= Finite-horizon invasion threshold of Tit for tat against unconditional defection
{title2=$\epsilon_c=(P-S)/[L(R-P)-(T+S-2P)]$}
In an $L$-round <iterated prisoner's dilemma>, <Tit for tat> against <always defect> gets $S$ once and $P$ thereafter; the defector gets $T$ once and $P$ thereafter. Their self-payoffs are $R,P$. At introduced frequency $\epsilon$, subtracting mixed-population <payoffs> gives $\epsilon[(R-P)-(T+S-2P)/L]-(P-S)/L$. The displayed positive threshold applies when its denominator is positive and the threshold is less than one. Taking $L\to\infty$ first removes this finite entry cost, so the horizon and rare-mutant <limits> cannot be casually interchanged.
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