4
$\begingroup$

I play a game called Summoner's War. It is RNG-based and as such, has a chance of success for each sample.

Data has shown that in a particular part of the game where one may 'summon' from a small pool of monsters, there is a success rate of 5.67% per sample for a certain type of monster.

Based upon this data, the Expected Value of obtaining this type of monster is believed to be one every ~17.64 samples (100/5.67).

We can now assume that after 17.64 samples that we have a 50% chance to obtain said monster because we know 17.64 is the average number of samples it takes to obtain said monster.

Since the success rate is 5.67% and one has a 50% chance to obtain said monster at 17.64, shouldn't we be able to also calculate the number of samples (17.64) from a binomial formula that equals 0.5, the average?

To reiterate, the formula 1-(1-P)^x = 0.5. Shouldn't x result in 17.64? A 50% chance to summon the monster.

It doesn't seem to work for me.

1-(1-0.0567)^(17.64) results in ~0.64 or 64% chance to summon the monster.

1-(1-0.0567)^x = 0.5 results in x = ~11.8749 or about 11.8749 samples to summon the monster.

What am I missing? Is the average value not 50%, but instead 64%?

$\endgroup$
  • 3
    $\begingroup$ You are evaluating $1-(1-p)^{1/p}$ . That ranges from a limiting value of $1-e^{-1} = 0.6321...$ at p = 0, increasing to 1 at p = 1. It never equals 0.5. $\endgroup$ – Mark L. Stone Aug 16 '16 at 0:58
  • $\begingroup$ @MarkL.Stone: Thanks for the response. Does that mean I could be calculating the average incorrectly? Is there another formula I should be using in order to simulate fails/successes over multiple trials that would return 0.5 as the probability of success? $\endgroup$ – nabakin Aug 16 '16 at 2:21
  • $\begingroup$ Tiny detail but 0.0576 in the title looks like a typo for 0.0567. $\endgroup$ – Nick Cox Aug 17 '16 at 15:03
2
$\begingroup$

I believe you have confused the meaning of expected value.

We can now assume that after 17.64 samples that we have a 50% chance to obtain said monster because we know 17.64 is the average number of samples it takes to obtain said monster.

This statement is incorrect, and doesn't make fundamental sense (you cannot do 17.64 trials)

An expected value of 17.64 in this case does not mean this, (I used 18 as approximation):

X~B(18, 0.0567)

P(X=1) = 0.5 or P(X>=1) = 0.5.

The expected value of 17.64 means that if you draw many many trials from this process, we expect 1 success with the average interval of 17.64 trials. Note that this says nothing about the probability of seeing success or not.

In fact to find the probability that in 18 trials you will observe 1 or more successes:

X~B(18, 0.0567)

P(X=1) = 0.378357850443551

P(X>=1) = 0.65029888269312

i.e. your chance of getting exactly one success for 18 trials is 38%, and getting 1 or more is 65%.


However if you are insistent on defining an "average" as number of trials needed for the cumulative chance of 50% to obtain at least 1 success or more, then General Abrial's answer is the perfect way to calculate it. i.e.

"Average" = What is the needed number of trials n, to have a 50% shot at getting at least 1 success?

Although, from a gamer perspective, why is it so important to use 50% cumulative chance? Personally I would rather go for something more significant, like 80~95% cumulative chance to get at least 1 success.

Say if the summon (presumably powerful) is needed to beat a boss (or win a PvP), would you rather calculate for 50% chance of making it or 80% chance of making it.

$\endgroup$
1
$\begingroup$

Most of your question is premised on a misunderstanding of how these probability models work. I'm not going to directly engage with that -- instead, I'll give examples of how to work out these problems correctly.

Suppose that we model this process as a binomial distribution. This model assumes that each trial is independent and has a finite, fixed integer number of trials and a fixed probability of success for each trial. All trials are attempted, and the number of successes are totaled up at the end.

The probability of obtaining the monster in a sequence of $n$ trials is $\text{Binom}(n, 0.0567),$ which has density $$ f(x;n,p)=\binom{n}{x}p^x(1-p)^{n-x} $$. The density formula gives the probability of obtaining exactly $x$ monsters in $n$ trials.

For example, the probability of obtaining it in one trial is 0.0567. But in the more general case, you need to be more specific -- when you say "obtain" do you mean obtain exactly once? Or at least once? If you mean at least once, the probability of obtaining the monster is $1-f(0;n, p)$, i.e. the complement of the event that you obtain the monster 0 times in all $n$ trials.

In the scenario where you take 17 trials, the probability of obtaining the monster at least once is $1-f(0;17,0.0567)=0.629279$.

On the other hand, the outcome I think you're really interested in might be better modeled as a negative binomial distribution. This model has a fixed probability of success for each trial, the trials are independent, and the trials continue until the desired number of successes are obtained (specifically, the last trial is a success).

So we want to know how many trials to carry out to have some probability of having one success at summoning the monster. We can do this computation in R using the qnbinom function. The quantile function takes a probability, i.e. a value of the CDF, and returns the corresponding value of the random variable (in the case of discrete r.v.s, the smallest value of the r.v. that includes the probability).

qnbinom(0.5, size=1, prob=0.0567)=11

This means that there's a 50% shot of summoning the monster on the 12th attempt: 11 failures followed by the successful summoning.

$\endgroup$
  • $\begingroup$ Then does that mean that 12 summons is the true average? Or does having a 50% chance of summoning the monster not have anything to do with the average? I thought that since 1 monster is summoned approximately every 17 summons, that means 17 has a 50% shot of summoning the monster. $\endgroup$ – nabakin Aug 17 '16 at 22:17
  • $\begingroup$ I'm saying that the average you've computed doesn't really have anything to do with the gameplay. Either you're summoning a fixed number of times or you're summoning until you get the monster. I outline sample computations for both. $\endgroup$ – Sycorax Aug 17 '16 at 22:41
1
$\begingroup$

I think you have mean and median mixed up.

The mean (or expected value) is, indeed about 17. But this value is skewed high by the small odds of taking a very large number of turns.

The formula you use looks like a continuous approximation of the discrete CMF (cumulative mass function) $$1-(1-p)^n$$

But solving the CMF = 0.5, does NOT give you the expected value. It gives you the median (equal chance of greater or lesser outcome) value, which is indeed about 12.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.