---
title: "Red bus, blue bus"
output: pdf_document
bibliography: chess.bib
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```

The red bus-blue bus problem was originally stated by
@davidson1970.  The other one was the holiday problem.

A traveller has a choice of going to work by car or taking a bus.  The
bus may be red or blue.  Here, we assume that the traveller has strong
(but unknown) views on cars vs buses.  If considering a car or bus,
the colour of the bus is of no importance to him.  
However, when offered the choice between a red bus or a blue bus, the
colour becomes important (having nothing else to guide his decision).

A sensible likelihood function is shown in figure \ref{likefuncred}.

\begin{table}[h]
\centering
\caption{Likelihoods\label{likefuncred} for red bus, blue bus problem}
\begin{tabular}{|l|l|}
\hline
Choice set & likelihood  \\ \hline
$\left\lbrace C,RB  \right\rbrace$ & $\left(\frac{p_C   }{p_C+p_{B}   }\right)^a \left(\frac{p_{B}}{p_C+p_{B}}\right)^b $ \\ \hline
$\left\lbrace C,BB  \right\rbrace$ & $\left(\frac{p_C   }{p_C+p_{B}   }\right)^c \left(\frac{p_{B}}{p_C+p_{B}}\right)^d $ \\ \hline
$\left\lbrace RB,BB \right\rbrace$ & $\left(\frac{1}{2}\right)^e 
                                      \left(\frac{1}{2}\right)^f
$ \\ \hline
$\left\lbrace C,RB,BB\right\rbrace$ & $
\left(\frac{p_C}{p_C + 2p_{B}}\right)^g
\left(\frac{p_B}{p_C + 2p_{B}}\right)^h
\left(\frac{p_B}{p_C + 2p_{B}}\right)^i
$\\ \hline
\end{tabular}
\end{table}

\begin{table}[h]
\centering
\caption{Likelihoods\label{likefuncredghost} for red bus, blue bus problem}
\begin{tabular}{|l|l|}
\hline
Choice set & likelihood  \\ \hline
$\left\lbrace C,RB  \right\rbrace$ & $\left(\frac{p_C   }{p_C+p_{B}   }\right)^a \left(\frac{p_{B}}{p_C+p_{B}}\right)^b $ \\ \hline
$\left\lbrace C,BB  \right\rbrace$ & $\left(\frac{p_C   }{p_C+p_{B}   }\right)^c \left(\frac{p_{B}}{p_C+p_{B}}\right)^d $ \\ \hline
$\left\lbrace RB,BB \right\rbrace$ & $\left(\frac{p_{B}+p_{RG}}{2p_{B}+p_{RG}}\right)^e 
                                      \left(\frac{p_{B}       }{2p_{B}+p_{RG}}\right)^f
$ \\ \hline
$\left\lbrace C,RB,BB\right\rbrace$ & $
\left(\frac{p_C       }{p_C + 2p_{B}+p_{RG}}\right)^g
\left(\frac{p_B+p_{RG}}{p_C + 2p_{B}+p_{RG}}\right)^h
\left(\frac{p_B       }{p_C + 2p_{B}+p_{RG}}\right)^i
$\\ \hline
\end{tabular}
\end{table}



 @debreu1960 offers:
 
Let the set U have the following three elements: $D_c$, a recording of
the Debussy quartet by the $C$ quartet, $B_F$, a recording of the
eighth symphony of Beethoven by the $B$ orchestra conducted by $F$,
$B_K$, a recording of the eighth symphony of Beethoven by the $B$
orchestra conducted by $K$.  The subject will be presented with a
subset of $U$, will be asked to choose an element in that subset, and
will listen to the recording he has chosen.  When presented with
$\left\lbrace D_c, B_F\right\rbrace$, he chooses $D_c$ with
probability $\frac{3}{5}$.  When presented with $\lbrace BF,
BK\rbrace$ he chooses $B_F$ with probability 1/2. When presented with
$\left\lbrace DC, BK\right\rbrace$ he chooses $D_c$ with probability
$\frac{3}{5}$.  What happens if he is presented with $\lbrace D_c,
B_F, B_K\rbrace$?  According to the axiom, he must choose $D_c$ with
probability 3/7.  Thus if he can choose between $D_c$ and $B_F$, he
would rather have Debussy.  However if he can choose between $D_c$,
$B_F$, and $B_K$, while being indifferent between $B_F$ and $B_K$, he
would rather have Beethoven.



THERE IS ALSO THE THING WITH THE HOLIDAY IN CALIFORNIA AND THE HOLIDAY
IN FLORIDA AND THE HOLIDAY IN FLORIDA PLUS TEN BUCKS


```{r loadlib}
library("hyper2")
2+4
```



# References
