Card Game: prob. of drawing 0 red cards in opening hand?

[nwonder]

I play Magic:The Gathering. Some of you may be familiar with the game, but I don't think you need to be well-versed to undestand the nature of my mathematics question.

In Magic, you typically play with a 60-card deck. You draw 7 cards at the beginning of the game. If you like these 7 cards, you can keep them and begin playing. If you find them unsatisfactory you can "mulligan" by shuffling your opening 7 back into the deck and then drawing 6 cards. You can keep the 6, or you can send them back to get 5. Keep your 5 or mulligan to 4 etc.. You can mulligan as many times as you like.

You are allowed to have only 4 of any one card in your deck. Here are my questions:

Assuming I have 4 red cards and 56 non-red cards in my 60-card deck, what is the probability of drawing 0 red cards in my opening hand of 7 cards? Exactly 1 card? Excactly 2? 3? All 4?

Now, how do I factor in the option of taking a mulligan? What is the probability of drawing at least 1 red card if I want to start the game with at least 5 cards (a maximum of 2 mulligans)? What are the chances I will end up never drawing a red card if I mulligan no more than twice?

It's been a long time since I have taken statistics and probability, so I appreciate any help you can give me! Thanks!

 

[galactus]

Re: Card Game

Assuming I have 4 red cards and 56 non-red cards in my 60-card deck, what is the probability of drawing 0 red cards in my opening hand of 7 cards?

I will get you started on this one for now. Getting late.

Let's do it 2 separate ways. See if we get the same thing.

You have 56 non-reds and 4 reds. Suppose you select a card from the 60 card deck and it isn't red. Then you selected one of the 56 from the 60.

You have 56/60. Continue in this fashion unitl all non-reds are drawn, without replacement .

(56/60)(55/59)(54/58)(53/57)(52/56)(51/55)(50/54)=58565/97527=60%
probability.

Now, combinations. The number of ways to choose 0 cards out of 4 times the number of ways to draw 7 cards out of 56, all divided by the number of ways to draw 7 cards from 60:

\(\displaystyle \L\\\frac{C(4,0)\cdot{C(56,7)}}{C(60,7)}=\frac{58565}{97527}=60%\)

Wow, same answer. The probability of drawing 7 non reds out of the deck is about 60%.

You try it with the others?.

 

[Denis]

Love writing simulators:
2 runs, out of a million each: 600279 and 599946
You're welcome, galactus :wink: