TLDR

I have created a webpage that combines the strengths of Oddsbitch and u/Obikin89 's approach to determining your summoning odds.

Webpage

I have written a webpage to help you determine the probability for your pulls.

Here: g-t.io/

What can you calculate?

You can calculate step-up banners (current or future) or even something completely unrelated to FFBE (oh, the horror). This is a general purpose binomial probability calculator with an FFBE origin.

What do I get out of this?

All feedback is welcome, constructive feedback is more welcome. Please be kind as this is a proof of concept rather than a fully-functional, bug-free masterpiece. The underlying code is freely available, but a first-pass mess. PM me if you have skills in web development and want to improve upon this. I’m happy for the tool to be credited to someone else.

Why did I do this?

I did this because Oddsbitch is limited to 2 types of tickets (3* and 4*) and Obikin’s posts are useful, but limited in their approach (brute force simulations) and not always applicable to individual redditors. Of course, the reverse could be said of this website where it is complex and not easily usable.

You can stop reading here unless you like to understand how something works

Maths

When calculating the probability of a success you can either give the expected value of units, a.k.a. the Mean (e.g. 0.35 Rainbows), or the probability of getting a specific number of units (e.g. 29.9% of 1 or more Rainbow) . Both of them represent different views of the same distribution. The ‘X or more’ represents the skew of the data better than the average and the skew is important for distributions with low number of occurrences and low success.

As an example, for a 10 1 pull you have a Mean of 0.35 rainbows (sum of 10*0.03 1*0.05). Of course, 0.35 Rainbows means very little to people as you can’t have part of a rainbow.

Therefore, we use binomial probability to calculate the chance of obtaining 1 or more rainbows. In this case, it is simply calculated by determining the probability of 0 rainbows and subtracting that from 1 (1 is equivalent to 100% or all pulls, ‘not getting 0’ rainbows is the opposite to ‘getting one or more’).

Using (1-0.03)¹⁰ *(1-0.05)¹ for 0 rainbows to give 0.7001. Then 1-0.7001 = 0.2994 or 29.94% to get 1 or more rainbows. Here, (1-0.03) is the chance of not obtaining a rainbow from a 3* ticket.

It becomes a bit more involved when you want to know “2 or more” or “X>1” rainbows as you need to calculated multiple paths to achieve the outcome. If you pull 3x3* tickets you have 8 (2³ ) outcomes.

The probability tree would have outcomes FFF, FFS, FSF, FSS, SFF, SFS, SSF, SSS (where S = success and F = failure). These can be grouped based on the number of successes: 0 = FFF; 1 = FFS, FSF, SFF; 2 = FSS, SFS, SSF; 3 = SSS;. So there are 1 outcomes with 0 Rainbows, 1 with 3 rainbows, 3 with 1 Rainbow and 3 with 2 rainbows. The probability of the paths is independent of the order in which the rainbow was obtained an only depends on the number of rainbows. You can determine the number of combinations of trees that give the same results by using the binomial coefficient (nCk). For example: 3C0 = 1; 3C1 = 2; 3C2 = 2 and 3C3 = 1 – the same as the paths along the probability tree.

You could get your rainbow from either the 3* tickets or the 4* tickets, so the paths need to be extended to account for this too, but this is an FFBE sub, not a math sub (nor I a mathematician).

The above is fine for 3* and 4* tickets. However, we now have many different types of tickets (Step-ups, 10% 5* ticket, 30% 5* ticket, guaranteed any rainbow, guaranteed banner rainbow, etc) and people are more interested in obtaining up to 4 units for STMR.

The different tickets are easy to add into the calculation based on their chance of a rainbow, similar to a 4*.

Step-up to the Clouds

What is the probability of 2 Clouds on the Step-up banner (note that I use Cloud as an example. The same would apply to Sephiroth, Lila, or Elfreeda).

What are the chances of success?

• 3* pulls = 47 = 0.005 (0.5% as 1% on banner and split banner)
• 4* pulls = 5 = 0.01875 (half of 75% of 5%)
• 5* @ 10% = 1 = 0.033 (assumed 1/3 of 5* chance, no-one seems to know the on banner rate for a 10%5* ticket)
• 5* guaranteed any = 1 = 1/59 (59 units in the pool, so 0.0169)
• 5* guaranteed on split banner = 1 = 0.5 (half of 100%)

Total pulls 55 (but wait, you included the 10% ticket so it should be 56 pulls. I am glad you are keeping track. However, I have omitted the guaranteed 4* from step 2 as you cannot have a success of a rainbow. If you were interested in 4* units then it would need to be included).

What does this give?

1 lap = 17.80%; (2 laps = 54.95%; 3 laps = 78.62%).

Coding

I actually started on an Excel spreadsheet with a single step-up, but it quickly became apparent that spreadsheet based calculations would not adapt to multiple laps easily (PM me if you would like the original).

A general code was required! So I wrote it.

I first coded the maths for the successes and found a cookbook snippet for the binomial coefficient. Then I coded the permutations of success that are possible.

I tested it against oddsbitch and Obikin's simulations and found that the maths was good. The only deficiency is that Obikini’s methods provides easy access to many outputs that I have yet to calculate, such as chance of getting 1 Cloud and 2 Elfreeda. I can calculate the majority of the same outputs, but it often requires multiplying the output from 2 calculations. At some point I may go back to make this part easier.