---
title: "Very simplified likelihood"
author: "Robin Hankin"
output: bookdown::html_document2
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
library("hyper2")
library("pracma")
library("magrittr")
```

```{r hexsticker, out.width='20%', out.extra='style="float:right; padding:10px"',echo=FALSE}
knitr::include_graphics(system.file("help/figures/hyper2.png", package = "hyper2"))
```

This follows on from `very_simplified_likelihood.Rmd`.  Consider some
form of comparison between a competitor of Bradley-Terry strength $a$
and $n$ competitors each of strength $b$; we require $a+b=1$.  NB we
have $n+1$ competitors in total, 1 of strength $a$ and $n$ of strength
$b$.  An observation is indexed by $r$, the number of $b$ clones
finishing ahead of $a$, so $0\leqslant r\leqslant n$.  The initial
field strength is $a+nb=1+b(n-1)$.


Suppose we split the $n+1$ competitors into two teams.  The team with
our bro has $r+1$ members and the other one has size $n-r$ members.
Then, using standard notation

\begin{equation}
\frac{a+rb;(n-r)b}{a+nb}
\end{equation}

Let us suppose that the teams are of equal size:  $1+r=n-r=m$, say:

\begin{equation}=
\frac{a+(m-1)b;mb}{a+(2m-1)b}=\frac{mb (a-b); mb}{2mb+(a-b)}=\frac{(m-2)b+1;mb}{(2m-2)b+1}
\end{equation}

Let's try $m=15$:

```{r}
m <- 15
a <- seq(from=0,to=1,by=0.001)
b <- 1-a
L_win <- ((m-2)*b+1)/((2*m-2)*b+1)
L_lose<- (m*b)      /((2*m-2)*b+1)
plot(a,L_win ,type='b',ylim=0:1,main="Focal competitor's team wins")
plot(a,L_lose,type='b',ylim=0:1,main="Focal competitor's team loses")
S_win <- log(L_win)
S_win <- S_win-max(S_win)
S_lose <- log(L_lose)
S_lose <- S_lose-max(S_lose)
plot(a,S_win,type='b',main="Win")
plot(a,S_lose,type='b',main="Lose")
logodds <- log(a/(1-a))
plot(logodds,S_win,main="win")
plot(logodds,S_lose,main="lose")
```


Now in week 4 Laura came in the second team of four (of four members
each).  I get the teams's strengths are $\beta=4b$ and
$\alpha=3b+a=1+2b$, so

\begin{eqnarray}
L &=& \frac{\alpha}{3\alpha+\beta}\cdot\frac{\beta}{2\alpha+\beta}\\
  &=& \frac{4b}{12b + (1+2b)}\cdot\frac{1+2b}{8b+(1+2b)}\\
  &=& \frac{4b}{1+14b}\cdot\frac{1+2b}{1+10b}
\end{eqnarray}


```{r plotsecondteam}
a <- seq(from=0,to=1,by=0.001)

f <- function(a){
    b <- 1-a    
    4*b*(1+2*b)/(1+14*b)/(1+10*b)
}
L <- f(a)
L <- L/max(L,na.rm=TRUE)
plot(a, L,type='b')
plot(a, log(L),type='b',ylim=c(-5,0))
abline(h= -2)
```
