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I have a series of problems that involve events that happen one after another, and where the usual formulas like $P(\frac{A}{B})=\frac{P(AB)}{P(A)}$ don't seem to help.


In a horse race there are $3$ horses. $E12$ means horse $1$ finishes before horse $2$. $E123$ means horse $1$ finishes before $2$, which finishes before $3,$ etc.

If $P(E13)=2/3$ and $P(E23)=1/2$ then $P(E123)=?$

How can I go about solving this kind of problems?

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@GeorgeS Please refrain from petty unnecessary edits on 4+ years old posts – Did May 11 '15 at 15:57
Sure. I was checking out older questions and it was bothering me. – George Simpson May 11 '15 at 15:58
up vote 4 down vote accepted

This is not a conditional probability question. Your particular example has insufficient information to solve it. Here $E123$ is the event of the horses finishing in the order 1, 2, 3. So (assuming the horses can't tie for a position) $$P(E13)=P(E132)+P(E123)+P(E213),$$ $$P(E23)=P(E123)+P(E213)+P(E231)$$ and of course $$1=P(E123)+P(E132)+P(E213)+P(E231)+P(E312)+P(E321).$$ Knowing $P(E13)$ and $P(E23)$ isn't enough to determine any one of the $P(Eijk)$s.

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