[Solved] Shining Revelry %

Horiet · March 27, 2025, 1:59pm

I just pulled a pack and got 5 different cards, but the site still says 100% chance to pull a new card. Isn’t that technically not true if I can pull the same 5 again?

marcel · March 27, 2025, 2:52pm

I think it’s probably more like 99.999% but rounded up to 100.

Horiet · March 28, 2025, 6:26am

I pulled some more and now it says 99.9%. does it round to the nearest tenth?

marcel · March 28, 2025, 6:40am

Yes that is correct.

user_sksk · March 30, 2025, 10:29am

I pulled 42/72 cards and the site still says 90% chance to pull a new card. I think that the % is too high or am I wrong?

Red · March 30, 2025, 1:48pm

Generally speaking, I think that sounds about right for (my limited understanding) of binomial probability:

My assumptions are for the 72 you are filtering the Shining Revelry for 1-4 Diamonds.
You own 42 out of 72 cards.
You are missing 30 cards.

The probability that a single card is a new (missing) card is:

P(new card) = number of missing cards / total number of card
P(new card) = 30 / 72 = 5 / 12 ≈ 0.4167 (41.67%)

Since each card is drawn independently with replacement logic, the probability that a single card is not new (a duplicate) is:

P(duplicate card) = 42 / 72 = 7 / 12 ≈ 0.5833 (58.33%)

Therefore, the probability that all 5 cards are duplicates:

P(all duplicates) = (7 / 12)⁵ ≈ 0.0580 (5.8%)

The complement of the above:

P(at least one new card) = 1 - P(all duplicates)
P(at least one new card) ≈ 1 - 0.0580 = 0.9420 (94.2%)

There is approximately a 94.2% chance that you will get at least one new card in a pack of 5.

Note that the actual percentage displayed on the site will be different to the chance percentage listed above, because my example does not take into account the pull rates of each of the cards for 1-4 Diamonds (or higher) for each of the card slots 1-5. But this should demonstrate that you would be very likely to pull at least 1 new card.

For comparison, a simpler example would be the chance of tossing 5 coins and landing at least 1 head:

Chance of getting all tails = 0.5 × 0.5 × 0.5 × 0.5 × 0.5 = 0.03125 (3.125%)
Chance of getting at least one heads = 1 - 0.03125 = 0.96875 (96.875%)

Although that being said, the number of times I have played a Misty card and landed tails is a different story… Hope this helps.

Shinjimx · March 30, 2025, 2:16pm

I suppose that Marcel needs to make a post explaining how the calculation works…

But I believe ( I am probably really mistaken) that you are not taking into consideration the probability that each card has to appear as a first, second, third, fourth and fifth card.

All those details are also in the pack information section.

But I am just writing out of ignorance.

By the way, there is also a way to improve your chances to get a head from Misty

Red · March 30, 2025, 2:25pm

Already made the call out that the actual probability will differ from the example above as there are different pull rates for each slots.

The example merely demonstrates that 90% is in the ballpark that it would be very likely to pull a new card. The actual probability calculation is a lot more complex than that.

marcel · March 30, 2025, 6:52pm

For anyone interested in the math, you can find it here in the “pullRate” function:

github.com/marcelpanse/tcg-pocket-collection-tracker

frontend/src/lib/CardsDB.ts

0e2f93eb1


      
          export const pullRate = ({ ownedCards, expansion, pack, rarityFilter = [], numberFilter = 1 }: PullRateProps) => {
            if (ownedCards.length === 0) {
              return 1
            }
          
            //probabilities
            // console.log('calc pull rate for', pack.name, ownedCards.length, rarityFilter)
          
            const cardsInPack = expansion.cards.filter((c) => c.pack === pack.name || c.pack === 'everypack')
            // console.log('cards in pack', cardsInPack.length) //79
            let missingCards = cardsInPack.filter((c) => !ownedCards.find((oc) => oc.card_id === c.card_id && oc.amount_owned > numberFilter - 1))
          
            if (rarityFilter.length > 0) {
              //filter out cards that are not in the rarity filter
              missingCards = missingCards.filter((c) => {
                if (c.rarity === 'Unknown' || c.rarity === '') return false
                return rarityFilter.includes(c.rarity)
              })
            }

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Shinjimx · March 31, 2025, 3:55am

Maybe open a FAQ with the most common issues that you encounter in the forums…

I have seen many people trying to figure out the math of the card pick…

I would try to create a post trying to explain, but i may not understand properly everything, So I suppose that you are the one that could explain how it works.