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

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Discussion Overview

The discussion revolves around the algebraic expression involving probabilities \( p \) and \( q \) where \( q + p = 1 \). Participants are attempting to show the equality of two expressions under specific conditions, particularly focusing on the implications of \( p \) and \( q \) being probabilities constrained between 0 and 1. The context includes references to the gambler's ruin problem.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents the expression to be proven and requests assistance in demonstrating its validity under the condition \( q + p = 1 \).
  • Another participant questions the necessity of showing the denominators are equal, suggesting that if the denominators are the same, the numerators must also be equal.
  • A participant expresses frustration in proving the equality of the numerators and claims that the problem may be irrelevant.
  • One participant assigns specific values to \( p \), \( q \), \( k \), and \( a \) to illustrate a counterexample, indicating that the expression does not hold true in that case.
  • Another participant notes that \( p \) and \( q \) being probabilities (between 0 and 1) may affect the validity of the expression.
  • It is mentioned that the expression can be equal if \( p = q = \frac{1}{2} \) or if \( a = k \).
  • A participant provides a specific case with values for \( p \) and \( q \) and concludes that the expression is not true under those conditions.
  • One participant suggests factoring the numerator and indicates that for the equality to hold, \( \left(\frac{q}{p}\right)^{k-a} \) must equal 1, implying a potential typo in the original question.
  • Context is provided regarding the gambler's ruin problem, linking the variables to the probabilities of winning and losing in a game scenario.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the algebraic expression, with some providing counterexamples and others suggesting conditions under which the expression might hold true. The discussion remains unresolved, with no consensus on the correctness of the expression.

Contextual Notes

Participants note that \( p \) and \( q \) must be probabilities, which introduces constraints that may affect the validity of the expression. There are also references to specific values that lead to conflicting conclusions about the expression's truth.

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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Since denominators are the same, WHY do you bother showing them?
(if a/x = b/x, then a = b)
 
you are right but that is irrevelant, I have tried to show numerators are equal and have failed.
 
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!
 
I perhaps should have mentioned that p and q are probabilities, so must be between 0 and 1. Does that make a difference?
 
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:
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 2^{k-a}- 2^k= 1- 2^a. And, if k=2, a= 1, that says 2- 4= 2 which is certainly not true. You cannot prove what you want, it is not true.
 
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.
 
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').
 

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