# How likely am I to have a key card in my starting hand in Brawl?

I'm playing the Brawl format (a 60-card singleton format), and I really want one specific card in my starting hand.

I'm willing to take two mulligans (leaving me with a starting hand of 5 cards after putting 2 on the bottom of my library).

What are the odds of drawing the card in my opening hand within 2 mulligans?

• You currently have a few questions peppered in there. For the avoidance of doubt, it would be good to remove all but the questions in the last paragraph. – Pureferret Dec 31 '19 at 13:29
• The question Aetherfox refers to is very different from this. In my opinion, this is not a duplicate.It is true that they are similar; however, this is a different probability problem, as this question asks about the probability of obtaining a specific card even AFTER making two mulligans - and therefore it is about "probability with re-entry in the initial sample".Furthermore, in the other question a very different example is made, in which the probability of having a particular card in the initial hand is requested, but of which there are 4 copies in the deck.I don't think that's the case... – ManoFromBerlin Jan 15 at 16:19
• @Aetherfox Re "Does this answer your question?", It does not. The linked question doesn't cover mulligans whatsoever. (It also about a different deck size and a different number of target cards.) – ikegami Jan 15 at 19:51
• Doh! I didn't want to unilaterally reopen the question – ikegami Jan 15 at 22:26

The first mulligan is free in Brawl[CR 903.11g], so it's 3 mulligans leaves you with 5 cards. I shall reply accordingly.

After removing the Commander, 59 cards remain in the deck. 58 of those aren't the card.

The probability that the 1st card drawn isn't the card is 58/59.
The probability that the 2nd card drawn also isn't the card is 57/58.
The probability that the 3rd card drawn also isn't the card is 56/57.
etc.

So the probability that you don't have the card in your opening hand
= (58/59)*(57/58)*(56/57)*(55/56)*(54/55)*(53/54)*(52/53)
= (58*57*56*55*54*53*52)/(59*58*57*56*55*54*53)
= 52/59

Since you draw seven card from a freshly shuffled deck when taking a mulligan, the probably of not getting the card in a mulligan is the same as for the initial hand.

So the probability that you don't have the card after up to three mulligans
= (52/59)^4
= 60%

The probability that you do have the the card after up to three mulligans
= 1 - 60%
= 40%

Probabilities are easy to get wrong, so I like to verify through experimentation.

``````#!/usr/bin/perl

use strict;
use warnings;

use List::Util qw( shuffle );

use constant NUM_TRIALS => 10_000;

my @deck = ( 1, (0)x58 );
my \$successes = 0;
for (1..NUM_TRIALS) {
for (7,7,6,5) {
if ( grep { \$_ } ( shuffle(@deck) )[0..6] ) {
++\$successes;
last;
}
}
}

printf("%.0f%%\n", \$successes/NUM_TRIALS * 100);
``````

Output:

``````\$ ./a
40%

\$ ./a
40%

\$ ./a
39%
``````

The current rules for mulligans:

103.4. Each player draws a number of cards equal to their starting hand size, which is normally seven. (Some effects can modify a player’s starting hand size.) A player who is dissatisfied with their initial hand may take a mulligan. First, the starting player declares whether they will take a mulligan. Then each other player in turn order does the same. Once each player has made a declaration, all players who decided to take mulligans do so at the same time. To take a mulligan, a player shuffles the cards in their hand back into their library, draws a new hand of cards equal to their starting hand size, then puts a number of those cards equal to the number of times that player has taken a mulligan on the bottom of their library in any order. Once a player chooses not to take a mulligan, the remaining cards become that player’s opening hand, and that player may not take any further mulligans. This process is then repeated until no player takes a mulligan. A player can take mulligans until their opening hand would be zero cards, after which they may not take further mulligans.

903.11g In any Brawl game, the first mulligan a player takes doesn’t count toward the number of cards that player will put on the bottom of their library or the number of mulligans that player may take. Subsequent mulligans are counted toward these numbers as normal.

(Emphasis mine.)

• This answer appears to use a mulligan rule different than given on the Wizards website: magic.wizards.com/en/game-info/gameplay/formats/brawl Card draws are 7, 7, 6, 5 as I read the rules. Not 7, 7, 7, 7 as this answer states. – John Jan 15 at 14:44
• But with the London mulligan rule you draw 7 and discard down to 6 or 5. So the odd can be calculated on 7 if you are only looking for one card. – Styxsksu Jan 15 at 14:48
• The question is specified in the sense that the player starts with 7 cards in hand; and then the first mulligan is made, drawing only 6;then, the second mulligan consists, based on what is requested by the application, of a third drawing of only 5 cards. The answers must be based on this data. – ManoFromBerlin Jan 15 at 16:23
• Thank you @Styxsksu, I wasn't aware of the London Mulligan rule. – John Jan 15 at 17:29
• @John, You should follow your own link :) That is not how mulligans work in MTG (anymore). Please read 103.4 (and 903.11g), now conveniently quoted in my answer. – ikegami Jan 15 at 19:32

Odds for non-Brawl using the pre-Core 2020 mulligan rules (according to Ikegami's editing)

The probability of having a desired card in the hand,at the first draw of seven cards, is, by removing the commander, (7/59) * 100 = 11.8644%

