[Northern boy]

here's my probability tree, roll with me.
note: P denotes "probability".
also, Each event in the tree is independent of each other.


Stock(X) trades at 2$. If pay-off A = B, than instinctively P(A)=P(B)....



Now, Stock (X) trades at either 3$ or 1$. Pay-off C ≠ D, than instinctively P(C) ≠ P(D) if the game is fare.

But if the market's development in (P)A = (P)B , and the directions are inverse of each other...

Than shouldn't the directions that are equally inverse to each other in C and D mean (P)C = (P)D?

That would imply that:

Pay-off A = B, and (P)A = (P)B

but

Pay-off C < D, and (P)C = (P)D




If this is wrong, please tell me why (P)C > (P)D

but

if this is right, than WHAT THE HELL ARE WE DOING???????????????????? ????????




(lastly, I thought one might say that starting point C is > than starting point A. But finishing point A is > than finishing point C. So although trajectories are inverse, the resistance they both meet are equal,

and P(A) = P(B) = P(C) = P(D) )

[Head2k]

I am a simple guy and not educated in statistics and probability at all. But I guess that P(C) > P(D) because although the absolute price change is the same, in case C the price declines 33%, while in case D it needs to advance 100%.

[Northern boy]

Nevermind, I'm an ass. C and D are irrelevant.... lol I'm going to bed.

[Northern boy]

this is actually only relevant to a particular approach and useless in general, so if a moderator could delete this thread that'd be cool