# If I really need a card on my start hand, how many mulligans make sense? [duplicate]

There is a funny modern deck where you play Treasure Hunt, but you really need to have it on the starting hand. How many mulligans should I take, if my deck contains 60 cards and contains 4 copies of Treasure Hunt?

• – doppelgreener Nov 23 '18 at 12:36
• The link doesnt help because it only tells the possibility for the start hand, but I have more factors, like start hand with fewer cards (mulligan) vs. drawing the card during play. – dan Nov 23 '18 at 12:48
• If you follow the "Hypergeometric Distribution" link in the linked answer, you will find a tool to calculate the odds for any starting hand including after any number of mulligans. Just put in "population size" 60 (cards in deck), "number of success in population" 4 (4 copies in the deck), "sample size" 7 (hand size, enter 6, 5, etc. for mulligans), "number of success in sample" 1 (you want at least 1 copy in hand). The lowest calculated field named "Cumulative Probability: P(X >= 1)" will tell you the odds of having at least 1 of the wanted card in your hand. – Hackworth Nov 23 '18 at 12:59
• Also, the question "how many mulligans should I take" doesn't really make sense here, because you have given no criterion other than "you really need the card in your starting hand". Well, if you need the card in your starting hand to win, then the answer is "you take mulligans until you have the card in hand or you run out of mulligans". – Hackworth Nov 23 '18 at 13:01
• What's the other (non-land) card in that deck? – Caleth Nov 23 '18 at 15:14

If you “need” the card in your starting hand and there is no other option—you take as many as you need until you have it, then stop. If you don't have it when you've mulliganed to 1, you concede and hope for better luck next game.

Granted you also need lands for mana, so you probably concede if you don't have it and an island when you mulligan to 2.

However this is a bit of a fragile fragile game plan: you're going to lose about 1 in 8 games simply for not having the card in your opening hand and conceding. 1 in 8 chances can also happen multiple games in a row.

# Odds of finding that card in your mulligans

From How do you calculate the likelihood of drawing certain cards in your opening hand?, we can calculate the odds of having it by each mulligan.

If you have a 60-card deck, the chance of starting with at least one 4-of in your opening hand is about 40%. That's from this calculation that finds the odds of not finding that one card:

(56*55*54*53*52*51*50) / (60*59*58*57*56*55*54) = ~0.6005 or ~60%

### Chance of finding the card in each separate hand

Let's run that calculation for each of your hands:

• 7 cards (first hand): 0.60 → 40% chance of finding the card
• 6 cards: (56*55*54*53*52*51) / (60*59*58*57*56*55) = 0.64 → 36% chance
• 5 cards: (56*55*54*53*52) / (60*59*58*57*56) = 0.699 → ~30% chance
• 4 cards: (56*55*54*53) / (60*59*58*57) = 0.75 → 25% chance
• 3 cards: (56*55*54) / (60*59*58) = 0.81 → 19% chance
• 2 cards: (56*55) / (60*59) = 0.87 → 13% chance
• 1 card: (56) / (60) = 0.933 → ~7% chance

### Chance of finding the card by a certain mulligan

To get this, we just multiply the mulligan results together.

• Chance of finding the card by your second hand of 6 cards: (0.60*0.64) = 0.38 → 62%
• By your third hand (5 cards): (0.60*0.64*0.699) = 0.265 → 73.5%
• By your fourth hand (4 cards): (0.60*0.64*0.699*0.75) = 0.20 → 80%
• By your fifth hand (3 cards): (0.60*0.64*0.699*0.75*0.81) = 0.16 → 84%
• By your sixth hand (2 cards): (0.60*0.64*0.699*0.75*0.81*0.87) = 0.14 → 86%
• By your seventh hand (1 card): (0.60*0.64*0.699*0.75*0.81*0.87*0.933) = 0.13 = 87%

In 87% of your games, you'll find your card in one of your mulligans if you mulligan all the way down to 1. In 13% of them, or 1 in 8 games, you'll just never find it and you'll concede.

