I'm not sure if I got this right.
So we have a 52 card deck.
We took out 2 cards.
And 2 other random non cards.

So we have 48 cards left in the deck and 11 from those are .

What is probability of drawing 2 or more cards out of 8 draws.

I put it somehow like this:

8C2 (11/48) * (10/47) * (37/46) * 37/45 * (37/44) * (37/43) * (37/42) * (37/41)

and did it for all 8C3, 8C4, 8C5, 8C7, 8C8.

Then I just summed the 8C2 + 8C3 + 8C4 + 8C5 + 8C7 + 8C8

Am I doing this right? Are there any simpler/faster way of doing this?

it's not exactly clear what you're meaning - do you mean that you pick eight cards out of the 48 and you want to know what the chance is that at least two are hearts?

It’s a lot easier if you calculate the probability of getting 0 or 1 heart and then subtracting from 1.0

Pr (0 or 1 heart) =[C(37,8) + C(37,7)*C(11,1)] / C(48,8) = 0.4024

Pr(2 or more hearts) = 1 – 0.4024 =0.5976

Yes, and there are only 11 hears left.

Thanks! Can you explain the whole thing with words a little bit cause I'm not very good at it.

Probability forum is underused imo. Used to be one of the best on the site.

How would we do it for the same problem but this time "at least 3 heart cards"?
I tried and I failed, got too many combos..

Same as statmanhal solution but substract all combinations of two hearts too, which is
https://www.wolframalpha.com/input/?...choose+8%29%29

26%

Ok so 3 or more is ~25% if I got that correctly.

Correct.

Also if you want to be sure about any particular probability in case you made a mistake, and you can program, you can do a little program to run a montecarlo experiment and verify the result. Or if you don't care about accuracy just do the montecarlo experiment.

Explanation. The complement of two of more hearts is drawing 0 or 1 heart. So first calculate that. Then 100% - Pr(0 or 1 heart) is the probability of two or more hearts.

Let E be an event – e.g. drawing x hearts when being dealt y cards

Then for card games,

Pr(E) = Number of ways E can occur/ Total number of possible ways (combos)

If there are n possible outcomes and if r are consistent with event E, then the number of ways for success is

C(n,r) = n! / [r! *(n-r)!] where x! = x*(x-1) * . . . *3*2*1 . e.g. 4! = 4*3*2*1

For our problem there are 37 non-hearts and 11 hearts. For getting 0 hearts when drawing 8 cards, we have to get the number of ways to draw 8 of 37 non-hearts, which is C(37,8).

For 1 heart, we have to draw 7 non-hearts from the 37 non-hearts in the remaining deck and 1 heart from the 11 hearts remaining. This is the C(37,7)*C(11,1) term.

Add the two terms together (there is no overlap so you can do that) and you have the numerator of the probability equation – number of successes/total number of combos.

For the denominator, you are drawing 8 cards from a 48 card stub deck which can happen in C(48,8) ways. Divide numerator by denominator, subtract from 1 and you have your answer.

Sorry for the long explanation with some math. If I didn’t include the math in the way I did, it would be even longer

This. It's like people don't know it exists. People be asking probability questions in BQ and Theory. Still is if I may say so myself! But obv not the same without Bruce. And some of the old posters seem to have retired from 2p2 or something (e.g. I can't remember when I last saw DarkMagnus or spadebidder).

@OP -- what statman said. But I'll comment on your math: You correctly decreased the 11 to 10 and the 48 to 47 to 46 etc, but for some reason didn't decrease the 37 to 36 and so on. Had you done that, your method would be correct (though not the preferred method).

Note that if you combine the numerators into one: 11*10*37*36*35*...
and combine the denominators into one: 48*47*46*...
Then your numerator is (11 P 2)*(37 P 6) and your denominator is (48 P 8).

Altogether it's C(8,2)*11*10*(37 P 6) / (48 P 8)
which is equal to C(11,2)*C(37,6) / C(48,8)

With your way you're counting order in the numerator and denominator and it cancels out. With statman's way, no order is counted anywhere.

More importantly, he subtracted instead of add, which for this problem was clearly easier. Had you asked what's the probability of at least 6 hearts, then adding would be better. Had you asked what's the probability of at most 6 hearts, I'd subtract the probability of at least 7.