A Serious Mathematical Problem - Cards PROBABILITY

[finaleshot2016]

If you will draw 9 cards, what is the probability of having "at least" 2 pairs?

I just want to clarify my answer if it's correct.
I use combination to solve the probability

My solution:

P = P of 1 pair + P of 2 pairs + P of 3 pairs + P of 4 pairs
P =

(13C1)(4C2) x (12C7)(4^7)  +   (13C2)(4C2)(4C2) x (11C5)(4^5)   +  
 (13C3)(4C2)(4C2)(4C2) x (10C3)(4^3)   +   (13C4)(4C2)(4C2)(4C2)(4C2) x (9C1)(4)  
_______________________________________________________________________________ ______
52C9

P = 77%

What's yours?



I hope there are mathematicians who can answer my probability problem. I'm struggling on this problem and makes me think all the day to look out the right answer.

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[benjamin07]

Do you put back the each card after you draw it or not?

[NeuroticFish]

I am not that good at math, but maybe you should check the numbers at http://www.durangobill.com/Poker_Probabilities_8_Cards.html (9 cards, no wildcard)

Just I can't tell how good those numbers also are. 
I made a small test program calculating the odds to get the chance for (2 pairs) OR (3 of a kind) OR (Full House) OR (4 of a kind)
For hand size 5 I get the same result as the sum from http://www.durangobill.com/Poker_Probabilities_5_Cards.html, 7.03%
For hand size 9 the sum is 55.46% and my result is 67.39% 


Another idea could be to try and count the number of possibilities for "worse than 2 hands" and see from there. However, I hope that its main page (http://www.durangobill.com/Poker.html) gives you some details and help you find out if your numbers are good.

[finaleshot2016]

Do you put back the each card after you draw it or not?

Nah, it won't. If you are playing poker then it will look like that, the difference is, 9 cards will draw.

I am not that good at math, but maybe you should check the numbers at http://www.durangobill.com/Poker_Probabilities_8_Cards.html (9 cards, no wildcard)

Just I can't tell how good those numbers also are.  
I made a small test program calculating the odds to get the chance for (2 pairs) OR (3 of a kind) OR (Full House) OR (4 of a kind)
For hand size 5 I get the same result as the sum from http://www.durangobill.com/Poker_Probabilities_5_Cards.html, 7.03%
For hand size 9 the sum is 55.46% and my result is 67.39%  


Another idea could be to try and count the number of possibilities for "worse than 2 hands" and see from there. However, I hope that its main page (http://www.durangobill.com/Poker.html) gives you some details and help you find out if your numbers are good.

Thanks man! My answer is correct, it's 77%.

You should add all the probability of getting 1 pair, 2 pairs, 3 pairs until 4 pairs. Why? because the problem stated "atleast" means there are chances that 2pairs-4pairs will show when we draw 9 cards. This is my Prelims Exam so I just wanted to know if my answer is correct and yeah my solution is right!

This is a proof also that I studied and memorized it until I got home    Because some people might judge me.

PS: LOCKED TOPIC