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Odds of hitting a home run is $80\%$.

You get $1$ at-bat per day.

How long can you expect it will it take to finally get a streak of $30$ home runs?

Does this relate to geometric probability?

barak manos
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JackOfAll
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  • If each bat is independent of the other, then getting a streak of 30 home-runs has a probability of .8^30, and the number of attempts at bat you've had before that is irrelevant. – Jonathan Hebert Mar 13 '15 at 13:48
  • @JonathanHebert: OP is asking about the expectancy, not the probability. – barak manos Mar 13 '15 at 13:49
  • Ok, so P(30 HR's) = .8^30 --> E(x) = 1/p --> so E(x) = $\frac{1}{.8^{30}}$ = 807 days ?? – JackOfAll Mar 13 '15 at 16:32
  • Call $t_n$ the mean time to get a streak of $n$ homeruns, then $t_1=1$ and, for every $n\geqslant1$, $$t_{n+1}=t_n+1+(1-p)t_{n+1},$$ with $p=.8$. Thus, for every $n\geqslant1$, $$t_n=\frac1{p^n}\frac{p(2-p)}{1-p}-\frac1{1-p}.$$ This proof does not require Mathematica. The result is not $1/p^n$. For instance, $$t_{30}\approx3872.$$ – Did Mar 14 '15 at 14:49
  • A new twist on the site: OP posts question, OP posts wrong answer to own question, OP accepts wrong answer. Convenient. – Did Mar 17 '15 at 21:56
  • Can you explain (or tell me what topic to research) to understand your method? – JackOfAll Mar 19 '15 at 01:05
  • Did, is this the same answer as what you formulated? http://math.stackexchange.com/questions/1218810/pbowling-a-strike-70-expected-number-of-trials-until-a-perfect-game-10-s – JackOfAll Apr 06 '15 at 16:04

1 Answers1

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$P(30 HRs) = .8^{30} $

$E(x) = \frac{1}{p} = \frac{1}{.8^{30}} = 807 days $

JackOfAll
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