Card Games

[Higgs_world]

Hi everyone,
I have no background in Probability and Statistics, but I do understand some Calculus, so I think I will be able to understand your answer if you so choose to answer.

The game is called Magic: The Gathering. In this game you build a deck of sixty cards or more, but sixty is the number I want to know about. In the beginning of the game you draw 7 cards. In the opening hand, it is crucial that I draw at least 2 cards of which there are 23 of in the 60 card deck.

After the opening hand, you draw one card per turn. So the number of cards in the deck is decreasing. I need a formula that takes this into account.

I am sure this question has been answered already, so links are much appreciated.

Thanks everyone

[matt grime]

formula for what? there is no question asked here is there?

[shmoe]

So in the opening hand you want any 2 (or more) cards from this set of 23? I'll work out the probability this happens for you. \binom{n}{k}=\frac{n!}{k!(n-k)!}=the number of ways to select a subset of size k from n distinct elements.

The total number of hands in the first round is \binom{60}{7}=386206920.

To find the number of hands with 2 or more, we'll find the number with none and the number with 1 and subtract from the total.

To have none, all 7 cards must be taken from the 37 others, so \binom{37}{7}=10295472

To have 1, we take 6 cards from the 37, and 1 card from the 23, so
\binom{37}{6}\binom{23}{1}=(2324784)(23)=53470032

So the total number of first round hands that give 2 or more of the 23 is 386206920-10295472-53470032=322441416. Divide by the total number of hands gives the probability of a hand with 2 or more from the 23:
322441416/386206920\approxeq .8345

You can figure out the probability that it takes you until round 2, 3, etc. to get your 2 of the 23. Before I go on, is this the sort of thing you were after?

edit-nuts, is LaTeX busted for everyone?

[arildno]

shmoe is right; the probability of getting 2(or more) land cards on your opening hand is approx. 83.5%
It might be of some interest to work out the probability given the same proportions but infinite deck size.
In that case, the probability is approx. 81.8%
Hence, it doesn't matter too much to use constant probabilities independent of the fact that the deck size decreases.

[Higgs_world]

Schmoe, you hit it right on the money. I took an error analysis class in Physics and we were shown how to find the probabilities of throwing dice three dice, but either I did not apply the equation correctly or it does work the same.