This question was previously asked in Probability of $5$ fair coin flips having strictly more heads than $4$ fair coin flips
I know that the answer is 0.5, but I'm not following OP's and the accepted answer's logic.
Using the notation that André Nicolas used, there is a probability, $p$, that player 1 or player 2 wins if each player has 4 tosses (this can be generalized to N tosses), so the probability of tying is thus $q=1-2p$. In case anyone's curious, for 4 tosses, $p=0.36328125$, and the chance of tying is $q=0.2734375$
Consider the situation where the players are tied after each player has tossed the coin 4 times. Say player 2 is the one who gets a 5th toss. Allowing a 5th toss will reduce the chance of a tie, and it will increase player 2's chance to win. This is intuitive and obvious.
I do not understand how the answer from the old post used the formula
$$ p + \frac{1}{2}(1 - 2p) = \frac{1}{2} $$
Could someone explain to me what this is doing?
This to me this is saying "with the addition of the 5th toss, we are reducing the chance of tying by half $\frac{1-2p}{2}$ and increasing player 2's probability of winning by the same." But the chance of tying does not become cut in half with the addition of the 5th toss. The chance of tying is $0.25390625$, about $93\%$ that of $q$.