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 Author Topic: A Serious Mathematical Problem - Cards PROBABILITY  (Read 148 times)
finaleshot2016
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 August 01, 2018, 09:35:50 AMMerited by paxmao (1)

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
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 August 02, 2018, 04:53:04 AM

Do you put back the each card after you draw it or not?
NeuroticFish
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 August 02, 2018, 09:28:57 AMMerited by paxmao (1)

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
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 August 02, 2018, 11:47:24 PMLast edit: August 09, 2018, 12:46:27 PM by 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.

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