Doubt about Mixed Strategy Nash Equilibrium

Question

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.

user28372 · Accepted Answer

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:
$BR_1(p_2) = begin{cases}
0 & text{ if } 2p_2 < 1-p_2 \
(0,1) & text{ if } 2p_2 = 1-p_2\
1 & text{ if } 2p_2 > 1-p_2 \
end{cases}$
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 $p^*_1in(0,1)$. 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 $p^*_2$.

pegasus · Answer

Notice that if a player is indifferent between two strategies, she gets the same payoff from either strategy. This means that anything is optimal (a best response): Playing either pure strategy or any mixed strategy. This includes the mixed strategy which makes her opponent indifferent.
Of course, if the player is not indifferent, she will choose either one pure strategy or the other. But this cannot be the case in a Nash equilibrium where both players use mixed strategies.

Henry · Answer

The question being investigated by the video is the existence of Nash equilibria, not the optimal choices by the players.
There are two obvious pure Nash equilibrium joint strategies, namely both play B or both play F, since in either case a deviation from the strategy by one of the players brings  a negative expected effect for that play is the other goes on with the strategy.
The question being addressed is whether there is also a mixed Nash equilibrium joint strategy.  It will be a Nash equilibrium if neither player can improve their own outcome by changing strategy while the other's strategy stays the same.  So if a player decides to find a mixed Nash equilibrium (rather than the more normal maximising expectation), then that player's approach is to find a way of making the result indifferent to the other player's strategy.
That is the answer to your question.  But it is not a particular good approach for that player from the start. The outcome is an expected gain of  $frac23$ for each player, while the coordinated strategies Nash equilibria give $2$ or $1$.  The one merit of this mixed strategy Nash equilibrium is that it reduces the risk of opposing strategies, which could be worse.

Giskard · Answer

why would a player want to balance out the payoffs of another player

I don't think anyone is saying that a player wants to do this. But in mixed equilibrium their strategy is such that this property holds. Without this property, any mixed strategy of the other player would be suboptimal.

Doubt about Mixed Strategy Nash Equilibrium

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