Opening Hand Probabilities (BSS)

By BigStupidJellyfish | 2023-October-08

Battle Spirits Saga is a 2023 remaster/remake of the Battle Spirits card game, originally released in 2008/2009.

For deckbuilding reference, let's go over some of the probabilities of seeing certain cards/combinations in your opening hand. Battle Spirits Saga has a mulligan, which complicates things a little bit. Here's how that works, per the comprehensive rules:

  • 5-2-5. Each player draws four cards from their deck to create their hand.
  • 5-2-6. Use a random method such as rock paper scissors to determine player one and player two. The winner decides who becomes player one. Player one will take their turn first.
  • 5-2-7. Starting with player one, each player decides if they will mulligan. Players may only mulligan once. Players who choose to mulligan must first return their hand to the bottom of their deck in any order, draw four cards, then randomly shuffle their deck. Lastly, their opponent either further shuffles the deck and/or cuts it. Players who choose not to mulligan draw one additional card.
  • 5-2-8. Player one begins taking their first turn.

Neither player skips the draw step of their first turn, so our tables work for both going first and second.

Let's start by looking at your odds of seeing at least one copy of a particular card, with a standard 50-card deck.

# Copies played No mulligan (first 4) No mulligan (all 6) Mulligan
18.0%12.0%18.0%
215.5%22.8%33.1%
322.6%32.4%45.6%
429.1%41.1%56.0%
535.3%48.7%64.6%
641.1%55.6%71.7%
746.4%61.6%77.5%
851.4%67.0%82.2%

These numbers assume the mulligan is only performed if needed. The odds represent the chance of either seeing it in your opening 4, or not seeing it there but seeing it after the mulligan.

Quantities greater than 4 are to account for multiple cards that fulfill the same role (e.g., Suppression+Infinity Shield). Let's see how much going up to a 60-card deck hurts:

# Copies played No mulligan (first 4) No mulligan (all 6) Mulligan
16.7%10.0%15.0%
213.0%19.2%28.0%
319.0%27.5%39.1%
424.7%35.1%48.8%
530.1%42.1%57.0%
635.1%48.4%64.0%
739.9%54.1%70.0%
844.5%59.3%75.1%

Back to 50 card decks. If you like redundancy, the odds of seeing at least two copies are:

# Copies played No mulligan (first 4) No mulligan (all 6) Mulligan
10.0%0.0%0.0%
20.5%1.2%1.5%
31.4%3.5%4.2%
42.8%6.6%7.9%
54.5%10.3%12.5%
66.6%14.6%17.8%
78.9%19.2%23.5%
811.5%24.2%29.5%

Still in 50, the odds of seeing exactly one copy:

# Copies played No mulligan (first 4) No mulligan (all 6) Mulligan
18.0%12.0%18.0%
215.0%21.6%31.9%
321.1%29.0%42.4%
426.4%34.5%50.4%
530.8%38.4%56.4%
634.5%41.0%60.7%
737.5%42.4%63.8%
839.9%42.8%65.8%

To pay some respect to longer games, here are some generic odds of seeing at least one copy of a card (1-4 copies in deck), for the first n cards drawn (4-12). Mulligans not considered.

# Copies 4 5 6 7 8 9 10 11 12
18.0%10.0%12.0%14.0%16.0%18.0%20.0%22.0%24.0%
215.5%19.2%22.8%26.3%29.7%33.1%36.3%39.5%42.6%
322.6%27.6%32.4%37.0%41.4%45.6%49.6%53.4%57.0%
429.1%35.3%41.1%46.4%51.4%56.0%60.3%64.3%67.9%

So, for example, by your fourth turn without mulligans (9 total draws), you'd have a 56.0% chance of seeing at least one of a 4-of.

And a short one on two-card combos (seeing at least one each of two different cards):

# Copies played No mulligan (first 4) No mulligan (all 6) Mulligan
2,21.9%4.5%5.4%
3,34.1%9.3%11.4%
4,46.9%15.1%18.5%

Probably better to rely on deep draw power for these.

Notes#

  • These numbers are largely presented without commentary. Use them when deckbuilding however you see fit.
  • When compared to a game like Yugioh, these stats are significantly less important. Games go on much longer than the first 3 turns, so you have many more opportunities to draw into cards and it's less critical to have combo pieces in your opening grip. It quickly becomes impossible to assign specific probabilities to longer matches. You can't get away without playtesting.
  • For further reading on how these probabilities are derived: Wikipedia. In short: for \( N \) cards in your deck, \( K \) desirable cards in your deck, \( n \) cards drawn, and \( k \) target desirable cards obtained, the odds of getting exactly \( k \) of them is: $$ P(X = k) = \frac{{K\choose {k}} \cdot {{N-K}\choose{n-k}}}{N\choose n} $$ And for getting at least \( k \) copies, subtract from 100% all probabilities of getting less than \( k \) copies (e.g., for \(k\ge 1\), calculate \(1-P(X=0)\)).