Trading Card Game Player needs Probability help please

[GtB]

Searched the forums and found two posts related to what I want to know but did not help me enough to answer my question...probably do not know enough about probability and do not know the right keywords.

The game I play people usually have 60 card decks, you can have more then 60 but 60 is the number of cards people usually stick with.

Lets say for this question my deck has 15 different cards, 4 of each card for a total of 60 cards in the deck.

To start the game I draw 7 cards into my hand.

I have two questions.

1. What is the chance of my drawing a specific card, I will refer this card as Card X, in my opening hand?

My understanding is this (assuming it is the last card I draw)

Chance to draw Card X in Opening Hand of Seven Cards= (4/60)+(4/59)+(4/58)+(4/57)+(4/56)+(4/55)+(4/54)
Chance to draw Card X in Opening Hand of Seven Cards = 49.2%

Is this correct?

2. In the game I play if we do not like our first 7 cards we can reshuffle our deck and draw another seven cards, called mulligan.

Knowing this, what is the chance that I will draw Card X in my Opening Hand of Seven cards in either my intital draw or my mulligan draw?

I know enough about this to know I am not sure what kind of math needs to be done.


On a side note, what kind of math would I need to study to learn about the kind of probabilites I would use for my card game? I am not a professional player but it is my main hobby and I would very much like to not only be able to build my decks knowing how many cards I should put in my deck but I would also like to know the probabilities during a match. Knowing the basics would really help me.

Thanks for any help that can be provided.
GtB
Rob

 

[Moo Of Doom]

1. Nope, that's not quite correct. You are adding the probabilities that the first card is X, ..., that the seventh card is X, but this only works if these are mutually exclusive events, which they clearly aren't in this case. A simple way to calculate the odds would be to calculate the odds of not drawing X, and subtracting that from 1.

There are 60 choose 7 possible opening hands, and 56 choose 7 opening hands that don't contain any of the four copies of X, so the probability of having no X's is (56 choose 7)/(60 choose 7). This is about 60.05%, so the chance of having at least one X in your opening hand would be a little less than 39.95%

2. Do we shuffle our initial seven cards back into our deck during the mulligan? If yes, then we can assume the two draws are independent, which means that the chance of getting no X's both times is just 60.05% of 60.05%, or 36.06%, so that the chance of getting at least one X one of the times is 63.94%.

If we put aside our initial X-less hand when taking a mulligan, things are slightly more complicated. Since our initial seven cards are gone, the second time, we're choosing out of a pool of 53 cards, so that there are 53 choose 7 hands, and 49 choose 7 unfavorable ones. This means there are 55.73% X-less second hands. So that the probability of having not drawn any X's either time is 60.05% times 55.73%, that is 33.47%. So 66.53% of the time, you'd have drawn at least one X.

Regarding your side note, this would fall under combinatorics, I believe.

 

[zli034]

2. Do we shuffle our initial seven cards back into our deck during the mulligan? If yes, then we can assume the two draws are independent, which means that the chance of getting no X's both times is just 60.05% of 60.05%, or 36.06%, so that the chance of getting at least one X one of the times is 63.94%.

I don't really agree with the quoted answer. Because that is basically a binomial model of 2 trials and which assume the mulligan is always drawn following the initial draw. So I will have to ask how likely the players will make a mulligan draw. And the model should be derived by conditioning on mulligan draw probability. Here goes my concerns.

 

[CRGreathouse]

I don't really agree with the quoted answer. Because that is basically a binomial model of 2 trials and which assume the mulligan is always drawn following the initial draw. So I will have to ask how likely the players will make a mulligan draw. And the model should be derived by conditioning on mulligan draw probability. Here goes my concerns.

MoD's analysis is consistent with my understanding of the OP: if the first hand does not have the desired card, mulligan; otherwise, keep the hand.

 

[DaveC426913]

Huh. Do people still play Magic??

 

[CRGreathouse]

Huh. Do people still play Magic??

The rules proposed in the OP don't quite match my recollection of Magic... .

 

[zli034]

MoD's analysis is consistent with my understanding of the OP: if the first hand does not have the desired card, mulligan; otherwise, keep the hand.

I suggest to use computational simulation to cross validate the result.

 

[CRGreathouse]

I suggest to use computational simulation to cross validate the result.

Of course the results of the simulation would depend entirely on the assumptions you used in building the simulation! If we used your assumption (always mulligan), you'd get one of the cards about 40% of the time; if we used MoD's assumption, 64% of the time. The problem is basic enough that I don't see any value in simulation, especially considering that the simulation would take longer than just solving the problem! (It took me about 8 seconds to solve, but I'm only a modest typist; a good typist could do the 29-character Pari command in half that time. Of course I only made 23 keypresses, thanks to tab completion.)

 

[DaveC426913]

The rules proposed in the OP don't quite match my recollection of Magic... .

It all sounds right to me except for the Mulligans, which the OP admits is their personal method.

