# Calculating Probability of Drawing Card Combos: A Closer Look at Counting

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• quasar987
In summary, when calculating the probability of getting at least 2 cards from category A, 1 card from category B, and 1 card from category C from a deck of 60 cards, the correct formula is (21 choose 2)(4 choose 1)(8 choose 1)(27 choose 3) divided by (60 choose 7). The incorrect formula (21 choose 2)(4 choose 1)(8 choose 1)(56 choose 3) is off by a factor of 3 because it counts the same hands multiple times. The correct formula takes into account the fact that the categories are mutually exclusive and that the order of the cards does not matter.
quasar987
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TL;DR Summary
compute a probability
I have a deck of 60 cards. I draw 7 cards. Among the 60 cards are 21 cards in category "A", 4 cards in category "B" and 8 cards in category "C". The categories are mutually exclusive. I want to the probability that my 7 cards contain at least 2 from category A, 1 from category B and 1 from category C. I count as follows:
$$\frac{\binom{21}{2}\binom{4}{1}\binom{8}{1}\binom{56}{3}}{\binom{60}{7}}$$
This is 0.482328 but computer simulations (as well as a sample of 100 hand drawn manually) show that this is off by a factor of 3. The real answer is close to 16%.

Why is my way of counting no good? Where's that factor of 3 coming from ?

What is the 56 / 3?

FactChecker and malawi_glenn
quasar987 said:
I want to the probability that my 7 cards contain at least 2 from category A, 1 from category B and 1 from category C.
Then you need to add all those possibilities. You have counted "exactly" 2 from A, exactly 1 from B and exactly 1 from C.

Btw, do you mean also "at least 1 from B" and "at least 1 from C"? It is important to be specific.
I will assume that you actually mean "exactly one from B and exactly one from C"

You need to add 3 from A, 1 from B & 1 from C; 4 from A, 1 from B & 1 from C
and so on

Or, just do the complementary probability.
Then you subtract that from 1.

Also the ## \binom{56}{3}## should be ## \binom{27}{3}## because you have only 27 cards that are neither in cathegory A, B, or C.

For instance, the probability that you have exactly 2 A, exactly 1 B and exactly 1 C and then exactly 3 "nor A, B, C" is calculated as
## \dfrac{\binom{21}{2}\binom{4}{1}\binom{8}{1} \binom{27}{3} }{\binom{60}{7}} ##

Last edited:
PeroK and FactChecker
malawi_glenn said:
Btw, do you mean also "at least 1 from B" and "at least 1 from C"? It is important to be specific.
I will assume that you actually mean "exactly one from B and exactly one from C"
That gives a probability of about 10%.

Assuming at least 2 from A, at least 1 from B and at least 1 from C gives a probability of about 16%.

I put all the options on a spreadsheet:

 A B C X n_A n_B n_C n_X n 2​ 1​ 1​ 3​ 210​ 4​ 8​ 2925​ 19,656,000​ 2​ 1​ 2​ 2​ 210​ 4​ 28​ 351​ 8,255,520​ 2​ 1​ 3​ 1​ 210​ 4​ 56​ 27​ 1,270,080​ 2​ 1​ 4​ 0​ 210​ 4​ 70​ 1​ 58,800​ 2​ 2​ 1​ 2​ 210​ 6​ 8​ 351​ 3,538,080​ 2​ 2​ 2​ 1​ 210​ 6​ 28​ 27​ 952,560​ 2​ 2​ 3​ 0​ 210​ 6​ 56​ 1​ 70,560​ 2​ 3​ 1​ 1​ 210​ 4​ 8​ 27​ 181,440​ 2​ 3​ 2​ 0​ 210​ 4​ 28​ 1​ 23,520​ 2​ 4​ 1​ 0​ 210​ 1​ 8​ 1​ 1,680​ 3​ 1​ 1​ 2​ 1330​ 4​ 8​ 351​ 14,938,560​ 3​ 1​ 2​ 1​ 1330​ 4​ 28​ 27​ 4,021,920​ 3​ 1​ 3​ 0​ 1330​ 4​ 56​ 1​ 297,920​ 3​ 2​ 1​ 1​ 1330​ 6​ 8​ 27​ 1,723,680​ 3​ 2​ 2​ 0​ 1330​ 6​ 28​ 1​ 223,440​ 3​ 3​ 1​ 0​ 1330​ 4​ 8​ 1​ 42,560​ 4​ 1​ 1​ 1​ 5985​ 4​ 8​ 27​ 5,171,040​ 4​ 1​ 2​ 0​ 5985​ 4​ 28​ 1​ 670,320​ 4​ 2​ 1​ 0​ 5985​ 6​ 8​ 1​ 287,280​ 5​ 1​ 1​ 0​ 20349​ 4​ 8​ 1​ 651,168​ 62,036,128​

