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 
$$\begin{array}{c|a|c} &amp; L_2 &amp; R_2&amp; \\ \hline L_1 &amp; (3,1) &amp; (0,1) \\ \hline R_1 &amp; (0,1) &amp; (4,1) \\ \hline \end{array}$$                 

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 

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

How many Nash's equilibria are possible in this ?
Well the intuition is quite straightforward: as you have mentioned yourself, player 2 is indifferent between playing any action in pure strategies, or randomizing in any possible way. So, for a mixed strategy equilibrium to exist, player 2 needs to play L w/ probability 4/7. If this is the case, it is irrelevant what Player 1 player---player 2's outcome is the same. As a result, in pure strategies the Equilibria are L,L and R,R and, in Mixed strategies, q=4/7 and p can take any value between 0 and 1. Hence, there exist infinite possible Nash Equilibria (p just has to obey the fundamental laws of probability).