Below is one way to break down the calculation. In Lorcana you roll with a 60‐card deck that contains 4 copies of “Belle” and 4 copies of “Pawpsicle.” At the start you draw 7 cards. Then—using your one mulligan opportunity—you may keep any “key” (i.e. Belle or Pawpsicle) cards you already have and put all the rest on the bottom of your deck, drawing new cards until you again have 7 cards in hand. (In other words, if you’re missing one or both of your key cards, you discard every “non‐key” card and draw that many replacements from the remainder of the deck.) Under the assumption that you follow this optimal strategy, we can split the outcomes into four cases.
Step 1. Find the chance you see each key in the initial 7 without mulligan.
Let’s define the events:
• A = “at least one Belle is in your 7‐card hand”
• B = “at least one Pawpsicle is in your 7‐card hand”Because there are 4 copies of a given key in a 60‑card deck, the chance to see no copy of (say) Belle in a 7‑card draw is
P(no Belle) = C(60 – 4, 7)⁄C(60, 7)
A quick sequential estimation gives about
(56/60)·(55/59)·…·(50/54) ≈ 0.60,
so
P(at least one Belle) ≈ 1 – 0.60 = 0.40.
The same logic gives P(at least one Pawpsicle) ≈ 0.40.
But note that these events overlap. Using inclusion–exclusion, one finds that the probability your initial 7 cards already contain at least one Belle and at least one Pawpsicle is about 15%. (A more exact calculation shows roughly 14.8%.)
Step 2. Split the initial-hand outcomes into four cases:
Both keys appear.
– Probability ≈ 15% (0.148).Only Belle appears (but no Pawpsicle).
– Since P(at least one Belle) ≈ 40% and the “both” case is 15%, the chance for “Belle only” is about 25% (0.253).Only Pawpsicle appears (but no Belle).
– By symmetry, also ≈ 25% (0.253).Neither key appears.
– The remaining chance is about 35% (0.346).
Step 3. Add in the mulligan drawing.
When you mulligan, you keep every key card that appears and discard the rest, drawing enough cards from the remaining deck to refill your hand to 7. (Since the initial 7 cards are out of the deck, your replacement draws come from the remaining 60 – 7 = 53 cards.)
For each case where you are missing a key, you have to “find” that key among your replacement draws:
- Case “Only Belle”:
You already have 1 Belle and are missing a Pawpsicle. In this situation you discard the 6 non‑Belle cards and draw 6 new cards from 53. None of the 7 initial cards had a Pawpsicle, so all 4 copies are still in the remaining deck. The probability that at least one Pawpsicle is drawn among 6 cards is1 – [C(53 – 4, 6)⁄C(53, 6)]
A sequential calculation shows this is roughly 1 – 0.61 ≈ 0.39 (39%).
Case “Only Pawpsicle”:
By symmetry, if you only have Pawpsicle initially, your probability to draw at least one Belle in your 6 replacement cards is also about 39%.Case “Neither key”:
Here you discard all 7 cards and draw 7 new ones from the remaining 53. Since you saw no keys, all 8 key cards (4 Belle + 4 Pawpsicle) remain in that reservoir. Now you need at least one Belle and at least one Pawpsicle among the 7. A hypergeometric calculation (or a sequential probability approximation) shows that the chance to draw “both keys” in 7 cards in this situation is about 18%.
Step 4. Combine the cases.
Weight each case by its probability:
Case 1 (both already present):
0.148 (or about 15%).Cases 2 & 3 (only one key present):
For each of these two cases the chance to “find” the missing key is about 39%, so each contributes0.253 × 0.39 ≈ 0.099,
and the two combined add about 0.198 (or roughly 20%).
- Case 4 (neither key present):
0.346 × 0.18 ≈ 0.062.Now, adding these up gives an overall chance of
0.148 + 0.198 + 0.062 ≈ 0.408
or about 41%.
Conclusion:
Under these assumptions and using the optimal mulligan strategy (keep any Belle or Pawpsicle you see, discard the rest, and then draw until you have 7 cards), you will end up with at least one copy of both Belle and Pawpsicle in your starting hand roughly 41% of the time.
It’s interesting to compare this result with what you’d have if you couldn’t mulligan. Without the mulligan ability, only about 15% of your initial 7‑card hands would hold both keys. By allowing you to selectively discard and draw replacements, the mulligan nearly triples (from ~15% to ~41%) your chances of seeing your “must‐have” cards early.
If you’re curious, you might also consider similar calculations for other card combinations or even explore what happens if you have multiple mulligan opportunities; the combinatorial trade‐offs can be quite rich and may even influence deckbuilding decisions.
This kind of probabilistic reasoning can help you better understand the consistency of your deck and plan your strategy accordingly.