Reference no: EM132843727
Let's suppose for a moment that the hot hand model is valid for Kobe.
During his career, the percentage of time Kobe makes a basket (i.e. his shooting percentage) is about 45%, or in probability notation,
P(shot 1 = H)=0.45
P(shot 1 = H)=0.45
If he makes the first shot and has a hot hand (not independent shots), then the probability that he makes his second shot would go up to, let's say, 60%,
P(shot 2 = H|shot 1 = H)=0.60
P(shot 2 = H|shot 1 = H)=0.60
As a result of these increased probabilites, you'd expect Kobe to have longer streaks. Compare this to the skeptical perspective where Kobe does not have a hot hand, where each shot is independent of the next. If he hit his first shot, the probability that he makes the second is still 0.45.
P(shot 2 = H|shot 1 = H)=0.45
P(shot 2 = H|shot 1 = H)=0.45
In other words, making the first shot did nothing to effect the probability that he'd make his second shot. If Kobe's shots are independent, then he'd have the same probability of hitting every shot regardless of his past shots: 45%.
Question:
What is the probability that kobe misses the second shot in a streak, if he has a cold hand and independent probabilities?
What is the probability that kobe misses the second shot in a streak, if he has a hot hand and dependent probabilities?
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