Last updated: 2020-12-17
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Knit directory: gacha/
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| Rmd | 81db14e | Lijia Wang | 2020-12-17 | 10-pull bonus (WIP) |
In this section, we will explore how making a “10-pull” instead of using single pulls changes the probability model we constructed for the generic single-pull mechanism. Most gacha games encourage 10x pulls in order to incentivize the players to purchase more in-game currency and make extra pulls than they normally would, since presumably a player only doing single pulls would stop as soon as they obtained the card they desire. Therefore, they offer various “10-pull bonuses”, where they either provide an extra free pull, or guarantee a card above a certain rarity.
We will continue to use our assumption regarding the i.i.d nature of each Bernoulli trial, and that our player would like an SSR card. However, most of these “10-pull” bonuses do not offer a guaranteed card of top rarity, but normally one above the second highest rarity. So the 10-pull bonus for our player could be an SR or an SSR card.
For most games, it is unclear the distribution in probability between the SR and SSR cards for the 10-pull bonus. We will assume that they follow the same ratio their non-bonus probabilities, only that they now add up to 1 for this bonus pull.
We’re unsure about how the gacha algorithm employed in the game actually works, but I will assume that in this scenario, the first 9 pulls are still i.i.d Bernoulli trials. During the last pull, a new set of probabilities, which we will refer to as “bonus probabilities”, will be applied regardless of whether you obtained any SR/SSR cards previously. Bonus probabilities are only assigned to SR and SSR cards only, and they add up to 1.
In any 10-pull, the probability of you NOT obtaininging an SSR card can be modeled as: you did not obtain any SSR cards in the first 9 pulls AND pulled an SR card for the bonus.
For each normal trial, let the probability of pulling any SSR card be \(p\), and the probability of pulling any SR card be \(q\). For the bonus probabilities, let that for an SSR card be \(bp\), and that for an SR card be \(bq\), where \(b\) is a constant and \(bp + bq = 1\)
Pr(no SSR in first 9 & SR in bonus) = \((1-p)^9 bq\)
Therefore, the probability of PULLING at least one SSR during any 10-pull is:
\[Pr(X\geq1) = 1-(1-p)^9 bq \]
Similarly, we will use Fate/Grand Order non-event pool as an example, and our player desire a 5-star servant (Note that in Fate/Grand Order, Servants/characters have different probabilities from Craft Essence/equipments). Since we only desire a subcategory of SSR cards (characters, not equipment), we need to subdivide the overall probability into character probability \(p_c\), and equipment probability \(p_e\). Their respective bonus probabilities can be denoted as \(bp_c\) and \(bp_e\).
Probabilities:
SSR character (\(p_c\)): 1.000%
SSR equipment (\(p_e\)): 4.000%
total SSR probability (\(p\)): \(p_c + p_e = 5.000 \%\)
SR character (\(q_c\)): 3.000%
SR equipment (\(q_e\)): 12.000%
total SR probability (\(q\)): \(q_c + q_e = 15.000 \%\)
b = 5
For each 10-pull:
\[Pr(X\geq1) = 1-(1-p_c)^9 (bq + bp_e) = 1-(1-0.01)^9(5*0.15+5*0.04) = 0.132 \]
As a quick comparison, the probability of obtaining more than 1 SSR for 10 single pulls are:
\[Pr(X\geq1) = 1- Pr(X=0) = 1-e^{-n*p} = 1-e^{-10*0.01} = 1-0.905 = 0.095\]
Which shows that the 10-pull bonus is indeed working in our favor.
Is there a good way to visualize this? Like the pdf in single-pulls?