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Title: A Serious Mathematical Problem - Cards PROBABILITY Post by: finaleshot2016 on August 01, 2018, 09:35:50 AM 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) + P = 77%(13C3)(4C2)(4C2)(4C2) x (10C3)(4^3) + (13C4)(4C2)(4C2)(4C2)(4C2) x (9C1)(4) _______________________________________________________________________________ ______ 52C9 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. Title: Re: A Serious Mathematical Problem - Cards PROBABILITY Post by: benjamin07 on August 02, 2018, 04:53:04 AM Do you put back the each card after you draw it or not?
Title: Re: A Serious Mathematical Problem - Cards PROBABILITY Post by: NeuroticFish on August 02, 2018, 09:28:57 AM 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. Title: Re: A Serious Mathematical Problem - Cards PROBABILITY Post by: finaleshot2016 on August 02, 2018, 11:47:24 PM 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 |