# Card Games

1. Sep 15, 2004

### 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.

Thanks everyone

2. Sep 15, 2004

### matt grime

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

3. Sep 15, 2004

### 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?

Last edited: Sep 15, 2004
4. Sep 16, 2004

### 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.

5. Sep 21, 2004

### 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.