How do you explain the "closer to success" paradox? Let's use tossing a coin as an example. If you toss a coin and get a head, then it is considered a success. If you have to toss it 5 times and finally get a head, that is also considered a success.
Now let's say, you tossed it 10 times, and all 10 times, it was a tail.
Now person A claimed: well, the 11th time, the 12th time, each time is independent, so at the 11th time you toss it, it is just likely that you get a head, or a tail. This is very logical, and understandable.
But person B claimed: to toss a coin 11 times and still get a tail and cannot succeed, the probability is 1/2048, which is 0.049%. And to toss it the 12th time and still do not succeed, it is 1 / 4096, which is 0.024%. Now in fact, your probability of success of at least one time getting a head is 99.976%. So try a few more time, it is almost guaranteed that you will succeed. You are getting closer to success, because your chance of still failing is getting closer and closer to infinitely small which is like a feather going through a wall which is possible in quantum mechanics. This argument is very logical and understandable as well.
In fact, let's say, if you toss it 20 times, the probability of success is 99.9999%, and people will start to say, toss it one more time, and if it is still tail, it is getting hard to believe, I'd have to think it is either magic or something supernatural going on. But person A will say, no, it is just quite likely (50% chance) that 21 times, and it is still a tail.
So argument 1 is: you are not getting closer to success and argument 2 is: you are getting closer to success. How can you explain why statement 1 and 2 both sound logical and true?
Update: maybe I can add a Person C:
Person C tossed it one time, and got a tail, and he said, "I am not going to try any more, because many people say that no matter how many times the coin is tossed, it is not getting closer to success, so I may as well stop now." So this sounds wrong, as people say, try again and you are closer to success, but it sounds true for the people who say, you are not getting closer to success.
 A: The possibility of throwing TTTTTTTTTTT is indeed $1/2048$ (if measured before you start throwing).
But the probability of throwing TTTTTTTTTTH is also $1/2048$.
The remaining $2046/2048$ of probability are cases where the first ten throws were not all tails. You already know you're not in one of those cases.
A: Person A is correct. No matter what the previous coin flips are, if you have not succeeded yet, you still have a 50% chance of succeeding on the next flip.
The fallacy that person B has made is called the gambler's fallacy, that prior outcomes should be "balanced out" by future outcomes. The probability that the coin landed tails the first 10 times is $\frac1{1024}$, but that does not affect the future flips in any way – they might as well be ignored.
A: The point is that Person B has selected a very, very specific (and wrong) branch of possibility-space in order to make their judgement: Person B has not conditioned on the past when attempting to predict the future, but has instead done something crazy.
Both people have noted that the past contained nine tails, but Person A has (correctly) conditioned the tenth throw on the information from the past, whereas Person B has taken the conjunction of the probabilities instead. When phrased like this, Person B's mistake is unfathomably and bafflingly wrong: they have not followed the correct procedure for predicting the future based on the past (instead, they followed the procedure for working out whether two things will happen together in the future, ignoring the past entirely), so of course their answer is nonsense.
A: Let's take $3$ coin tosses to demonstrate what is going on.
There is a total of $8$ possible outcomes:
$$\{TTT,TTH,THT,HTT,THH,HTH,HHT,HHH\}.$$
From this we can observe that there is only $1/8$ chance that $H$ won't appear at least once. Thus, before we begin the tosses, we know that there is $7/8$ chance that we will succeed.
Now, observe the following table:
$$\begin{array}{c|c|c}
1 & 2 & 3\\ \hline
\bf\color{red}T & \bf\color{red}T & T\\
\bf\color{red}T & \bf\color{red}T & H\\ \hline
\bf\color{blue}T & \bf\color{blue}H & T\\ 
\bf\color{blue}T & \bf\color{blue}H & H\\ \hline
\bf\color{green}H & \bf\color{green}T & T\\ 
\bf\color{green}H & \bf\color{green}T & H\\ \hline
\bf\color{yellow}H & \bf\color{yellow}H & T\\ 
\bf\color{yellow}H & \bf\color{yellow}H & H\\
\end{array}$$
We have listed all the possible outcomes of $3$ coin tosses. However, after we've completed the first two tosses, not the whole table is relevant for the third toss, we are now either in red, blue, green or yellow case. Whatever the color, for the third toss the probability is $1/2$. 
Imagine that it's not the case, that in the red case probability for heads is $7/8$. Let's apply the same logic to the yellow case. Probability of getting three heads in $1/8$, so in the yellow case the probability of heads should be $1/8$.
How does that make sense? The coin should have some kind of "memory" of its previous states. However, the key mathematical term here is independence of each toss. Coin does not know whether it's in the red, blue, green or yellow case, and it won't be affected by it for the third toss. Each row in the table has probability of $1/8$ and thus, events of the same color are equally likely, i.e. the probability is $1/2$ for heads or tails on the third toss.
In conclusion, even though before we begin there was $7/8$ chance to succeed, after we observed two failures ($TT$), our chances to succeed dropped to only $1/2$.
