# How many clicks are required to reach a 50/50 chance of winning the Powerbloop?

In the 0.25 Clicker Heroes patch, the Powerbloop, which is essentially a click lottery, was introduced (likely a reference to the Powerball lottery held around the same time).

Participation in the Powerbloop is done by clicking a Bloop, and each click "draws" a random number between 1 and 292,201,338. Winning is done by drawing exactly 292,201,338, thus every click is a 1/292,201,338 chance of winning.

How many times do I, on average, have to click to reach a certain (say 50 %) chance of winning, and how long would it take?

# TL;DR

50 % chance - ~202,538,534 clicks (~58 days at 40 clicks per second).

90 % chance - ~672,818,450 clicks.

99 % chance - ~1,345,636,901 clicks.

x % chance - ~`log(1 - x) / log( 292,201,337 / 292,201,338 )` clicks.

Consider the two following questions:

• How many times will I on average win if I click n times?
• What is my chance to have won, given I click n times?

They may seem similar at first, but the answers differ. The first one is quite simple. If you click 292,201,338 times, you will on average win one time. If you click 146,100,669 times, you will on average win 0.5 times.

Now, it may seem intuitive at first, but the chance to have won at all after 146,100,669 clicks is lower than 50 % (it's actually around 39.3 %).

This is easier to realize for even larger numbers. Say you clicked 292,201,338,000 times (1000 times as many). The answer to the first question is now 1000 times, but the second one is still not quite 100% (it's very close though). There will always be some small probability that you never drew the winning number (assuming independent draws).

# Maths

To avoid confusion from the possibility of winning several times, I will focus on probability of not winning at all, since that is easy to work with and (in my opinion) aligns well with what most people actually want to know.

For simplicity, let `N` be the total amound of drawable numbers:

`N = 292,201,338`

There is only one winning number. The chance to win at least once ("at least" makes no difference for one click, but is included for consistency with later formulae) in any individual click, `P(1)` is thus:

`P(1) = 1/N ~ 3.42e-9`, or 0.000000342 %.

The probability of not winning in a single click, `P´(1)` is:

`P´(1) = 1 - P(1) = 1 - 1/N = (N - 1)/N`.

Now, assuming the chance to win on each click is independent of earlier clicks (I have no official source that this is the case, so I will also mention what happens if there is a dependency), the probability of not winning with any of `n` clicks is:

`P´(n) = P´(1)*P´(1)*P´(1)*... = P´(1)^n = ((N - 1)/N)^n`.

The chance of having won at least once, `P(n)`, during `n` clicks is thus:

`P(n) = 1 - P´(n) = 1 - ((N - 1)/N)^n`.

Solving for `n` gives us:

`n = log(1 - P(n)) / log((N - 1)/N)`.

To answer the question given any specific percentage, we just insert that as `P(n)` in the equation (as a fraction, so 45 % = 0.45).

Doing this for 50 %:

`n = log(1 - 0.5) / log((N - 1)/N) ~ 202,538,535`.

We see that this number is considerably higher than the average amount of clicks per 0.5 wins (discussed in the above section).

### Depleting set

If you cannot draw the same number twice, we say that the draws are dependent and that the set of possible drawable numbers is depleting. This means that:

`P´(n) = P´(1)*P´(1)*P´(1)*... = P´(1)^n = ((N - 1)/N)^n`

is no longer true.

Instead, since the amount of non-winning possible numbers is reduced by 1 for every non-winning draw, the probability of not winning decays as:

`P´(n) = (N - 1)/N * (N - 2)/(N - 1) * (N - 3)/(N - 2) * ...`,

which simplifies to:

`P´(n) = (N - n)/N`

(This only applies to `n <= N`, which makes sense, since after N draws the set of numbers is fully depleted).

We see that this probability is linear with `n`. Thus, the chance to have won after `n` clicks becomes:

`P(n) = n/N`.

This is much better than the independent case for high numbers, but for small amounts of clicks (millions), the difference is barely visible.

For example, clicking 10 million times, you have a ~3.364 % chance of winning in the independent case, and a ~3.422 % chance in the depleting case.

The following figure illustrates the difference between the two cases: We can see that both start off almost linear, but the depleting case starts to get significantly ahead around the 100,000,000 mark.

# Remember the gambler's fallacy!

Don't think of this as something you can make "progress" towards. Your chance of success going forward is not affected by the amount of clicks you have made so far. So no matter if you have never clicked before or if you have several billion clicks, you have the same (miniscule) chance of winning on your next click. Or the next million.

# Time

The highest numbers I hear people mention for autoclicking is about 40 per second. To having had a 50 % chance of winning, you would then need to leave it running for 202,538,534 / 40 ~ 5,063,463 seconds or about 59 days.

Needless to say, clicking by hand is mostly a hopeless venture (mostly).

• I was about to make a rant about how that's not the Gambler's fallacy, but then I realized I had confused it with Gambler's ruin. So many wondrous things happening in gambler's heads... Jan 28, 2016 at 15:55
• @Ordous Either way, it usually means bad things for the gambler :) Jan 28, 2016 at 17:32
• Sniff the API then replicate the input with a script :D Jan 28, 2016 at 21:39