Doubt about Mixed Strategy Nash Equilibrium Here's the video I will be referring too.
Now I am a complete beginner in game theory, so sorry if this sounds like a stupid question, but why would a player want to balance out the payoffs of another player through his strategy.
Like it doesn't make much intuitive sense to me as to why he would want to do that. How exactly is this the best response to the situation?
Edit:
Sorry for not making the question self contained. So the game concerned is the Battle of the Sexes game, and this is the payoff matrix.
Suppose player $i$ plays the mixed strategy $\mathbb{P}_i(B)= p_i$, and assume for now that the support of $\mathbb{P}_i$ is $\{B,F\}$ (i.e. player 1 plays a fully mixed strategy). For both $B$ and $F$ to be in 1's support, he must obtain the same expected payoff from either strategy (otherwise, he would put all the weight on the strategy with the higher payoff!).
Now, the expected utility of player 1 from playing $B$ is:
$\mathbb{E}[u_1(B,.)] = p_2u_1(B,B)+(1-p_2)u_1(B,F) = 2p_2$
Similarly, the expected utility of player 1 from player in $F$ is:
$\mathbb{E}[u_1(F,.)] = p_2u_1(F,B)+(1-p_2)u_1(F,F) = 1-p_2$
The important point to note here is that the expectation is over the actions of player 2 - since player 1 knows the distribution used by player 2 (in equilibrium), but not the realised action.
You can write the best response function of player 1 as follows:
$BR_1(p_2) = \begin{cases}
0 &amp; \text{ if } 2p_2 &lt; 1-p_2 \\
(0,1) &amp; \text{ if } 2p_2 = 1-p_2\\
1 &amp; \text{ if } 2p_2 &gt; 1-p_2 \\
\end{cases}$
Since we assumed that 1 uses a fully mixed strategy, the $BR_1$ function dicates that this can happen only when $2p^*_2 = 1-p^*_2 \implies p^*_2 = \frac{1}{3}$.
In other words, $p^*_2$ is the unique probability that is consistent with player 1 mixing over both her strategies. Whether this forms an equilibrium is still not clear - for that you need to calculate $BR_2$ (using the same steps) and see if $p^*_1\in(0,1)$. In that case, both players are best responding to each other - and hence playing a Nash Equilibrium.
P.S - for instance, if you found out that $p^*_1 = 1$ (i.e. 1 would like to play pure strategy $B$), then our starting assumption is wrong! So we need to redo the calculation for $p^*_2$.