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I am perplexed by an exercise problem I came across yesterday:

In a sequence of consecutive throws of a die, find the probability that six will show before a one or a two.

A.1/6

B.1/2

C.2/3

D.5/6

E.1/3

I am quite confused about this problem.I think A(if I denote the event in the problem as A) is sorta kinda complex, by which I mean, for example, it's probable that both 612 and 12612 are in A.I don't know how to solve this. My question is:if I change the problem into “the first emerging 6 is before the first emerging 1 or 2”, what is the answer?And if I consider the more complex problem I mentioned above,how should I fix it? Thanks a lot!

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The answer is simple. Look at the first time, one of the numbers $1,2,6$ occurs. Then it is clear that the probability of a $6$ is $\frac {1}{3}$. It does not matter that we do not know how many throws are necessary. The correct answer is therefore E.

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  • $\begingroup$ I just upvoted a nice answer which was deleted at the same time. $\endgroup$ – Peter Aug 30 '14 at 9:57
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Less elegantly, the probability that none of 1, 2, 6 occur on a particular throw is $\frac{1}{2}$. So the probability that we have $n$ throws with none of 1, 2, 6 occurring, followed by a 6 on the $n+1$st throw is $\frac{1}{2^n}\frac{1}{6}$. So the chance that a 6 occurs first is $\frac{1}{6}(1+\frac{1}{2}+\frac{1}{4}+\dots)=\frac{1}{3}$.

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