Math and Logic Puzzles: Redux

[Dessew]

Hey, haven't been to this site in ages :D

I wanted to ask my crush out for a date, but her brother is very protective of her, and gave me a condition I first have to fulfil unless I wish him to give me a purple nurple. I'll play 2n+1 games of chess (he hasn't decided exactly how many yet, but he'll tell me before we start) against him and his sister, always alternating. So for example if I first play against my crush, then I'll have to play against the brother next, then against her and so on, and in this case my last possible game would be against her. Only when I prove myself to be a colossal nerd by winning two games in a row, am I allowed to ask his sister out.

Now, she's very smart, and I can't even figure this problem out on my own, so my chances of winning against my crush are slim. Unfortunately, I can't tell her to lose against me on purpose, though, because then both my nurples will be purple, and that's where the limits of my ever-lasting love lie. On the other hand, her brother is a brute, and beating him should be a walk in the park.

Can you help me? Is there a way for me to maximise the probability of my crush's brother allowing me to ask his sister out? He's already letting me always play with the white pieces.

[Dessew]

These are very creative solutions, but if the brother thinks I'm cheating or up to some trickery, he'll purple the nurple till I burple the purple. That wouldn't be a pretty sight.

HINT 1:

This is a maths problem.

HINT 2:

Perhaps there's an apect of these series of games that hasn't been explicitly specified yet, and where one option gives me a clear mathematical advantage over the other(s).

[Dessew]

And yes, the games are played sequentially.

[Dessew]

Thanks to your help, managed to win two games in a row, and asked my crush out for a date. As it turns out, she hates my guts, and wants nothing to do with me. She broke my legs and my heart.

Spoiler:

For the argumentation you don't necessarily need induction (in fact, I'm not sure if that works). Consider that the difference of the probabilities you succeed with the teo different orders is some kind of (not constant) polynomial of p and q that's monotonic in both variables, and the only time the order doesn't matter is when p=q. So by continuity arguments you can set p and q to anything as long as their relative values are correct.

EDIT:

Small correction: I don't think the polynomial is monotonic.

[Dessew]

Let's try another one. I generate a random number p between 0 and 1 with uniform distribution, then I mint a coin such that the probability that it lands on heads when I toss it is p. What is the probability that I get exactly as many tails as heads if I toss the coin 2n times?

[Dessew]

Yes, correct!

Spoiler:

Alternatively, we can also observe that a coinflip with probability p is equivalent to taking the indicator function of a random number uniformly generated between 0 and 1 being less than p. So we have 2n+1 iid random numbers, and we want to know if the first generated is in the middle. (We don't get pairwise distinct numbers with 0 probability, so there's no need to elaborate on that any further.