I'm struggling to make sense of a statement in a paper. This is from McKelvey & Palfrey (1992) paper on the Centipede game. It is a two player game in which a player can choose either TAKE (which stops the game) or PASS in which case the game continues. The game has 6 rounds (each player plays 3 times). The proportion of rational players is given by $q$. The likelihood that a rational player chooses TAKE in round $n$ is given by $p_n(q)$.
Now we are interested in computing the likelihood of observing all possible outcomes T, PT, PPT, PPPT, PPPPT, PPPPPT, PPPPPP. This is how the authors describe it:
Sentence from McKelvey & Palfrey (1992) paper
They give $s_1$ and $s_2$ explicitly. What are $s_3$ ... $s_6$? I fail to understand the logic of $s_2$ to deduce what $s_3$ and following should be.
Take $s_{1}=qp_{1}$ (I am dropping dependence on $q$ which is understood). This is probability of $T$. You have two cases to consider. The first (red) player is altruist with probability $1-q$ and then chooses $P$. Or the red player is selfish, with $q$, and then chooses $T$ with $p_{1}$.
Take $s_{2}=q^2(1-p_{1})p_{2}+q(1-q)p_{2}$. This is probability of $PT$. You have four cases to consider.
Case 1: two altruists, hence $(1-q)^2$. Then the probability of $PT$ is zero.
Case 2: two selfish, hence $q^2$. Then the probability of $PT$ is $(1-p_{1})p_{2}$.
Case 3: red altruist, blue selfish, hence $q(1-q)$. Then the probability of $PT$ is $p_{2}$.
Case 4: red selfish, blue altruist, hence $q(1-q)$. Then the probability of $PT$ is $(1-p_{1})0$.
Does this suffice for you to calculate $s_{3}$ and beyond?