 

[zli034]

2. Do we shuffle our initial seven cards back into our deck during the mulligan? If yes, then we can assume the two draws are independent, which means that the chance of getting no X's both times is just 60.05% of 60.05%, or 36.06%, so that the chance of getting at least one X one of the times is 63.94%.

I think I should have to explain how to get the 63.94% result first.
63.94% = BINOMDIST(1,2,0.3995,false) + BINOMDIST(2,2,0.3995,false) --Excel formula
.3995 is the prob. to get a X card in one single draw. There are 2 Bernoulli trails, so that corresponds to 2 draws the initial and mulligan. In order to get 63.94% result, we need to always to do perform 2 Bernoulli trails, the initial and mulligan.

 

[zli034]

Of course the results of the simulation would depend entirely on the assumptions you used in building the simulation! If we used your assumption (always mulligan), you'd get one of the cards about 40% of the time; if we used MoD's assumption, 64% of the time. The problem is basic enough that I don't see any value in simulation, especially considering that the simulation would take longer than just solving the problem! (It took me about 8 seconds to solve, but I'm only a modest typist; a good typist could do the 29-character Pari command in half that time. Of course I only made 23 keypresses, thanks to tab completion.)

If we always do mulligan, that is making 2 draws, the initial and mulligan. We expect to see higher probability to obtain X card than only use the single initial draw. More likely to have the X card in 2 draws than in 1 draws. So you say always mulligan could get one X cards about 40% of the time, doesn't make any sense.

 

[DaveC426913]

It is safe to say that the Mulligan is only implemented if the player does not draw card X in the first 7-card draw. Can we all agree this is the intended idea?



Question however: are the two discrete drawings relevant to the probabilistic outcome? It seems to me it's simply a matter of whether card X is in the first 14 cards drawn. Regardless of how the player gets there - i.e. whether he has to Mulligan or not - ultimately he has a probability of starting the game with card X in his hand.

 

[zli034]

It is safe to say that the Mulligan is only implemented if the player does not draw card X in the first 7-card draw. Can we all agree this is the intended idea?



Question however: are the two discrete drawings relevant to the probabilistic outcome? It seems to me it's simply a matter of whether card X is in the first 14 cards drawn. Regardless of how the player gets there - i.e. whether he has to Mulligan or not - ultimately he has a probability of starting the game with card X in his hand.

I totally agree with you. And this post can serve as a counter example to disprove that if we allow mulligan the desired probability is 63.94%

 

[CRGreathouse]

If we always do mulligan, that is making 2 draws, the initial and mulligan. We expect to see higher probability to obtain X card than only use the single initial draw. More likely to have the X card in 2 draws than in 1 draws. So you say always mulligan could get one X cards about 40% of the time, doesn't make any sense.

Sure it makes sense.

"Mulligan if you don't get the card":
~40% chance: Draw card initially, then stop.
~24% chance: Fail initially, then draw the card after the mulligan.
~36% chance: Fail twice.
Result: get card ~64%.

"Mulligan always":
~16% chance: Draw card initially, mulligan, then draw the card.
~24% chance: Draw card initially, mulligan, then fail.
~24% chance: Fail initially, then draw the card after the mulligan.
~36% chance: Fail twice.
Result: get card ~40%.

 

[CRGreathouse]

I totally agree with you. And this post can serve as a counter example to disprove that if we allow mulligan the desired probability is 63.94%

If you didn't shuffle the chance would be 1 - (56 choose 14) / (60 choose 14) =~ 66.54%, but the OP said to shuffle so it's 1 - ((56 choose 7) / (60 choose 7))^2 =~ 63.94%.

 

[zli034]

Sure it makes sense.

"Mulligan if you don't get the card":
~40% chance: Draw card initially, then stop.
~24% chance: Fail initially, then draw the card after the mulligan.
~36% chance: Fail twice.
Result: get card ~64%.

"Mulligan always":
~16% chance: Draw card initially, mulligan, then draw the card.
~24% chance: Draw card initially, mulligan, then fail.
~24% chance: Fail initially, then draw the card after the mulligan.
~36% chance: Fail twice.
Result: get card ~40%.

Your mistake of "Mulligan always" situation is obvious. You should use 1-36%, which 64% is the correct result. Because 36% is the chance not getting the card. 1-prob(not getting the card)= prob(getting the card).

Here is a hint for the other situation, use mulligan to partition the probability space.

 

[CRGreathouse]

Your mistake of "Mulligan always" situation is obvious. You should use 1-36%, which 64% is the correct result. Because 36% is the chance not getting the card. 1-prob(not getting the card)= prob(getting the card).

Your mistake is confusing the cases.

 

[GtB]

Should have known people would get busy on this when Winter break was over.

When we mulligan we take the 7 cards we did not like and shuffle them back into the deck so when we draw our 2nd hand it comes from the original 60 card deck.

Thanks everyone for the help

The game I player is World of Warcraft Card Game btw.