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A and B take alternate turns at kicking a football. Their probability of scoring the goal are $p_1$ and $p_2$ respectively. The first scorer allows the second one more kick. If the other scores, the game is drawn otherwise the first kicker wins. A begins the game.

If $p_1=p_2=1/3$, find the expected values of kicks in the game.

Can somebody please tell me how to find the expected values of kicks using a geometric summation?

And why is number of kicks = $1/p +1$?(in this case) Shouldn't it just be $1/p$ since it's geometric distribution?

mathnoob123
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  • What have you tried? As for the mean of a geometric distribution, there are multiple definitions for the geometric distribution; depending on which one you choose, you'll get $\frac{1}{p}$ or $\frac{1}{p}-1$ for the mean. – Batman May 21 '17 at 19:52
  • @Batman The basic idea is integrating xf(x) where f(x) is the probability density function. Here I can't figure out that function. – mathnoob123 May 21 '17 at 19:55
  • The +1 in the expected number of kicks is due to the first scorer allowing one more kick. Thus if $X$ is geometric, the number of kicks is $N = X+1$. – knrumsey May 21 '17 at 20:20
  • Got it. can you also tell me how to solve this by setting up a geometric summation? – mathnoob123 May 21 '17 at 20:22

1 Answers1

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Maybe it is because it's the number of success before the first failure: you have to add the failure bace to the denominator. I also find this confusing.

Thunder
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