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

Assign values, say q=2, p=-1, k=2, a=1

left numerator = -6, right numerator = 3

So STOP!

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Equal (both = 0) if p = q = 1/2I perhaps should have mentioned that p and q are probabilities, so must be between 0 and 1. Does that make a difference?

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.

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- Jan 26, 2012

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

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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Yes that is why I deleted it immediately (so why are you bringing it up)?. How did you recieve that?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.

---------- 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.

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

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