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

[quasar987]

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 ?

 

[DaveC426913]

What is the 56 / 3?

 

[malawi_glenn]

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
$\dfrac{\binom{21}{2}\binom{4}{1}\binom{8}{1} \binom{27}{3} }{\binom{60}{7}}$

 

[PeroK]

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 $0.16$.

 

[PeroK]

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.

 

[malawi_glenn]

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

 

[PeroK]

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

Something to do over breakfast!

 

[quasar987]

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.

 

[malawi_glenn]

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 $\dfrac{\binom{4}{1}\binom{6}{2} +\binom{4}{2}\binom{6}{1} + \binom{4}{3}\binom{6}{0}}{\binom{10}{3}}$

 

[quasar987]

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.