What is the probability of getting r heads before s tails?

[Eclair_de_XII]

So is it $\frac{p}{q}(\frac{1-q^{s-1}}{p})=\frac{1-q^{s-1}}{q}$, then...?

 

[StoneTemplePython]

So is it $\frac{p}{q}(\frac{1-q^{s-1}}{p})=\frac{1-q^{s-1}}{q}$, then...?

I'm not sure how you're getting to this? You should be able to telescope this quite easily. I wrote out the derivation in a couple of lines but I decided not to post it yet. You need to show some work here.

 

[Eclair_de_XII]

This is how I'm getting it: $\sum_{k=1}^{s-1} pq^{k-1}=\frac{p}{q}\sum_{k=1}^{s-1} q^{k}=\frac{p}{q}(\frac{1-q^{s-1}}{1-q})=\frac{p}{q}(\frac{1-q^{s-1}}{p})$.

 

[StoneTemplePython]

This is how I'm getting it: $\sum_{k=1}^{s-1} pq^{k-1}=\frac{p}{q}\sum_{k=1}^{s-1} q^{k}=\frac{p}{q}(\frac{1-q^{s-1}}{1-q})=\frac{p}{q}(\frac{1-q^{s-1}}{p})$.

Two fundamental issues:
1.) You still have not shown your work for the telescoping process
2.) $\sum_{k=1}^{s-1} q^{k}\neq (\frac{1-q^{s-1}}{1-q})$
this obviously cannot be true. Selecting $q = 0$ doesn't work because $0 \neq 1$. Now select some small $\delta \gt 0$ and set $q := \delta$. The left hand side is approximately 0 while the right hand side is approximately 1. This is not an equality.

 

[Eclair_de_XII]

I don't know what you mean by telescoping process. Do you mean that the finite sum is a telescoping series?

 

[StoneTemplePython]

I don't know what you mean by telescoping process. Do you mean that the finite sum is a telescoping series?

I mean show your work such that you telescope a series with $s-1$ terms in it and convert it into a fraction that has much less terms. I have not seen your work here. My sense is you're just guessing or otherwise skipping steps.

 

[StoneTemplePython]

it's killing me here because your post 25 and 27 are so close and I think you're within shouting distance of the finish line. All I'm getting at is

we have $p \in (0,1)$ and $q = 1-p$, referencing the left side of post 30

$y:= \sum_{k=1}^{s-1} pq^{k-1} = p\big( 1+ q + q^2 + ... q^{s-2}\big)$

I created $y$ for convenience. What you actually want is $y \cdot P(E|H_1) = P(E|H_1)\cdot p\big( 1+ q + q^2 + ... q^{s-2}\big)$, so we'll just multiply by $P(E|H_1)$ at the end.

the magic of telescoping is, we find some pattern in the sum and exploit it.

In this case, multiply each side by $(1-q)$ noting that $(1-q) \gt 0$, so you get

$(1-q)y = (1-q) p\big( 1+ q + q^2 + ... q^{s-2}\big) = p\big( 1 - q^{s-1}\big)$

divide each side by $(1-q)$ and get

$y =\frac{ p\big( 1 - q^{s-1}\big)}{1-q} = \frac{ p\big( 1 - q^{s-1}\big)}{p}= 1 - q^{s-1}$

so you what you're looking for is
$P(E|T_1) = y \cdot P(E|H_1) = P(E|H_1) \big(1 - q^{s-1}\big)$
- - - -
If you swap this into your second line on post 16, and solve the system of equations good things should happen...

 

[Eclair_de_XII]

First of all, I'm so sorry for being so clueless about finite sums and geometric series; Calculus II was so very long ago for me, and from what I remember of the class, I covered mostly infinite series. Anyway:

$P(E|H_1)+(p^{r-1}-1)P(E|T_1)=p^{r-1}$
$(q^{s-1}-1)P(E|H_1)+P(E|T_1)=0$

$A= \begin{pmatrix} 1 & (p^{r-1}-1) \\ (q^{s-1}-1) & 1 \\ \end{pmatrix}$

$det(A)=1+(1-q^{s-1})(p^{r-1}-1)$

$A^{-1} = \frac{1}{det(A)}\begin{pmatrix} 1 & (1-p^{r-1}) \\ (1-q^{s-1}) & 1 \\ \end{pmatrix}$

$\begin{pmatrix} P(E|H_1) \\ P(E|T_1) \\ \end{pmatrix} =\frac{1}{det(A)} \begin{pmatrix} 1 & (1-p^{r-1}) \\ (1-q^{s-1}) & 1 \\ \end{pmatrix} \begin{pmatrix} p^{r-1} \\ 0 \\ \end{pmatrix} =\frac{1}{1+(1-q^{s-1})(p^{r-1}-1)}\begin{pmatrix} p^{r-1} \\ p^{r-1}(1-q^{s-1}) \\ \end{pmatrix}$

So $P(E)=\frac{p^r+qp^{r-1}(1-q^{s-1})}{1+(1-q^{s-1})(p^{r-1}-1)}$