Probability Question.

[ceustice]

If I run 6 of card A, 4 of card B, and 7 of card C what is the chance of having 1 of each in my opening hand in a 60 card deck.
I can't find anywhere I can plug this in or a formula for something along these lines.

[lordofthepit]

If I run 6 of card A, 4 of card B, and 7 of card C what is the chance of having 1 of each in my opening hand in a 60 card deck.
I can't find anywhere I can plug this in or a formula for something along these lines.

Exactly one of each or at least one of each?

In the former case, I believe it's 6*4*7*C(43, 4)/C(60,7). I'm on a phone though so I'll need to do a sanity check.

In the latter, 6*4*7*C(57, 4)/C(60,7).

Edit: there's a mistake with my second expression. I'll fix it when I get a chance.

[Tempus]

If I remember correctly it be

1 - C(43, 7)/C(60, 7)

for the at least one case.
The exactly one the formula should be correct.

[ceustice]

Yes atleast one of each. Thanks guys for your quick replies.
In the first combination one of you used (43,4) and the other did (43,7) what is the second number representing?

Edited twice hoping for further clarification. Thanks for you help in advance.

[lordofthepit]

If I remember correctly it be

1 - C(43, 7)/C(60, 7)

for the at least one case.
The exactly one the formula should be correct.

This isn't correct; it gives the probability of NOT having seven blanks (i.e. at least one A, B, or C).

For example, if your deck had 17 As but no Bs or Cs, you would get the same solution using this expression when clearly there would be no probability if having A, B, and C. Hope that clarifies things!

[sippingcider]

In the latter, 6*4*7*C(57, 4)/C(60,7).
*where the latter was at least one of each

I am trying to follow what you did. C(60,7) is all the number of possibilities of combinations, which makes sense. However, I am a little confused with the numerator. Wouldn't 6*4*7*C(57,4) give you all the possibilities of a starting hand assuming the first three cards were one copy of A, B, and C?
Wouldn't this exclude all the possibilities of drawing those cards in a different order? (like what if the last 3 of the 7 drawn are the cards A, B, and C?).

(I am still learning this subject so you are probably right)

[Phoenix Ignition]

Shameless plug:
http://www.mtgthesource.com/forums/showthread.php?26618-Mathemagic-%96-Breaking-Down-Probabilities

[lordofthepit]

*where the latter was at least one of each

I am trying to follow what you did. C(60,7) is all the number of possibilities of combinations, which makes sense. However, I am a little confused with the numerator. Wouldn't 6*4*7*C(57,4) give you all the possibilities of a starting hand assuming the first three cards were one copy of A, B, and C?
Wouldn't this exclude all the possibilities of drawing those cards in a different order? (like what if the last 3 of the 7 drawn are the cards A, B, and C?).

(I am still learning this subject so you are probably right)

I'm still posting from a phone so my response will be brief.

You are interested in the probability of having a "combo hand", so you can use total combinations of desired hands (with A, B, and C) divided by total combinations of possible hands. Or you can use total permutations desired divided by total permutations possible (a permutation "counts" the ordering in which the cards appear). Since you don't care what order you seven card hand is dealt, it's easier to just calculate using combinations.

Edit: Everything I said above is true, but the expression 6*4*7*C(57, 4) I gave in a previous post was incorrect because it double-counted some of the combinations. I don't know if that's what you were getting at with your question, but please let me know if you did (and your reasoning).

Essentially, the probability of getting in a 7 card hand exactly a copies of card A (of which 6 copies exist), b copies of card B (of which 4 copies exist), and c copies of card C (of which 7 copies exist in a 60 card deck), given the numbers above comes out to the following:

C(6, a) * C(4, b) * C(7, c) * C(43, 7 - a - b - c) / C(60, 7)

Unfortunately, I don't know an easy way of calculating everything with a >= 1, b >= 1, c >= 1 without individually adding up all 34 possibilities. But it turns out there are 39498249 ways of satisfying those conditions, and you can divide by C(60, 7) to find that the probability of getting at least one copy of each card type is approximately 10.22%.

[sippingcider]

Edit: Everything I said above is true, but the expression 6*4*7*C(57, 4) I gave in a previous post was incorrect because it double-counted some of the combinations. I don't know if that's what you were getting at with your question, but please let me know if you did (and your reasoning).

Essentially, the probability of getting in a 7 card hand exactly a copies of card A (of which 6 copies exist), b copies of card B (of which 4 copies exist), and c copies of card C (of which 7 copies exist in a 60 card deck), given the numbers above comes out to the following:

C(6, a) * C(4, b) * C(7, c) * C(43, 7 - a - b - c) / C(60, 7)

Unfortunately, I don't know an easy way of calculating everything with a >= 1, b >= 1, c >= 1 without individually adding up all 34 possibilities. But it turns out there are 39498249 ways of satisfying those conditions, and you can divide by C(60, 7) to find that the probability of getting at least one copy of each card type is approximately 10.22%.

That looks a lot better. Yeah I was just confused woth your last expression, I did not understand it.

It would be interesting to see if anyone knows how to find the probability with at least one of each.

[PirateKing]

Magic Workstation had a probability calculator where you could build a deck and request a set of cards to appear in your opening hand and it would give you percent of that. Even went on to chance of having those cards at turn 2 and onward. Take a look and see if that helps your problem.