive as they are similar. Familiar with the sad story of Isaac and Gottfried, these two brothers were determined to avoid a fluke outcome. So instead of competing once a year, they competed once an hour! (This is 876,600 events!) Each brother wins any event with probability 1/2, independent of any other events. Amazingly, Hal was ahead for the last 50 years of his life! Should we have expected this?

  3. A Random Walk Let xkdenote the difference in victories after k events. We can view x0, x1, x2, … , xnas a “random walk” that depends on the outcomes of the individual events.

  4. Theorem: The number of sequences in which xk is never negative is equal to the number that start and end at 0, for even n.

  5. Proof: Put the two sets in one-to-one and onto correspondence.

  6. Lemma: Suppose that Y is an integer random variable in [0,m] such that Then Proof:

  7. Lemma: For odd n,the number of sequences that last touch 0 at position i is equal to the number that last touch 0 at position n-i-1. Proof: Put these sets in one-to-one and onto correspondence.

  8. Theorem: Let Y be the length of the longest suffix that does not touch 0. For odd n, Proof: Y is a random variable in the range [0,n-1]. By previous lemma,

  9. A seemingly contradictory observation: The expected length of the final suffix that does not touch 0 is (n-1)/2, but the expected number of times the sequence touches 0 in the last n/2 steps is

  10. Berthold’s Triangle The number of sequences with k1’s and n-k0’s in which the number of 1’s is greater than or equal to the number of 0’s in every prefix. Examples: 111 110 101

  11. 0 0 0 5 4 1 0 0 0 5 9 5 1 0 0 0 0 14 14 6 1 Notice that the n’th row sums to 1 0 1 0 1 1 0 0 2 1 0 0 2 3 1

  12. Theorem: Proof: By inductionon n. Base cases: Inductive step:

  13. Another strange fact… Let Z be the number of steps before the sequence returns to 0. Let Z’ be the number of steps before the sequence drops below 0. This looks like trouble! E[Z’] isn’t finite!

  14. The moral of this story is... When it comes to expectations, you never know what to expect!

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