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# I need help calculating the probability of drawing a certain combination of cards from a non-standard deck.?

I've read a lot online about probability, but most of the sites I've seen have only discussed it in regards to a standard deck of playing cards. So a lot of their calculations revolve around ranks and suits.

In my hypothetical deck I have 14 cards, broken into 4 groups: A cards, B cards, C cards, and D cards. The quantities are as follows:

A Cards: 5

B Cards: 4

C Cards: 3

D Cards: 2

How do I calculate the probability of drawing A-B-C in a 5-card hand? I've been racking my brain with this but can't seem to get anywhere and it's hard for me to check if my work is correct. Any help would be much appreciated. Thanks in advance.

Edit: Sorry for being vague, I meant what are the odds of drawing at least one A, one B, and one C in the hand of 5. The other 2 cards don't matter. Thanks for the responses, hope this helps narrow it down.

### 3 Answers

- JamesDLv 47 years agoFavorite Answer
What does drawing A-B-C mean?

1) A, B, C, D, D in that order?

2) One of each A, B and C? (i.e. A, B, C, D, D in any order)

3) A, B, C, X, X in that order where X can be anything?

4) Or at least one A, one B and one C? (i.e. A, B, C, X, X in any order)

I'm assuming you mean drawing a sequence of A => B => C in that order and the other two card's don't matter?

Assume the fourteen cards are distinct ('numbered' 1-14) but still grouped in A-D in some 5-4-3-2 way.

Total outcomes = 14P5 = 14 x 13 x 12 x 11 x 10 = 240240.

Possible outcomes for A=>B=>C:

1) for A-B-C-X-X:

5P3 = 5 x 4 x 3 possible permutations (colloquially 'combinations') to get A-B-C, 11 cards left meaning there are 11P2 = 11 x 10 possible permutations for X-X.

Total = 5 x 4 x 3 x 11 x 10 = 6600.

for 2) X-A-B-C-X and 3) X-X-A-B-C, the total permutations are the same.

So total number of ways to get an A => B => C sequence = 6600 x 3.

Probability is 6600 x 3 / 240240 = 15/182 or 8.24%.

EDIT: For finding odds of getting A-B-C-X-X in any order where X is any card.

There are different numbers of each type cards so it's quite tricky.

There are 10 outcomes possible that are favourable, we find the probability of each one happening and add them all up.

[XCY = X!/(X-Y)!Y! is the number of ways of choosing Y objects from a total of X objects. (where order of the Y objects doesn't matter)]

ABCAA, P(AAABC) = (5C3 x 4 x 3)/14C5

ABCBB, P(ABBBC) = (5 x 4C3 x 3)/14C5

ABCCC, P(ABCCC) = (5 x 4 x 3C3)/14C5

ABCDD, P(ABCDD) = (5 x 4 x 3 x 2C2)/14C5

ABCAB, P(AABBC) = (5C2 x 4C2 x 3)/14C5

ABCAC, P(AABCC) = (5C2 x 4 x 3C2)/14C5

ABCAD, P(AABCD) = (5C2 x 4 x 3 x 2)/14C5

ABCBC, P(ABBCC) = (5 x 4C2 x 3C2)/14C5

ABCBD, P(ABBCD) = (5 x 4C2 x 3 x 2)/14C5

ABCCD, P(ABCCD) = (5 x 4 x 3C2 x 2)/14C5

So total probability is = 1190/14C5 = 1190/2002 = 85/143 = 59.44%

- Divide By ZeroLv 77 years ago
Success = at least one of each of the A, B and C

Probability of success = 1 - (probability of failure)

By inclusion-exclusion, here are the ways for a failure:

(9C5) + (10C5) + (11C5) - 5C5 - 6C5 - 7C5 = 812

(The first 3 terms are the ways to choose five non-A, plus the ways for 5 non-B cards, etc.)

(The last 3 terms were subtracted because they were counted twice by the added terms. They're the ways to choose all C&D, or all B&D, or all A&D.)

# Successful combos = 14C5 - 812 = 1190

Probability = 1190 / 14C5 = 85/143 = 59.44%

But give me a moment to check my answer using a more painstaking method.

Edit -- nevermind, JamesD already did that and got the same answer.

Now you see how inclusion-exclusion can save a lot of work!

- Anonymous7 years ago
Do you mean the chance of drawing an A, then a B, then a C? Or the chance of having an A B and C in your hand?