Hearthpwn

i am not good with these complicated probability problems, can someone teach these basic math to me? thanks!

here are the scenarios:

scenario 1: I have 2 copies of a card in my deck...so what are my odds of drawing at least 1, if the first hand doesn't have any card, mulligan all

scenario 2: I have 4 1-mana cards in my deck...so what are my odds of drawing at least 1, if the first hand doesn't have any card, mulligan all

Scenario 3: I want 2 specific cards only(mirror image+mana wyrm), each has 2 copies in my deck, if i get 1 of them in the opening hand, i will keep it then draw the rest...

Scenario 4: I have 6 cards that I can mulligan for...what are my odds of getting at least 1? what are my odds of getting 2 of them?

oh maybe there's already a guide for this already...please provide link if there's one already, thanks!

Quote from Khennen »

Hey pimpemup,

I actually asked myself the same question yesterday. So basically you can use this Hypergeometric calculator to help you do the math.

How it works ?

Population size: number of cards in your deck, don't forget to change it when you mulligan (you can't draw a card you just mulligan).

Number of success in population: number of copy of your card in your deck

Sample size: number of card you draw

Number of success in sample (x): how many copy of that card you want to draw (x=1 or 2)

P(X = x): drawing that exact number

P(X > x): drawing at least that number

I will help you doing some math but keep in mind everything depends if you have the coin or not, not everything but having the coin will slightly increase the probability since you can draw more cards. If i'm no lazy I will do it all ... if not finish by yourself. I will change a little bit your scenario 1 and 2, giving more stats.

Scenario 1:

First hand:

Probability of drawing at least 1 of this card: P'1(no coin)=0.193 (for math noob P=0.193 means 19.3% "chances" of drawing).

Population size=30 / Number of success in population=2 / Sample size=3 / Number of success=1

And P'1(coin)=0.253.

Mulligan:

Still considering at least 1 card, P2(no coin)=0.214 and P2(coin)=0.289

For no coin: Population size=27 (can't draw card you just mulligan) / Number of success in population=2 / Sample size=3 / Number of success=1

Here is the answer of your scenario 1, if your first hand was empty your probability of drawing at least one specific card when you mulligan is 21.4% without coin and 28.9% with coin.

Xtra:

Ptotal(no coin)= [1-P1(no coin)]*P2(no coin)=0.173

Ptotal(coin)= [1-P1(coin)]*P2(coin)=0.216

So statistically having an hand without this card and drawing at least one when you mulligan all, will happen approximately 17.3% of time without coin and 21.6% of time with coin.

But it's better to consider your probability of drawing at least one of this card, first hand + mulligan everything if not in first hand.

Ptotal(no coin)= P'1(no coin) + [1-P'1(no coin)]*P2(no coin)=0.366

Ptotal(coin)= P'1(coin) + [1-P'1(coin)]*P2(coin)=0.469

Here you simply add your chance before mulligan to chance after mulligan corrected by "deducting" your success before mulligan...it's not clear but make sense to me.

To sump up your probability of having a specific card in your hand when you have two copy in your deck and mulligan all is 36.6% without coin and 46.9% with coin.

Scenario 2:

Same as previous but 4 success in population instead of 2

First hand:

P1(no coin)=0.360 and P1(coin)=0.454

Mulligan:

P2(no coin)=0.395 and P2(coin)=0.511

Here is the answer of your scenario 2, if your first hand was empty your probability of drawing at least one card when you mulligan everything is 39.5% without coin and 51.1% with coin.

Xtra:

Same as previously your probability of drawing at least one card that cost 1 mana, having 4 in your deck, mulligan all is Ptotal(no coin)=0.613 and Ptotal(coin)=0.733 so respectively 61.3% and 73.3% without and with coin.

Scenario 3:

Now event are dependent from each other:

First hand: 

P1(no coin)=0.36 and P1(coin)=0.454. Your probability of drawing at least one card from the 2 Mirror Image and 2 Mana Wyrm in your opening hand.

Mulligan:

P2(no coin)= 0.145 and P2(coin)=0.221. Your probability of drawing at least 1 of the other card if you kept a card in your hand.

Final:

Ptotal(no coin)=P1(no coin)*P2(no coin)=0.052 and Ptotal(coin)=P1(coin)*P2(coin)=0.100

So your probability of drawing at least Mirror Image or Mana Wyrm, mulligan the other cards and draw at least one of the other is 5.2% without coin and 10.0% with coin.

 Scenario 4:

It starts to get difficult but considering you mulligan everything:

First hand:

At least 1: P1(no coin)=0.501 and P1(coin)=0.612

Getting two of them on first hand: P'1(no coin)=0.089 and P'1(coin)=0.151

Mulligan:

At least 1: P2(no coin)=0.545 and P2(coin)=0.676

At least 1 if you had one on first hand and kept this card: P"2(no coin)=0.342 and P"2(coin)=0.488

Drawing two of them: P'2(no coin)=0.107 and P'2(coin)=0.191

Final:

Having at least one is the same as usual formula:

Ptotal(no coin)=0.773 and Ptotal(coin)=0.874 so a probability of 77.3% without coin and 87.4% with coin.

Drawing two of them well:

Ptotal(no coin)= P'1(no coin) + [1-P'1(no coin)]*P'2(no coin) + P1(no coin)*P"2(no coin)=0.186

Ptotal(coin)= P'1(coin) + [1-P'1(coin)]*P'2(coin) + P1(coin)*P"2(coin)=0.336

So finally (I hate math ... I have no idea why I'm doing that) a probability of approximately 45.9% of drawing two card without coin and 64.9% with coin if you mulligan everything (expect the case your first draw was one of your card of interest).

Edit: Man... this is the biggest comment of my life I guess, Please correct me if I'm wrong about math

 

wow, thanks for your hardwork bro

i'll have to slowly absorb all this math

you're the best! thank you once again!

Quote from Khennen »

The correct functionality of mulligans is as follows: You cannot get back the same card you just mulliganed. You can get back a card you mulliganed as your first draw, and you can mulligan into a second copy of a card you have in your deck, but you cannot get back the same card you just mulliganed back into your opening hand. The card you mulligan are indeed still set aside, new cards are drawn from the deck, and the mulliganed cards are shuffled back into the deck. 

If there are instances where mulligans are not functioning as stated above, those can be considered bugs and should be reported on the Bug Report Forums with as much information as you can provide. 

hm interesting, is there any specific reason why in most (a lot!) cases i draw a card in T1 that was the last card which i cliced to mulligan?