What's wrong with this coin?

Question

Probability theory

Once, at an interview, I was asked the following problem: there is a crooked coin, it falls on heads with a probability of 0.49, and on tails with a probability of 0.51, how to use such a coin to conduct a fair draw, that is, to get two events, the probability of which is the same. Of course, I am not a big expert in the field, but the solution was obvious: in a full group of two throws, the probability of a series of OR and RO is the same, that is, we throw twice and wait for something to appear first.
I think everything is clear with the task itself, but after about a year (that is, now) I accidentally remembered her and thought about her answer. So I specifically wrote "twice", is it really that important "by"? That is, should we throw a coin exactly in series of two throws, or can we just throw it, writing down all the results, and the sequence of events that will meet first in the general row and be victorious? Again, my knowledge tells me that yes, of course, all this about "has no memory" obviously indicates that it does not matter at all how we throw it, and the result of the PPO determines the winner of the person who chose the PO. And if my reasoning up to this point is correct, then my intuition tells me that with a given advantage towards tails, a person with an RO event is obviously more likely to win..
Let's make the scatter stronger, 1 out of 100 for heads and 99 out of 100 for tails, the probability of occurrence of OR and RO events is set to be very small, but still equal. Now almost every toss will be a tail, and with a probability of 0.99 we will start our streak with a tail. The probability of falling heads is very small, but we will have to wait for it, because we need to draw lots, and someday the heads will still fall, but with a 99 percent probability it will be after tails, right? I understand that with such an interpretation, the conditional probability already seems to be gone, but after all, there is no memory for a coin, can I consider these cases separately? That is, I tossed a coin twice, the second toss is heads, but this heads does not depend on the first toss, which has a tails with a probability of 0.99.
It is clear that I am mistaken in something, in fact, this is the question - what is it? What laws explain what is happening?

Answer 1

It seems like a trivial solution: toss a coin an even number of times, changing the winning side by odd or even?

Answer 2

I think so,
we throw the first time:
if the Eagle - the winner is determined. less likely.
if Tails - we throw further, because most likely she fell out because she has a higher probability, and it would be unfair to declare her the winner.
We throw the second time (it means that Tails fell on the first throw):
if the Eagle - the winner is determined, because. the probability of his appearance in 2 throws = 0.49 * 2= 0.98, and he has already appeared whole.
if it's tails - here's the question, should we keep throwing or is the tails already winning?
And here I think tails is already winning, and the third roll would make sense if the probability of tails was > 0. (6), the fourth roll would make sense when the probability of tails > 0.75, etc.
Here you need to understand that the more likely it is for any of the parties to fall out, the more throws may be required to determine the winner.