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Determination of probabilities 9/29

Given knowledge.
Boy .513, girl .487. Although symmetry?
Symmetry.
Simulation.

Probability that Stanford will beat Cal?

or What are the odds?.

If you know about betting, you may believe for instance that it is much more likely that Stanford will win, you might accept a ``2 to 1'' odds bet, ie you pay 2 dollars if Stanford wins and you receive 1 dollar if they win. You think the probability of winning is 2/3.

r to 1 odds means that E occurs? This means you beleive it is r times more likely that E occurs than not.


\begin{displaymath}P(E)=rP(E^c) \mbox{ and } P(E)+P(E^c)=1=(r+1)P(E^c)\end{displaymath}

So that $P(E)=\frac{r}{r+1}$.

If we generalise this to an event E whose odds are r to s, written (r:s), then

\begin{displaymath}P(E)=\frac{r/s}{r/s +1}=\frac{r}{r +s}\end{displaymath}

If P(E)=p then the odds in favor of E are: r:s where r/s=p/(1-p).

An extension to property 3 to 3 events is done by writing:

\begin{eqnarray*}P(E\cup F \cup G) &=&P((E\cup F)\cup G)=
P(E\cup F)+P(G)-P(E\cu...
...(FG)-P(EGFG) \}\\
&=& P(E)+P(F)+P(G)-P(EF)- P(EG)-P(FG)+P(EGFG)
\end{eqnarray*}


Inclusion-Exclusion Principle for n events:

\begin{displaymath}P(E_1\cup E_2 \cup \cdots E_n)=
\sum P(E_i)-\sum_{i_1<i_2}P(E...
...i_2}..E_{i_r})
+\ldots
+(-1)^{(n+1)}P(E_{i_1}E_{i_2}..E_{i_n})
\end{displaymath}


next up previous index
Next: Infinite Sample Spaces 9/29 Up: Properties of Probabilities 9/28 Previous: Tree Diagrams
Susan Holmes
1998-12-07