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?

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    "In a real life example, say, each time when we catch a Pokemon..." :D I was not expecting that. – Sam Jul 23 '18 at 14:52
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    This is a false statement: *"However, if we let 1 person try it 1,000,000 times, the Law of Large Numbers in fact says that the number of times he will succeed is 200,000."* The Law of Large Numbers says that the number will *probably be close* to 200,000, which is a totally different fact. – Daniel R. Collins Jul 23 '18 at 15:56
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    @DanielR.Collins Right, the probability that it will be exactly 200,000 is pretty low. – JimmyJames Jul 23 '18 at 16:00
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    @JimmyJames: Indeed, around 0.1% chance using a normal approximation to the binomial distribution. – Daniel R. Collins Jul 23 '18 at 16:18
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    @DanielR.Collins This is one of the most counter-intuitive things about statistics. When I've tried to explain to otherwise knowledgeable people that the chance you will end up even in a 50-50 odds game actually decreases the more you play, I get the sense that people think I'm some sort of crank. – JimmyJames Jul 23 '18 at 17:26
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    I have an issue with your "Real Life" example. You are comparing "chance of at least 1 shiny" against "Expected Value of Shinyness". These are not comparible. To see why, scale it down: at a 1-in 2 chance tapping 3 Pokémon, there is a 7 in 8 chance (87.5%) of getting at least 1 shiny, but if you have 6-people tapping then you tap 18 Pokémon, giving a 262,143 in 262,144 chance (99.9996%) chance of at least 1 shiny - but your "expected value" of shinies increases from 1.5 shinies to 9 shinies (6-times higher) – Chronocidal Jul 23 '18 at 22:40
  • Do you stop counting on the first success, or do you count 6 successes as 6 successes? That makes all the difference. – user253751 Jul 23 '18 at 22:41
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    “Six times the number of successes” ≠ “six times the success **rate**” – ShreevatsaR Jul 24 '18 at 02:38

9 Answers9


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.

Mark Bennet
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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.

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    The last line of this definitely needs emphasising. There are 6-times as many *trials*, no matter whether it 6 people trying 1 million times or 1 person trying 6 million times - so, there are 6-times as many chances for success than with only 1 million trials. – Chronocidal Jul 23 '18 at 22:26

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%.

Martin Argerami
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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.

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  • I know how to calculate this, but I just feel weird that if 6 people do it, the number of successful outcome is indeed 6 times, but the probability that 1 person can succeed is not 6 times – nonopolarity Jul 23 '18 at 13:21
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    @太極者無極而生 Probabilities are not cumulative; only expected values are. – RayDansh Jul 23 '18 at 13:28
  • @RayDansh And the reason *why* that's the case is because in cases where $n$ people succeed, the *expected number of successes* goes up by an amount proportional to $n$, but the probability that *at least one person will succeed* goes up by the same amount it would have if there'd been only one success. – Ray Jul 23 '18 at 21:09

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):

Probability for at least one success with probability of 0.004, for 1 to 1000 tries (straight approximation, actual curve)

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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 %.

  • How did you get that the probability of catching is $70\%$? I think there is no info about catch probabilities... – Green.H Jul 23 '18 at 14:16
  • You are right. Replace every where I say "catch" with "catch a shiny pokemon". – Dhalsim Jul 23 '18 at 14:28

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}.

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To flesh out Mark Bennet’s answer with a little more detail, there are three questions here:

  1. If a group of six people all try something they’ll succeed at one time in five, how often will at least one of them succeed?

  2. If a group of six people all try something they’ll succeed at one time in five, a million times over, what is the expected number of successes?

As you know, the answer to question 1 is the chance that not all six will fail. $$1 - \left(\frac{4}{5}\right)^6 = \frac{11529}{15625} \approx 0.738$$

Question 2, you already answered: $\frac{1}{5} \cdot 6 \cdot 10^6 = 1.2 \cdot 10^6$. This is not only greater than 737,856, it’s greater than a million.

There’s no contradiction here. We can derive both from the probability of getting exactly one success on six attempts, two successes, and so on up to six.

What are those? Let’s say we have Alice, Bob, Clara, David, Emily and Frank. There’s exactly one success if Alice succeeds (1 time in 5) and each of the other five fails (4 times in 5, to the power of 5), or if the same happens to Bob, Clara, David, Emily or Frank.

There are exactly two successes if Alice and Bob succeed (1 time in 5, to the power of 2) and the other three fail (4 times in five, to the power of 3), and the same for any other pair of people: Alice and Clara, David and Emily, and so on. How many pairs of people are there? There are six people who could be first, and for each of them, five other people they could be paired with. Then, because picking Alice and David is the same as picking David and Alice, divide by two. As you probably know, this formula is written $\binom{6}{2}$ and pronounced “six choose two.” It’s equal to 15 (AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF, EF).

So, the odds of getting more than six or fewer than zero are nonexistent. Otherwise, the odds of getting exactly $i$ successes is $p(i) \equiv \binom{6}{i} \left( \frac{1}{5} \right)^i \left( \frac{4}{5} \right)^{6-i}$. That is, there are $\binom{6}{i}$ sets of $i$ people chosen from 6, a $\left( \frac{1}{5} \right)^i$ probability that all $i$ of them will succeed, and a $\left( \frac{4}{5} \right)^{6-i}$ probability that none of the remaining $6-i$ people will.

This allows us to compute the probabilities of every possible number of successes:

$ \begin{array}{ccccccc} p(0) & p(1) & p(2) & p(3) & p(4) & p(5) & p(6) \\ \frac{4096}{15625} &\frac{6144}{15625} &\frac{768}{3125} &\frac{256}{3125} &\frac{48}{3125} &\frac{24}{15625} &\frac{1}{15625} \end{array} $

We can see that these probabilities add to 1. Furthermore, we can see that the probability that the number of successes is greater than 0 is equal to $1 - p(0)$. What is the expected number of successes? If we conducted a million random trials and counted the number of successes on each, we would add zero times the number of trials with no successes, plus one per the number of trials with exactly one success, two per the number of trials with exactly two successes, and so on up to six. The formula for the expected number of successes per round is therefore

$$ E = 0p(0) + 1p(1) + 2p(2) + ... + 6p(6) = \sum_{i=0}^6 i\cdot p(i) $$

Plugging in the numbers from our table, we see that this is

$$ 0\frac{4096}{15625} + 1\frac{6144}{15625} + 2\frac{768}{3125} + 3\frac{256}{3125} + 4\frac{48}{3125} + 5\frac{24}{15625} + 6\frac{1}{15625} = \frac{6}{5} $$

So, the expected number of successes per round of six people flipping coins is confirmed to be 1.2. On average, that’s 1.2 million successes per one million rounds.

If you’d like to learn more of what mathematicians have discovered about processes like this, you can search for Bernoulli distribution and geometric distribution.

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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 $1-(1-p)^n$.

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