How do we understand 6 people trying something is not 6 times the success rate? Let's say if a task has a success rate of $20\%$, or $0.2$, meaning if a person tries it, then there is a $20\%$ chance he can succeed.
One example is, if we generate a random number from 1 to 10, and getting the number 9 or 10 is considered to be a success.
Now, if we let 6 people try it, and one person succeeding is considered a success, we cannot say the success rate is $6$ times as much, because then the success rate is $20\% \times 6 = 120\%$, and probability cannot be greater than $100\%$.  So the success rate is not 6 times as much.
However, if we let 1 person try it $1,000,000$ times, the Law of Large Numbers says that the number of times he will succeed is $200,000$. And if we let 6 people try $1,000,000$ times each, then the number of success is indeed $200,000 \times 6 = 1,200,000$ which is $6$ times.  How can we understand this?
In a real life example, say, each time when we catch a Pokemon, let's say there is a special type of Pokemon that when you tap on it, it can be "shiny", and the probability is $1/256$.  Now if one player try to tap on $300$ Pokemon, the probability of getting at least one shiny is not $1$, but less than $1$.  If we let 6 people, each try to tap on $300$ Pokemon (and a Pokemon can be non-shiny for player 1 but is shiny for player 2, meaning it is independent), then the probability of getting at least one shiny is not $6$ times. Now, however, if we let all 6 players, each tap on $3,000,000$ Pokemon, then the number of shiny Pokemon they will get is in fact $6$ times  if we only allow 1 player to play.  How can we understand this "6 times yes and no" dilemma?
 A: The probability at least one person succeeds out of $6$ equals $1$ minus the probability that all of the $6$ fail. So if the success rate is $p$, then the probability at least one person succeeds out of $n$ people is $1-(1-p)^n$. 
Going to your example of $20$%  success and $6$ people, we get $1-(1-0.20)^6=0.739$.
Probabilities aren't cumulative; only expected values are.
A: The probabilities work because there is a chance that more than one person is successful at the same time, even though there is also a chance that none are successful. The average number of successes for six people is six times the average for one person, but this average covers the case where all succeed at the same time (for example) as well as the cases where two out of the six succeed.
A: If six people each try $1,000,000$ times, the total number of success is approximate $1,200,000.$  The success rate is approximately $$
\frac{1200000}{6000000}=.2$$
You seem to have overlooked the fact that there are six million trials.
A: I think that the easiest way for you to grasp where the 'missing' successes went is actually mostly in multiple successes.
If one event has 0.2 probability of success, twice the event has probabilities:


*

*0.64 -> no success at all (0.8*0.8)

*0.04 -> two successes (0.2*0.2)

*0.32 -> one success


Therefore, relatively high probabilities and small number of trials means success rate is not at all being proportional to number of trials.
On the other hand, if you have a very low probability of success, then probability of at least one success is approximately proportional to  number of trials.
The graph below maybe illustrates this better than my words, but since probability of success is very low (for instance, 0.4%) then probability of two or more successes is A LOT lower (0.0016%, 0.000064%, etc., so basically negligible) so if several experiments are repeated total success rates will closely match the number of times, on average that 'one' success occurs.
Probability for at least one success with probability of 0.004, for 1 to 1000 tries (straight approximation, actual curve):

A: I think the situation is easier to see in a simpler example. Drop a coin, "heads" is success. The probability of success is 50%. This does not mean that if you drop the coin three times your probability of getting one head is 150%.
The probability of getting at least one head is one minus the probability of getting three tails, so it's  $1-1/8=7/8$. In percentage, that would be around 87%. 
A: The problem is that you cannot add probabilities when dealing with events that are not "disjoint". What you can do is multiply them, but only if the events are independent.
So, in your example, 6 people are given 300 attempts each to catch a pokemon each. First assume these are independent of each other (i.e. whatever one person does, that doesn't affect anybody else's chances of catching a pokemon). Now, the probability that one particular person catches at least 1 pokemon is about 70 % if each attempt has a 1/256 probability of success. 
So say you want the probability that each person catches at least one pokemon. Then you can't say 6*70 %, but you can multiply 70% with itself 6 times. That gives you about 11 %. 
A: The problem is your third paragraph, where you've confused the expected number of successes in 6 tries (1.2) with a percentage chance (120%). This is one reason whey probability students are encouraged to work in decimals/fractions instead of percentages.
If you let six people try it, you can expect 1.2 successes on average. If you're looking for the probability that at least one succeeds out of 6, you have (as noted elsewhere) {1-P(all six fail)} which would be {1-0.8^6}.
A: You've currently phrased this is terms of whether the probability is multiplicative (does having six times the number of trials give six times the probability of success), but we can equivalently ask whether it's additive (is the probability of success over two trials equal to the sum of probabilities for each individual trial).
The reason it's not additive is that on the first flip, everyone is eligible to get a success. If you have 20 people flipping coins, then all of them could get heads, and on average you're going to have 10 of them getting a success.
But on the second flip, while everyone has a chance of getting heads, only the ones who got tails are eligible to be new successes. If someone gets heads the first time, then you've already counted them, and counting them if they get heads the second time is double-counting.
For the number of successes to be additive, you would have to have the same number of new successes each trial. But you don't: to get the number of new successes, you have to multiply the probability of getting a success on that trial by the number of people who haven't gotten a success yet. 
In your case, after one trial, 20% will get a success on the first trial, so 80% won't have a success after one trial. During the second trial, 20% will get a success, but only 80% of them will be new successes. So after the first trial, you'll have 20% success from the first trial, plus 20%*80% from the second trial, giving a total of 36% success, rather than the 40% you would expect from taking 20% and multiplying it by 2. The missing 4% represents people who got successes both times, but should be counted only once.
If you work through the math, you'll find that the percentage of people who don't have any successes after $n$ trials will be $(1-p)^n$, and thus the percentage with at least one success will be $1-(1-p)^n$.
