Pointless probability

Question

POINTLESS is a BBC game show. Each night, four teams compete. If a team does not win, it comes back for a second night; but not a third night. Each night has 1, 2, 3 or 4 new teams. There are never 0 new teams because the previous night's winner does not come back.
1. What is the chance that all 4 teams are new?
2. If I replace 4 by $n$, what is the chance that all $n$ teams are new?
My Try
There are $n$ states, given by the number of new teams.
They form a linear Markov chain, with two states adjacent if they add up to either $n$ or $n+1$. If it weren't for the winner, new teams would become old; and old teams replaced by new, so they add to $n$; if a new team won, there is an extra new team.
All the nights with $n$ new teams are preceded by nights with 1 new team, but only one-in-$n$ of the nights with $1$ new team are followed by a night with $n$ new teams - when the single new team wins. So $1$ new team must happen $n$ times as often as $n$ new teams.
Similar arguments at each transition show that states $n,1,n-1,2,n-2,3,...$ appear with relative frequency $$1:n:n(n-1):{n\choose2}(n-1):{n\choose2}{n-1\choose2}:...$$ So state $m$ appears with relative frequency ${n\choose m}{n-1\choose n-m}$. The sum from $1$ to $n$ is clearly ${2n-1\choose n}$. Comment: Is there a simpler reason that $2n-1\choose n$ appears?
Question 1: What if an old team is $k$ times as likely to win as a new team?
Question 2: What if teams stayed for up to three nights unless they won?

Answer 1

We can use a markov chain to compute the long range probabilities.
Let A = 4 new, B = 3 new, C = 2 new, D = 1 new

The transition matrix is

$\;\;\;\; A\quad B\quad C\;\quad D$
$A\;\frac04\quad\frac04\;\quad\frac04\quad\;\frac44$

$B\;\frac04\quad\frac04\;\quad\frac34\quad\;\frac14$

$C\;\frac04\quad\frac24\;\quad\frac24\quad\;\frac04$

$D\;\frac14\quad\frac34\;\quad\frac04\quad\;\frac04$

The long range corresponding probabilities, $a,b,c,d$ can be worked out from

a = d/4,
b = c/2 + 3d/4,
c = 3b/4 + c/2,
d = a+b/4,
a+b+c+d = 1,

which yields $a = \frac1{35}$, but please check the numericals !

You can work out similarly for any other number of competitors after forming the appropriate transition matrix.

Answer 2

Answering my own question:
When old teams are $k$ times more likely to win, the relative likelihood of $m$ new teams is $$k{n-1\choose m-1}{n-1\choose m}+{n-1\choose m-1}^2$$ except for $k=0$. (When $k=0$, the number of new teams alternates each night.) The denominator of these likelihoods is $k{2n-2\choose n}+{2n-2\choose n-1}$. For example, with 4 teams, likelihoods are $$\frac{3k+1}{15k+20},\frac{9k+9}{15k+20},\frac{3k+9}{15k+20},\frac1{15k+20}$$

The results for $3$-night teams are not so neat. With two teams, one is new; likelihoods for the age of the other team are $2/7,4/7,1/7$. With 4 teams, there are ten age possibilities, with relative likelihoods $$9, 352, 188, 740, 1848, 274, 144, 1096, 752, 36$$