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A fair die is thrown repeatedly until a six appears for the first time. Let $A_n$ be the event that a six appears for the first time on the $n$-th throw. Let $B_r$ be the event that a five appears $r$ times before a six appears for the first time.

a) Explain why $$\operatorname{Pr}(B_r)=\sum_{n=r}^{\infty}\operatorname{Pr}(A_{n+1})\operatorname{Pr}(B_r \mid A_{n+1})$$ b) Using a), show that $$\operatorname{Pr}(B_r)=\left(\frac{1}{2}\right)^{r+1}$$

I have tried to expand the summation, but I keep on getting $(\frac{1}{6})^{r}$ and the expansion is something like $$\binom{n}{r} \left(\frac{1}{6}\right)^{r} \frac{1}{6}+ \binom{n+1}{r} \left(\frac{1}{6}\right)^{r} \frac{1}{6} \frac{4}{6}+ \cdot \cdot \cdot$$ but I have no idea how to simplify.

I am totally stuck on b), any help is appreciated thank you.

vitamin d
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popping900
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    An alternative approach is to say if $R$ is the number of fives then using memorylessness and symmetry, $\Pr(R=0)=\frac12$ and $\Pr(R=r\mid R\gt r-1)=\frac12$ – Henry Mar 31 '21 at 08:30

1 Answers1

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Well, I am not sure how you got your summations. But, $Pr(A_{n+1})=(\frac{5}{6})^n\frac{1}{6}$ because the first n throws needs to be non-sixes, and the last is six. On the other hand, $Pr(B_r\mid A_{n+1})=$probability that on 5-sider dice falls on $n$ throws exactly $r$ times number five. Therefore, $Pr(B_r\mid A_{n+1})=(\frac{1}{5})^r(\frac{4}{5})^{n-r}$${n}\choose{r}$ . Together,

$$ \sum Pr(A_{n+1})Pr(B_r\mid A_{n+1}) = \sum(\frac{5}{6})^n\frac{1}{6}(\frac{1}{5})^r(\frac{4}{5})^{n-r}\binom{n}{r} =\sum \binom{n}{r}\frac{4^{n-r}}{6^{n+1}} $$ This is equal to

$$ \frac{1}{6^{r+1}}\sum_{n=r}^{\infty} \binom{n}{r}(\frac{2}{3})^{n-r} = \frac{1}{6^{r+1}}\sum_{i=0}^{\infty} \binom{r+i}{r}(\frac{2}{3})^{i} $$ This sum is indeed equal to $3^{r+1}$ and therefore it is $1/2^{r+1}$ what is your result (using https://www.wolframalpha.com/input/?i=sum+n%3Dr+to+infty+4%5E%28n-r%29%2F6%5E%28n%2B1%29+\*+%28n%21%29%2F%28%28n-r%29%21*r%21%29 )

Albert Paradek
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