(Resolved) Computing Odds

[applescript]

I seem to have forgotten everything how to compute complex odds, sad b/c I took college math classes just two years ago.

Anywho, here's what I want to fiqure out how to do...

If the odds of Event A are x1 in y1, and the events of Event B are x2 in y2, how can I fiqure out what the odds are that Event A will occur before Event B?

Event A = x1 in y1
Event B = x2 in y2

Event A is less likely to occur than Event B.

How do you fiqure out what the odds are that Event A will occur before Event B?

[krtxmrtz]

What are y1 and y2?

[applescript]

For the purposes of discussion, let's say that:

Event A = 6 in 36 (1 in 6)
Event B = 3 in 36 (1 in 12)

[Ecniv]

um this is a guess...

a:1 in 6
b:1 in 12

b has 5/12 chances - cause on the sixth go the 'a' will get it.


But, I was thinking of factorials for some reason.

1/6 1/5 1/4 1/3 1/2
1/12 1/11 1/10 1/9 1/8


Dunno... I took Stats ten years ago and failed (E at A level standard - pure n stats)


Vince

[wossname]

I bet Carol Vordermann would know. :drool:

[opus]

My Stat class wa also long time ago, but i'll give it a try:

Event A has probability p(A) (from example 1/6)
Event B has probability p(B) (from example 1/12)
Event C is not having A or B, it has probability p(C)

p(a)+p(B)+p(c)=1

The prob that A happens before B in the first two draws is p(A).

The prob that A happens before B in the first 3 draws is:
first draw (A or C are possible) p(A)+p(C)
second draw (A or C are possible if A happened in first draw, A must be drawn if C happened in first draw) p(A)+p(C) ; p(A)
Total prob p(A)*(p(A)+p(C))+p(C)*p(A).

For the prob in case of four draws you can continue yourself.