Probability Theory: Counting in Terms of ProportionsA Probability DistributionThe Descendants Of AdamSlide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Unbiased Binomial Distribution On n+1 Elements.As the number of elements gets larger, the shape of the unbiased binomial distribution converges to a Normal (or Gaussian) distribution.Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Again, a Normal or Gaussian distribution! Let’s look after n stepsSlide 26Slide 27Slide 28Slide 29Probabilities and countingSlide 31How many n-bit strings have an even number of 1’s?Slide 33Slide 34Slide 35Slide 36Slide 37How many n-trit strings have even number of 0’s?Some puzzlesSlide 41Actually, 6 and 7 are equally likelyAnother viewSlide 44Slide 45Slide 46So, sometimes, probabilities can be counter-intuitiveLanguage Of ProbabilityFinite Probability DistributionSlide 50Sample spaceProbabilityProbability DistributionEventsSlide 55Slide 56Uniform DistributionSlide 58Slide 59Slide 60Using the LanguageUsing the Language: visuallySlide 63Slide 64PictureSlide 66Slide 67Slide 68Slide 69Slide 70Slide 71Another way to calculate Pr(no collision)More Language Of ProbabilitySlide 74Slide 75Independence!Slide 77Slide 78Slide 79Slide 80Monty Hall problemSlide 82why was this tricky?Random walks and electrical networksSlide 85Random walks come up all the timeProbability Theory:Counting in Terms of Proportions Great Theoretical Ideas In Computer ScienceSteven Rudich, Anupam Gupta CS 15-251 Spring 2005Lecture 20 March 24, 2005 Carnegie Mellon UniversityA Probability DistributionHEIGHTProportion of MALES50% of the male population75% of the male population50% of the male populationThe Descendants Of AdamAdam was X inches tall.He had two sonsOne was X+1 inches tallOne was X-1 inches tallEach of his sons had two sons …XX-1X+111X-2X+2XX-3X+3X-1 X+1X-4X+4X-2X+2X156655101015201X-1X+111X-2X+2XX-3X+3X-1 X+1X-4X+4X-2X+2X1566551010152011111X-2X+2XX-3X+3X-1 X+1X-4X+4X-2X+2X1566551010152011111112X-3X+3X-1 X+1X-4X+4X-2X+2X1566551010152011111112113 3X-4X+4X-2X+2X1566551010152011111112113 3114461566551010152011111112113 31144615665510101520In nth generation, there will be 2n males, each with one of n+1 different heights: h0< h1 < . . .< hn. hi = (X – n + 2i) occurs with proportionUnbiased Binomial Distribution On n+1 Elements.Let S be any set {h0, h1, …, hn} where each element hi has an associated probabilityAny such distribution is called an Unbiased Binomial Distribution or an Unbiased Bernoulli Distribution.As the number of elements gets larger, the shape of the unbiased binomial distribution converges to a Normal (or Gaussian) distribution.MeanAs the number of elements gets larger, the shape of the unbiased binomial distribution converges to a Normal (or Gaussian) distribution.¡2mm¡t¢¡2mm¢≈e¡(t2=m)11446Coin Flipping in ManhattanAt each step, we flip a coin to decide which way to go.What is the probability of ending at the intersection ofAvenue i and Street (n-i) after n steps?11446Coin Flipping in ManhattanAt each step, we flip a coin to decide which way to go.What is the probability of ending at the intersection ofAvenue i and Street (n-i) after n steps?Coin Flipping in ManhattanAt each step, we flip a coin to decide which way to go.What is the probability of ending at the intersection ofAvenue i and Street (n-i) after n steps?11446Coin Flipping in ManhattanAt each step, we flip a coin to decide which way to go.What is the probability of ending at the intersection ofAvenue i and Street (n-i) after n steps?11446Coin Flipping in ManhattanAt each step, we flip a coin to decide which way to go.What is the probability of ending at the intersection ofAvenue i and Street (n-i) after n steps?11446Coin Flipping in Manhattan2n different paths to level n, each equally likely. The probability of i heads occurring on the path we generate is: 11446n-step Random Walk on a lineStart at the origin: at each point, flip an unbiased coin to decide whether to go right or left. The probability that, in n steps, we take i steps to the right and n-i to the left (so we are at position 2i-n) is: 11446n-step Random Walk on a lineStart at the origin: at each point, flip an unbiased coin to decide whether to go right or left. The probability that, in n steps, we take i steps to the right and n-i to the left (so we are at position 2i-n) is: 11446n-step Random Walk on a lineStart at the origin: at each point, flip an unbiased coin to decide whether to go right or left. The probability that, in n steps, we take i steps to the right and n-i to the left (so we are at position 2i-n) is: 11446n-step Random Walk on a lineStart at the origin: at each point, flip an unbiased coin to decide whether to go right or left. The probability that, in n steps, we take i steps to the right and n-i to the left (so we are at position 2i-n) is: 11446n-step Random Walk on a lineStart at the origin: at each point, flip an unbiased coin to decide whether to go right or left. The probability that, in n steps, we take i steps to the right and n-i to the left (so we are at position 2i-n) is: 11446Again, a Normal or Gaussian distribution!Let’s look after n steps23pn50% oftimeAgain, a Normal or Gaussian distribution!Let’s look after n steps68% oftimepnAgain, a Normal or Gaussian distribution!Let’s look after n steps95% oftime2pnAgain, a Normal or Gaussian distribution!Let’s look after n steps99.7% oftime3pnProbabilities and Countingare intimately related ideas…Probabilities and countingSay we want to count the number of X's with property P One way to do it is to ask "if we pick an X at random, what is the probability it has property P?" and then multiply by the number of X's. (# of X with prop. P)(total # of X)Probability of X with prop. P =Probabilities and countingSay we want to count the number of X's with property P One way to do it is to ask "if we pick an X at random, what is the probability it has property P?" and then multiply by the number of X's. (# of X with prop. P)(total # of X)Probability of X with prop. P =×How many n-bit strings have an even number of 1’s?If you flip a coin n times, what is the probability you get an even number of heads? Then multiply by 2n.Say prob was q after n-1 flips.Then, after n flips it is ½q + ½(1-q) = ½.(# of X with prop. P)(total # of X)Probability of X with prop. P =×11446Binomial distribution with bias p Start at the top. At each
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