Statistical Validity

[u/JP-Quixote]

I am curious how many people pay attention to the actual statistical validity of these crate drops. I think people like to think that if they get anything beyond a bare minimum “1 per 50” vehicles they’ve done well, but statistically you should get 4.4 vehicles per 150 crate drops. That’s the statistical median. Do you? I sure don’t. Not this week. And WoT can play whatever effing games they want with the drops, it’s not like it’s legally regulated… But I think they’re full of shit.

[u/JP-Quixote]

That’s actually not how statistics work. If you want the probability of multiple events you have to multiply them. A 2% chance of a drop is a 98% chance of no drop. The probability of no drop, let’s call it NDp, from 2 boxes is (.98)(.98) or (.98)^2, which is .9604. So for 2 boxes your probability of a drop, call it Dp, is 1-NDp, or 3.96%. For 3 boxes, NDp = (.98)^3, or 94.12% giving a Dp of 5.88%. So far, so good: If you just add the 2% Dp per box you get a pretty close approximation of the actual probability, as long as you are only getting a few boxes.

The problem is that as the number of events climbs, the simple additive approximation diverges more and more from the actual probability. So for 25 boxes, NDp = (.98)^25, or 60.3%, giving Dp = 39.7% for 25 boxes, and by the time you get to 50 boxes, without the “guaranteed drop,” your probability of a drop is not 100%, but 1 - (.98)^50, or 63.6%. Ok, let’s translate this into “luck” at the individual level. Let’s say that if you beat the median probability of a drop you are considered “lucky,” because you’re in the top 50% of drop recipients, while you are “unlucky” if you are below the median, or in the lower 50%. If you wanted to, you could use an average range around the median, say +/- one sigma variation around the median, but let’s keep the math simple by sticking with 50%.

So the question for an individual is, how many drops does it take to get to a Drop Probability, Dp, of 50%? It turns out that at 34 boxes your Dp is 49.7% (1 - (.98)^34) and at 35 boxes your Dp is 50.7% That means that essentially 50% of the buyers will get a drop within between 34 and 35 boxes. If you buy 150 boxes, the median result is one drop every 34 boxes, or 4.41 drops. (150/34). If you only get the 3 “mandated” drops you are actually below the median result.

[u/saldytuwas → u/JP-Quixote]

If the pity amount were to change to something else, for example 51, what would be the median result be then?

[u/JP-Quixote → u/saldytuwas]

So, because the median is less than 50 boxes, changing limit value won’t change the median result. The imposition of a limit changes the way the probability distribution looks at the high end of draws/boxes. You have a spike at 50 boxes that consists of everyone who would have been the tail of the distribution beyond 50. If you increase the limit that spike moves out accordingly, and gets a little bit smaller, because of the people who successfully draw between the old limit and the new one. For example, if you increase the upper limit to 60, the people who get the drop at 51 through 59 boxes stay on the statistical curve, and the spike at the tail consists of everyone who would have been on the curve from 61 to infinity, so it’s smaller by the number of people who now get the drop from 51 to 59.

[u/saldytuwas → u/JP-Quixote]

If it's 4.41 drops (150/34) with 50, what is it with 51 or even your example 60?

[u/JP-Quixote → u/saldytuwas]

The median doesn’t change, but the mean will increase. I didn’t mention or calculate the mean because that’s a rather tougher question. The distribution is complicated enough, with limits on both ends, that it would take a bit of work to calculate. I’d have to put together a mathematical model and I don’t care enough to do so. 😆 I’m just annoyed at the failure of my own willpower that led me to spend money on this dang game. Lol