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[SOLVED] Show that question

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Poirot

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Feb 15, 2012
250
show that $\frac{(\frac{q}{p})^{k-a}-(\frac{q}{p})^k}{1-(\frac{q}{p})^k}=\frac{1-(\frac{q}{p})^a}{1-(\frac{q}{p})^k}$
where q+p=1

Thanks
 

Wilmer

In Memoriam
Mar 19, 2012
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Since denominators are the same, WHY do you bother showing them?
(if a/x = b/x, then a = b)
 
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Poirot

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Feb 15, 2012
250
you are right but that is irrevelant, I have tried to show numerators are equal and have failed.
 

Wilmer

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Mar 19, 2012
376
you are right but that is irrevelant, I have tried to show numerators are equal and have failed.
Looks to me like your whole problem is irrelevant !

Assign values, say q=2, p=-1, k=2, a=1
left numerator = -6, right numerator = 3
So STOP!
 
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Poirot

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Feb 15, 2012
250
I perhaps should have mentioned that p and q are probabilities, so must be between 0 and 1. Does that make a difference?
 

Wilmer

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Mar 19, 2012
376
I perhaps should have mentioned that p and q are probabilities, so must be between 0 and 1. Does that make a difference?
Equal (both = 0) if p = q = 1/2
Also equal if a = k, of course.

Sir Poirot, just noticed I received this compliment from you:
"Have I not stated that p+q=1, which you, in your eagerness to be contemptous, have ignored."
May I humbly defend myself by reminding you thay my post says: q=2 and p=-1 : 2 + (-1) = 1.
 
Last edited:

HallsofIvy

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MHB Math Helper
Jan 29, 2012
1,151
If q= 2/3, p= 1/3, so that p and q are both between 0 and 2 and add to 1, the numerator becomes [tex]2^{k-a}- 2^k= 1- 2^a[/tex]. And, if k=2, a= 1, that says [tex]2- 4= 2[/tex] which is certainly not true. You cannot prove what you want, it is not true.
 

Jester

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MHB Math Helper
Jan 26, 2012
183
Just an added note. If you factor the numerator then you have

$\left(\dfrac{q}{p}\right)^{k-a}\left(1 - \left(\dfrac{q}{p}\right)^a\right)$

Comparing with the right hand side give that

$\left(\dfrac{q}{p}\right)^{k-a}=1$ for your equality to be true. I suggest that you check the question again. There might be a typo somewhere.
 
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Poirot

Banned
Feb 15, 2012
250
Equal (both = 0) if p = q = 1/2
Also equal if a = k, of course.

Sir Poirot, just noticed I received this compliment from you:
"Have I not stated that p+q=1, which you, in your eagerness to be contemptous, have ignored."
May I humbly defend myself by reminding you thay my post says: q=2 and p=-1 : 2 + (-1) = 1.
Yes that is why I deleted it immediately (so why are you bringing it up)?. How did you recieve that?

---------- Post added at 15:30 ---------- Previous post was at 15:26 ----------

If it helps. the context is gambler's ruin problem. If you are familiar, then p is the probability of moving up one, q is the probability of moving down one, 'a' is the inital money of player A, and b=k-a is the inital money of B. I am trying to derive the probability of A winning the game, which is apparently the RHS of my equation. The probability B is ruined is the LHS so I thought they would be the same.
 

Wilmer

In Memoriam
Mar 19, 2012
376
Yes that is why I deleted it immediately (so why are you bringing it up)?. How did you recieve that?
I am not at liberty to divulge such, since the incident is presently under investigation.
 

Poirot

Banned
Feb 15, 2012
250
By the thought police?
 

Poirot

Banned
Feb 15, 2012
250
Wilmer, surely you are allowed to say why you brought it up. You could see I had deleted it. (by the way who on earth is 'Sir Poirot').