• You don't mulligan to 1. The chances of the top card in your library being Trasure Trove when you mulliganed to 2 are higher than the chances of mulliganing to one and getting it, specially when you get a free scry. So you hope to draw it. – Martin Epsz Nov 23 '18 at 21:29
• @MartinEpsz But after mulliganing to 1, you can still draw the top card of your library after that. So wouldn't mulliganing to 1 still give you more chances? – Brilliand Nov 23 '18 at 21:49
• @Brilliand You are right, I don't know how I didn't realize that. – Martin Epsz Nov 24 '18 at 14:42
• @Martin Epsz this is basically what my question is about, I think I didn´t make that clear. My question was really like "Do I mulligan to 1, or just to 2 or 3"? – dan Nov 24 '18 at 19:17
• You really didn't make that clear, yeah. :( You asked how much to mulligan if you need it in your start hand, and that's a very specific thing: you either have it in your start hand or you don't. I suggest asking a new question: if you depend on having a certain combo piece early, in a tournament setting, what point does it make sense to mulligan to versus just hoping for extra draws to give it to you early? – doppelgreener Nov 24 '18 at 19:46

If you really need a card on your starting hand to win, and otherwise you lose, then you have to take mulligans until the card shows up or you run out of mulligans. The chance to have Treasure Hunt in your hand by turn 2 with at 6 mulligans is 87.5%

When the alternative to taking a mulligan is losing the game, then taking another mulligan that draws you at least 1 card is always the right choice. Basically, you stop taking mulligans until you find the card you need, or you are down to 1 hand card.

Therefore, calculating mulligan probabilities is somewhat besides the point. If you still want to know the odds, then in the top answer to the question How do you calculate the likelihood of drawing certain cards in your opening hand? there is a link to a calculator that you can use to calculate a hypergeometric distribution.

You put in the following numbers, from top to bottom:

Population size: 60 (for the deck size),

Number of successes in population: 4 (number of copies of the card you need),

Sample size: 7 (hand size, adjust for mulligans), and

Number of successes in sample (x): 1 (you want at least 1 copy in hand).

Then you can calculate the odds for any mulligan. On the play on your initial hand, you have a ~60% chance to NOT have the card in hand. After 1 mulligan, the chance is ~64% and so on. To calculate the odds of NOT having the card after n mulligans, you calculate those individual n mulligan probabilities, and multiply them. For example, after 2 mulligans, the chance to not have the card is 0.6*0.64 = 38.4%, so the chance to have the card is 100%-38.4% = 61.6%.

Since you can cast Treasure Hunt only on turn 2, you can effectively consider the last mulligan to be 1 card larger than actual, because you will draw another card before you really need the Treasure Hunt in your hand. So for example, if you take 1 mulligan and keep the next hand regardless, then the probability of having the card in your hand by turn 2 is really 1 - .6*.6 = 64%.

The maximum chance to see the card is achieved by taking all mulligans down to 1 card. The probabilities to NOT have the card after each starting hand draw are, in order:

60%, 64.8%, 69.9%, 75.3%, 81%, 87%, 93.3%

So the chance to NOT have the card in hand after 6 mulligans is:

0.6 * 0.648 * 0.699 * 0.753 * 0.81 * 0.87 * 0.933 = 13.5%

The chance to draw at least 1 copy after 6 mulligans is

1 - 13.5% = 86.5%

As discussed before, the last mulligan is effectively 1 card larger:

0.6 * 0.648 * 0.699 * 0.753 * 0.81 * 0.87 * 0.87 = 12.5%

For a total chance to draw the card by turn 2 after 6 mulligans:

1 - 12.5% = 87.5%

Note that since you can scry 1 after your last mulligan means that you can practically eliminate another unwanted card from your deck, which makes the probability a little bit higher still.

• "Therefore, calculating mulligan probabilities is somewhat besides the point." But ... you go on to calculate Mulligan probabilities. 🤔 We both do. I don't understand this statement. Is this "this is beside the point but let's do it anyway"? – doppelgreener Nov 23 '18 at 14:12
• Yes it's besides the point because it's not strictly the answer to the question of "how many mulligans make sense", to which the answer I gave is "as many as it takes". As I wrote, I gave the probabilities "If you still want to know the odds"., because otherwise the answer would be a little bare-bones. – Hackworth Nov 23 '18 at 14:23
• Ok, fair, makes sense to me. :) – doppelgreener Nov 23 '18 at 14:24