# How infinite Nash equilibria are possible in a game?

I was studying games when one of the players seems to be indifferent between two or more pure strategies because he gets the same payoff with each strategy. We say that there are infinite Nash equillibria in the game but I am unable to calculate it and get an intuitive idea of how is it happening. For example in the following game $$\begin{array}{c|a|c} & L_2 & R_2& \\ \hline L_1 & (3,1) & (0,1) \\ \hline R_1 & (0,1) & (4,1) \\ \hline \end{array}$$

Seemingly, P2 should not even bother about anything even if he is playing pure strategy or randomizing. But here if we assume that P2 plays L with probability q and P1 plays L with probability p then making them indifferent we see that we cannot calculate p but q comes out to be 4/7. How is it possible or what is the interpretation ? If I am doing it wrong then what is the correct way to determine equilibria in this case ?

Another example is

$$\begin{array}{c|a|c} & L_2 & R_2& \\ \hline L_1 & (1,1) & (0,0) \\ \hline R_1 & (0,0) & (0,0) \\ \hline \end{array}$$

How many Nash's equilibria are possible in this ?

The second game has two equilibria, $(L_1, L_2)$ and $(R_1,R_2)$. It is straightforward to note that these two are Nash equilibria. On the other hand, if one player plays $R_i$ with probability strictly greater than 0, then it is in the other's best interest to play $L_j$ with probability 1. This last observation shows that no other equilibrium exits.
Finally, there is nothing wrong with having infinitely many equilibrium points. Your simplest example is $$\begin{array}{|c|c|c|} \hline & L_2 & R_2 \\ \hline L_1 & 0,0 & 0,0 \\ \hline R_1 & 0,0 & 0,0 \\ \hline \end{array}$$ for which every strategy profile is a Nash equilibrium.