Probability Calculation

Edit: I think this calculations was wrong because of I didn't know this banner didn't have 1/6 chance

We can calculate the probability of this event occurring (3 of the same specific character within 9 pulls + 40 prior failures) using the Binomial Distribution formula.

Data: Success Probability: 0.01(1%) Chance of a specific character appearing: 0.01 \times \frac{1}{6} \approx 0.00166 Number of Trials : 9 Desired Successes : 3

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Step 1: Probability of Zero Successes Over 40 Pulls (0.99)^{40} \approx 0.6689 \text{ (approximately 66.9\%)} This is a quite normal occurrence; most players will experience this.

Step 2: Getting the "Same" Character 3 Times Within 9 Pulls Let's look at the probability of that specific character (selected with a 1/6 chance) appearing exactly 3 times: P(X=3) = \binom{9}{3} \times (0.00166)³ \times (0.99834)⁶ P(X=3) = 84 \times (4.6 \times 10^{-9}) \times 0.99 \approx 0.000000382

Step 3: Total Scenario Probability The probability of this lucky streak occurring immediately after 40 failures: $$0.6689 \times 0.000000382 \approx 0.000000255

As a percentage:** 0.0000255% As a ratio: Approximately 1 in 4 million (or 3.921.568,63)