Calculating the number of pulls needed for a three star XXL unit

I was curious how many pulls I need on average to get a limited XXL buddy to three stars. So I asked ChatGPT, using the example of Orem. I am highly sceptical of the results. Please comment if you’re good with statistics.

My premise: You need five total copies of the XXL buddy to get him to 3 stars. (Please correct me if I‘m wrong. I didn‘t include shards from the special ticket shop.)

Here‘s what ChatGPT gave me:

Key Details:

• Orem pull rate: 1.2% (0.012)
• Other characters’ pull rate: 0.6% each (0.006)
• Combined rate of getting one of the four characters: 3% (0.03)

Effective probability for Orem:

1. On a single pull:

   • The chance of triggering the 3% event is 0.03.
   • Once the event triggers, there is a 40% chance it will be Orem, and a 20% chance it will be one of the other three characters.

So, the probability of pulling Orem from one 3% trigger is:

$$ P(\text{Orem on trigger}) = 0.40 \times 0.03 = 0.012 $$

2. If you pull one of the other three characters (which happens with probability 0.20 of the 3% trigger), the next pull will definitely be Orem (probability 1). The probability of this sequence is:

$$ P(\text{Other character then Orem}) = 0.20 \times 0.03 \times 1 = 0.006 $$

Thus, the total probability of getting Orem on any given pull is:

$$ P(\text{Orem}) = 0.012 + 0.006 = 0.018 $$

Step 1: Average pulls to get one Orem

The average number of pulls to get Orem is:

$$ \text{Average pulls} = \frac{1}{0.018} \approx 55.56 \text{ pulls} $$

Step 2: Considering the Pity System

We need to incorporate the pity system, which guarantees that if you don’t get any of the four characters in 79 pulls, the 80th pull will be guaranteed to be one of them.

• The probability of pulling at least one of the four characters within 80 pulls is:

$$ P(\text{At least one character in 80 pulls}) = 1 - (1 - 0.03)^{80} $$

$$ P(\text{At least one character in 80 pulls}) = 1 - (0.97)^{80} \approx 1 - 0.046 = 0.954 $$

Thus, in 95.4% of cases, you’ll get Orem within 80 pulls, and in 4.6% of cases, you’ll need the pity pull (at the 80th pull).

Expected pulls to get one Orem with pity system:

• 95.4% of the time, you will get Orem within 55.56 pulls (from the previous calculation).
• 4.6% of the time, you will need the pity pull, which happens on the 80th pull.

Thus, the average number of pulls to get one Orem is:

$$ \text{Average pulls with pity} = 0.954 \times 55.56 + 0.046 \times 80 = 52.97 + 3.68 = 56.65 \text{ pulls} $$

Step 3: Getting Orem Five Times

To get Orem five times, we multiply the average number of pulls for one Orem by 5:

$$ \text{Average pulls to get 5 Orem} = 56.65 \times 5 = 283.25 \text{ pulls} $$

Step 4: Range of Results (95% Confidence Interval)

Next, we calculate the range of results using the standard deviation approach.

Standard deviation for one Orem:

We estimate the standard deviation for the number of pulls using the formula for a negative binomial distribution:

$$ \text{Std. dev. for one Orem} = \sqrt{\frac{1 - P(\text{Orem})}{P(\text{Orem})^2}} = \sqrt{\frac{1 - 0.018}{0.018^2}} = \sqrt{\frac{0.982}{0.000324}} \approx 54.96 \text{ pulls} $$

Standard deviation for 5 Orem pulls:

For 5 Orem pulls, the standard deviation scales by $\sqrt{5}$:

$$ \text{Std. dev. for 5 Orem pulls} = 54.96 \times \sqrt{5} \approx 122.88 \text{ pulls} $$

Range for 5 Orem pulls:

Using the 95% confidence interval (1.96 times the standard deviation):

$$ \text{Lower bound} = 283.25 - 1.96 \times 122.88 \approx 283.25 - 240.46 = 42.79 \text{ pulls} $$

$$ \text{Upper bound} = 283.25 + 1.96 \times 122.88 \approx 283.25 + 240.46 = 523.71 \text{ pulls} $$

Final Conclusion:

To pull Orem five times, you can expect:

• Average pulls: ~283.25 pulls
• Range of pulls (95% confidence): between 42.79 and 523.71 pulls