# They See Me Rollin’

In Dragalia Lost right now there is a new banner, and something not done yet. It’s a Gala banner (meaning limited (recurring) thing, and double SSR rates), but this time it’s a dragon. Mars, specifically. And he’s really damn good. This isn’t powercreep, it’s a powerleap. Now, honestly, I don’t care. I mean, sure, it’d be good to get him, since my strongest fire dragon is a SR. But, double rates for every other SSR are definitely tempting. Now, this is kinda bait, since we currently have a limited banner for the colab with FEH, and part two of that colab later on (the first part is a rerun from last year’s FEH colab), but I don’t really care about FEH at all. (I think it’s weird to collaborate with other gacha games, but it happens all the time so whatever.)

I also have a lot of resources for rolling, since I’ve been saving. Before rolling I have 126 single tickets, 11 ten-rolls, and 106,098 crystals. At 120 crystals per roll, that gives me a total of 1120 rolls available. Now, I did say I don’t really care about Mars, and I don’t. But I do have a lot of single tickets, and those will be annoying to roll all at once (since whenever I roll, I use those first), and the more I get, the more annoying they are. So, the plan is to use 100, see what I get, and post that. Just for fun, I’ll also post which ticket got me which thing, just to see the distribution.

But that would be boring (it’s only 100 rolls, and there’s no drama if I don’t have a target)! So I’m going to do some math. While I might be a janitor now, my training is as a teacher, so today I’m going to teach about probability, since we’re kinda sorta gambling here. The gambling fallacy is that, as time goes on not winning, that just means it becomes more likely that the win is just coming up: the longer we lose, the closer the win will be. But of course that isn’t true; each event has the same probability, every time.

HOWEVER, that doesn’t mean cumulative probability is the same. And it’s relatively easy to figure out in this case: either win (SSR) or lose (not SSR), so there’s only two possible outcomes for each event. But as I just said, what really matters is the cumulative probability. Sure, I only have a 6% (.06) chance of winning each roll (and thus a .94 chance of losing), but cumulatively I can figure out odds for any number of successes. The thing to remember is: if you say AND, that’s multiply; and if you say OR, that’s addition.

Let’s just take the chance of not getting a single win. This means we just lose, right? .94 chance? But if that were true, no one would ever pay money for gacha. And of course, it’s not true. It’s .94 for each chance, but there are 100 events (in my case). So, the first roll has a .94 chance of failure. But the second has to also fail, and the third, and the fourth. AND. So, .94 AND .94 AND .94 AND so on, a hundred times. Multiply it all, and it comes out to .94^100, which is… .00205, or a 0.205% chance of not getting a single success. Sounds good, right? Seems you’ll almost certainly get your SSR!

Thus, let’s look at the case of getting a single SSR. That means you have one success (.06) AND 99 failures (.94^99). Multiply that together, and that gets… .000131, or 0.0131% chance. But wait, you say, how can that be? If the chances of getting none are so low, how can the chances of getting more than one be even less? And of course, the more successes, the less likely it gets! (Since replacing a .94 with a .06 only makes the probability smaller every time we do it.) Well, we only looked at one case – we didn’t say whether the success was the first, or the 50th. And that matters; or rather, it doesn’t matter which draw was the winner, as long as we drew one. So, it could have been the first, or the second, or the third, or… OR. Remember, with OR we add. So we have to add the probability of each possible group. Since there are only 100 possible groups, and each group has the same probability, we can just multiply that first answer by the number of groups, which gives us .0131, or 1.31%. Much better!

You can of course do this for the case of two SSR, or three, or any number, up to the total number of rolls (in this case, 100). The math to figure all this out is called binomial distribution. The equation goes like this:

P(k out of n) =  n!/(k!(n-k)!) pk(1-p)(n-k)

Where the total probability P of k wins (with a probability of p) out of n trials is all that business. The first part is the number of groups you’ll get, and the second part is the probability of each group.

Now, this can be annoying to calculate out with a calculator, so fortunately you can easily figure this out in any spreadsheet with a simple command. Which is what I did. And doing it this way makes graphing it easy too. So I did. You’ll see I did two graphs: one of each individual case by themselves, to see which is most likely; and the other is how likely it is to get at least that number. The first one is self-explanatory, so I won’t talk about it. The other is a little more involved. Basically, it’s just another OR thing. If you get at least zero SSR, you’ll get zero OR one OR two OR three OR… so you add up each individual probability. Or, you can just do what I did, which is reverse it. Since the total probability is 1, you can just subtract the total probability of all the ones less than what you’re investigating from 1. So, since you can’t get less than zero, than the probability of at least zero is 100%. The probability of one is 1-P(0), or in this case .998; of two is 1-P(0)-P(1) which is .985, and so on. Hopefully your spreadsheet functions also have a way of figuring out that cumulative total as well.