Is this algebraic expression true for q+p=1?

  • MHB
  • Thread starter Poirot1
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In summary, the conversation begins with a mathematical equation and question regarding the equal values of the numerators. One participant suggests that the denominators are irrelevant and assigns values to prove their point. However, the other participant states that the values used are not applicable and the equation is not true. The conversation then turns to discussing the context of the problem, which is related to the gambler's ruin problem. Ultimately, it is determined that the probability of A winning the game is equal to the right hand side of the equation, while the probability of B being ruined is equal to the left hand side.
  • #1
Poirot1
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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
 
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  • #2
Since denominators are the same, WHY do you bother showing them?
(if a/x = b/x, then a = b)
 
  • #3
you are right but that is irrevelant, I have tried to show numerators are equal and have failed.
 
  • #4
Poirot said:
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!
 
  • #5
I perhaps should have mentioned that p and q are probabilities, so must be between 0 and 1. Does that make a difference?
 
  • #6
Poirot said:
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:
  • #7
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.
 
  • #8
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.
 
  • #9
Wilmer said:
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 receive 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.
 
  • #10
Poirot said:
Yes that is why I deleted it immediately (so why are you bringing it up)?. How did you receive that?
I am not at liberty to divulge such, since the incident is presently under investigation.
 
  • #11
By the thought police?
 
  • #12
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').
 

FAQ: Is this algebraic expression true for q+p=1?

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