The importance of the numbers depends on time discounting and what solution concepts you'd like to use. Your intuition is good.

You mention tit for tat. This and other punishment strategies exist to ensure cooperation. However, you may be interested in finding punishment strategies that are credible, introducing refinements to the Nash equilibrium concept common in the economic literature. Credibility results don't necessarily hinge on the gain from defecting.

In your example, tit for tat is credible.

Here's my work showing why tit for tat is credible.

Let there be a discount factor $\delta\in(0,1)$. We can consider the average payoffs to player 1 in the four states given the tit for tat protocol:

When both players cooperate, player 1 earns $V_{1}(w_{CC})=100$.

When both defect, $V_{1}(w_{DD})=1$.

When only 1 defects, $V_{1}(w_{DC})=(1-\delta)101+\delta V_{1}(w_{CD})$.

When only 2 defects, $V_{1}(w_{CD})=(1- \delta )0+\delta V_{1}(w_{DC})$.

Then, $V_{1}(w_{DC})=\frac{101}{1+\delta}$ and $V_{1}(w_{CD})=\frac{\delta101}{1+\delta}$.

Then, EE is a Nash eq. in the state CC if $100\geq(1-\delta)101+\delta\frac{\delta101}{1+\delta}$, requiring $\delta\geq\frac{1}{100}$.

Now we check that once one player has deviated, they will find it in their interest to continue playing tit for tat. We compare average payoffs from different strategies.

So in states CD and DC, cooperating is a best response if $\frac{\delta101}{1+\delta}>(1-\delta)1+\delta=1$. This is satisfied if $\delta\geq\frac{1}{100}$

And defecting is a best response if $\frac{101}{1+\delta}\geq(1-\delta)3+\delta3=3,$ which is satisfied for any $\delta$ in $(0,1)$.

Following in this pattern, you would see that tit for tat is not credible for a game with payoffs T=4, R=3, S=-1, P=1.

For an academic reference, see

Mailath, G. J., & Samuelson, L. (2006). Repeated games and reputations: long-run relationships. OUP Catalogue.

The importance of the numbers depends on time discounting and what solution concepts you'd like to use. Your intuition is good.

You mention tit for tat. This and other punishment strategies exist to ensure cooperation. However, you may be interested in finding punishment strategies that are credible, introducing refinements to the Nash equilibrium concept common in the economic literature. Credibility results don't necessarily hinge on the gain from defecting.

In your example, tit for tat is credible.

Here's my work showing why tit for tat is credible.

Let there be a discount factor $\delta\in(0,1)$. We can consider the average payoffs to player 1 in the four states given the tit for tat protocol:

When both players cooperate, player 1 earns $V_{1}(w_{CC})=100$.

When both defect, $V_{1}(w_{DD})=1$.

When only 1 defects, $V_{1}(w_{DC})=(1-\delta)101+\delta V_{1}(w_{CD})$.

When only 2 defects, $V_{1}(w_{CD})=(1- \delta )0+\delta V_{1}(w_{DC})$.

Then, $V_{1}(w_{DC})=\frac{101}{1+\delta}$ and $V_{1}(w_{CD})=\frac{\delta101}{1+\delta}$.

Then, CC is a Nash eq. in the state CC if $100\geq(1-\delta)101+\delta\frac{\delta101}{1+\delta}$, requiring $\delta\geq\frac{1}{100}$.

Now we check that once one player has deviated, they will find it in their interest to continue playing tit for tat. We compare average payoffs from different strategies.

So in states CD and DC, cooperating is a best response if $\frac{\delta101}{1+\delta}>(1-\delta)1+\delta=1$. This is satisfied if $\delta\geq\frac{1}{100}$

And defecting is a best response if $\frac{101}{1+\delta}\geq(1-\delta)3+\delta3=3,$ which is satisfied for any $\delta$ in $(0,1)$.

Following in this pattern, you would see that tit for tat is not credible for a game with payoffs T=4, R=3, S=-1, P=1.

For an academic reference, see

Mailath, G. J., & Samuelson, L. (2006). Repeated games and reputations: long-run relationships. OUP Catalogue.