And dividing that total of 62,036,128 by ##\binom {60}{7}## gives a probability of approx ##0.16##.

malawi_glenn
quasar987 said:
Why is my way of counting no good? Where's that factor of 3 coming from ?
You're not off by exactly a factor of 3. You had several mistakes (as explained above) that gave an answer of approximately three times the correct answer.

@PeroK
Yeah I did not have the time to calculate what OP actually meant.

malawi_glenn said:
@PeroK
Yeah I did not have the time to calculate what OP actually meant.
Something to do over breakfast!

malawi_glenn
malawi_glenn said:
Then you need to add all those possibilities. You have counted "exactly" 2 from A, exactly 1 from B and exactly 1 from C.

Btw, do you mean also "at least 1 from B" and "at least 1 from C"? It is important to be specific.
I will assume that you actually mean "exactly one from B and exactly one from C"

You need to add 3 from A, 1 from B & 1 from C; 4 from A, 1 from B & 1 from C
and so on

Or, just do the complementary probability.
Then you subtract that from 1.

Also the ## \binom{56}{3}## should be ## \binom{27}{3}## because you have only 27 cards that are neither in cathegory A, B, or C.

For instance, the probability that you have exactly 2 A, exactly 1 B and exactly 1 C and then exactly 3 "nor A, B, C" is calculated as
## \dfrac{\binom{21}{2}\binom{4}{1}\binom{8}{1} \binom{27}{3} }{\binom{60}{7}} ##
Indeed, I meant "at least 2 from A, at least 1 from B and at least 1 from C", which is why I use the term ## \binom{56}{3}## instead of ## \binom{27}{3}## and don't add all the possibilities. I figured my way of counting bypasses that; the numerator reads: "number of ways of choosing 2 from 21, 1 from 4, 1 form 8 and the last 3 cards can be chosen from any of the 56 cards left in the deck", which includes the case where the last 3 cards are in category A, B or C. But apparently not?? Why? This is what I'm trying to understand.

quasar987 said:
Why? This is what I'm trying to understand.
Let's do a simpler example
Let's assume that we have a jar with 4 white marbles and 6 black marbles.
You take 3 marbles, without putting them back.
What is the probability that you get at least one white?

It is not ## \dfrac{\binom{4}{1}\binom{9}{2}}{\binom{10}{3}} ## it is ## \dfrac{\binom{4}{1}\binom{6}{2} +\binom{4}{2}\binom{6}{1} + \binom{4}{3}\binom{6}{0}}{\binom{10}{3}} ##

But I realize now that I'm counting the same hands multiple times over. For instance my way of couting considers the hands

A1, A2, B1, C1, A3, Y, Z

and

A1, A3, B1, C1, A2, Y, Z

as different. But they're not.

Every couple of years I find myself trying to count something similar and end up falling into the same trap.

malawi_glenn

## What is the purpose of calculating the probability of drawing card combos?

The purpose of calculating the probability of drawing card combos is to determine the likelihood of obtaining a specific combination of cards in a game or activity. This information can help players make strategic decisions and improve their chances of winning.

## What factors influence the probability of drawing card combos?

The factors that influence the probability of drawing card combos include the number of cards in the deck, the number of cards in the player's hand, and the rules of the game. Additionally, the probability can also be affected by any previous cards that have been drawn or discarded.

## How is the probability of drawing card combos calculated?

The probability of drawing card combos is calculated by dividing the number of desired outcomes (i.e. the specific combination of cards) by the total number of possible outcomes. This is typically represented as a fraction or percentage.

## What is the difference between theoretical probability and experimental probability?

Theoretical probability is based on mathematical calculations and assumes that all outcomes are equally likely. Experimental probability, on the other hand, is based on actual data collected from repeated trials and may differ from theoretical probability due to chance or other factors.

## How can understanding probability of drawing card combos be useful in real life?

Understanding probability of drawing card combos can be useful in real life situations such as playing card games, making strategic decisions in business or finance, and even in everyday activities like choosing lottery numbers. It can also help improve critical thinking skills and decision making abilities.

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