Nash equilibrium: Can I delete weakly dominated strategies in this case?

Question

As far as I know, an equilibrium can involve a weakly dominated strategy, but cannot involve a strictly dominated strategy. Is there a general rule for when/if you can safely delete a weakly dominated strategy?

In order to better illustrate my question, I will work through an example.

• The pure strategy nash equilibria are $(m,c)$ and $(d,l)$.

In order to proceed with finding mixed strategies, I am going to delete $\mathcal r$ and $\mathcal u$ as they seem to be weakly dominated.

After solving a few simple linear equations:

$\left[ \pi^{*},\rho^{*},\sigma^{*},\tau^{*} \right]$=$\left[ 0,\frac{7}{17},\frac{4}{9},\frac{5}{9} \right]$

$\:$

Question: Is my approach valid?

Answer 1

In general, if you eliminate weakly dominated strategies you might miss the equilibrium (think of a case where the row player always have the same payoff (hence every row is weakly dominated)).

Answer 2

I think the general idea is to consider separately the cases when weakly dominated strategy gives exactly the same payoff as the dominating one.

In your case: $l$ weakly dominates $r$. If $p_m<1$ then $l$ is strictly better than $r$, so we need to consider special case $p_m=1$. But if $p_m=1$ it is better to play $c$ for the second player. So $r$ may be safely removed.