If, at the first draw, the desired card has not arrived, then, according to what is stated in the question, you will draw again using the mulligan; and then the probability of reaching this point, which is obviously greater than 7/59, is as follows -

• first of all, this probability is due to the probability of not drawing the desired card to the first set of seven cards, which is: (52/59) * 100 = 88.1356%

• this value is then multiplied by the probability of drawing the desired card now, which, having returned the 7 first cards in the deck, but now having to draw 6, is (6/59) * 100 = 10.1695%

• Consequently, the probability of one mulligan, and then of finding the card sought in the six cards of the second draw is (52/59) * (6/59) * 100 = 8.9629%

If the card you are looking for hasn't arrived at this point, then, as specified in the question, you will mulligan for the second time.

So, this probability is obtained by multiplying the value of the probability of not drawing the card among the first seven (52/59), multiplied by the probability of not drawing the card even the second time (53/59), multiplied by the probability of draw the desired card now, but drawing only 5, or 5/59.

Consequently, the probability of mulligan twice, and then of finding the card sought among the five cards of the third draw is: (52/59) * (53/59) * (5/59) = 6.7%

At this point, the total probability of finding a desired "key-card", being able to make two mulligans, is given by the sum of the three possibilities of finding the card:

(possibility of finding the card after the first draw) + (possibility of finding the card after the first mulligan) + (possibility of finding the card after the second mulligan) = 11.8644+ 8.9629+ 6.7 = 27.5273%

Well ... that's all ... at least I think.
I hope there are no typing and/or math errors !!!

Odds for Brawl using the Core 2020 mulligan rules

The result shown above works only if, in making Mulligan the first time, the player draws 6 cards (not 7), and, in making Mulligan the second time, if the player draws only 5.
In addition, the result is valid only without looking at the cards that are placed at the bottom of the deck.
But, as I suspected, this does not accord with the current Mulligan rule...

I think I will leave this answer here just the same, also because, even if it refers to a mulligan rule that is not now applied, its procedure is correct.
To prove it, I will verify that his calculations are correct, rewriting them taking into account my own procedure, but considering the correct Mulligan rule.
The probability of having a desired card in the hand, at the first draw of seven cards, is, by removing the Commander, (7/59) * 100 = 11.8644%.

The probability of having the card at the first mulligan is:
(52/59)*(7/59)*100 = 10.4567652973%.

The probability of having the card at the second mulligan is: (52/59)(52/59)(7/59)*100 =9.21613212646%.

(This means that the player is able to, according to the current Mulligan rules, look at 7 cards, and choose for himself the one that will be placed at the bottom of the deck, according to his needs. And it is always for this reason that the calculations agree based on the fact that, making Mulligan, for the purpose of searching for a "key card", it is as if the player was still drawing seven cards).

The probability of having the card at the third Mulligan - (or when, granting the answer to the fact that the asker claims to want to stay,at most,with 5 cards in hand, after the Mulligans, which, however, are 3, and therefore not 2) - is: (52/59)(52/59)(52/59)*(7/59)*100=8.12269272162%

(And this also means that the player is able to, according to the current Mulligan rules, look at 7 cards, and choose for himself the two that will be placed at the bottom of the deck, according to his needs).

At this point, the total probability of finding a desired "key-card", being able to make three mulligans, is given by the sum of the four possibilities of finding the card. And is:

11.8644% + 10.4567652973% + 9.21613212646% + 8.12269272162%= 39.6590927216%.

This result is absolutely close to the correct one, already calculated in the right answer, higher.

• If so, you are right Ikegami. Even if this really leaves me perplexed ... I can't believe it: can a player see all seven cards even in the two mulligans after the first draw, and choose the 5 or 6 ones to hold in his hand? That's why this game has lost charm ... in this way it is practically always worth taking Mulligan ... – ManoFromBerlin Jan 16 at 15:55
• Anyway, I think Ikegami's 40% is really too high ... It is equal to take almost 24 cards,but in only one draw...while here we are talking about taking 21 cards,in three draws,and in the best case! But...Ok...Ikegami has right: it is equal to take 28 cards,in 4 draws, and because the first mulligan means to draw 7 cards, just as it was the first time. It is this way...but it is also unbelievable...These new and honeyed in syrup rules are killing me ...Mulligan should be an exception, but this way it becomes the rule instead, because it is too convenient to do it! – ManoFromBerlin Jan 16 at 16:49
• But no!Your calculation is absolutely right, Ikegami.But mine is too,and you can still leave it here with the new editing, because now the steps are well specified and can be followed well by anyone in this way.And then, if the current Mulligan rule were to change, there would be all the procedure specified step by step. Anyway, you are right, Ikegami:you calculated everything before and correctly, because you know the rules of the game much better than me! You have to admit, however, that kind of Mulligan is the rule,not the exception.You first say:-"Ok, players:make your hand as you please!" – ManoFromBerlin Jan 17 at 15:36
• Re "But mine is too", Well, now they are. – ikegami Jan 18 at 0:59
• I'm confused. Who said it wasn't a good attempt? And why would anyone care? – ikegami Jan 28 at 11:26