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 05-18-2017, 12:43 PM #1 pokerfresse newbie   Join Date: Dec 2015 Posts: 42 How to calculate winning 4 of 5 times with AA vs. KK Hello, I don't want this thread to be another bad beat thread, so I've made this new thread and reformulated my question: How can I calculate the probability to win 4 of 5 with AA vs. KK? If am only interested in this short term (5 times). Do I need the bionomal coefficient for the calculation? I'm sure i had this already at college or university and that it's pretty simple
 05-18-2017, 02:40 PM #2 statmanhal Pooh-Bah   Join Date: Jan 2009 Posts: 3,894 Re: How to calculate winning 4 of 5 times with AA vs. KK To win AT LEAST 4 out of 5 times if the constant win probability is 80% : Exactly 4 wins: 5*(0.8^4)*0.2 = 0.4096 (5 is the number of ways (combos) for 4 wins out of 5 trials) 5 wins: 0.8^5 = 0.3277 Sum = 0.7373 Yes, these are binomial probabilities.
05-18-2017, 03:13 PM   #3
pokerfresse
newbie

Join Date: Dec 2015
Posts: 42
Re: How to calculate winning 4 of 5 times with AA vs. KK

Quote:
 Originally Posted by statmanhal To win AT LEAST 4 out of 5 times if the constant win probability is 80% : Exactly 4 wins: 5*(0.8^4)*0.2 = 0.4096 (5 is the number of ways (combos) for 4 wins out of 5 trials) 5 wins: 0.8^5 = 0.3277 Sum = 0.7373 Yes, these are binomial probabilities.
Thank you!

And what would be the calc. for winning max. one time with AA vs. KK in 9 cases? So when I have 9 times a prob. to win of 0,8 but i win only 1 time in 9 cases?

EDIT:
Which of these formulas is it

Last edited by pokerfresse; 05-18-2017 at 03:19 PM.

 05-18-2017, 04:03 PM #4 statmanhal Pooh-Bah   Join Date: Jan 2009 Posts: 3,894 Re: How to calculate winning 4 of 5 times with AA vs. KK Those formulas do not directly address the question you have. You can get the answer yourself with a little thought and math. For 0 successes in 9 trials with success prob. = 0.8, that means you fail 9 of 9 times. Fail prob = 0.2. Fail 1 of 1 = 0.2: Fail 2 of 2 = 0.2*0.2 = 0.2^2=0.4 . . . fail 9 of 9 = 0.2^9 =.000000512 For exactly one success, consider the case of 8 fails first and then a success. Prob = (0.2^8) * 0.8 =0.000002048. But there are a total of 9 positions for the success, so the total is 9*(0.2^8)*0.8 =0.000018432. Since you asked for a max of one success,you add the two results together to get 0.000018944. The best way to do these problems is learn the binomial distribution. There are free calculators on the Internet. You can also use Excel’s BINOMDIST. If you are looking to show a site is rigged disregard my answer
05-18-2017, 04:13 PM   #5
pokerfresse
newbie

Join Date: Dec 2015
Posts: 42
Re: How to calculate winning 4 of 5 times with AA vs. KK

Quote:
 Originally Posted by statmanhal Those formulas do not directly address the question you have. You can get the answer yourself with a little thought and math. For 0 successes in 9 trials with success prob. = 0.8, that means you fail 9 of 9 times. Fail prob = 0.2. Fail 1 of 1 = 0.2: Fail 2 of 2 = 0.2*0.2 = 0.2^2=0.4 . . . fail 9 of 9 = 0.2^9 =.000000512 For exactly one success, consider the case of 8 fails first and then a success. Prob = (0.2^8) * 0.8 =0.000002048. But there are a total of 9 positions for the success, so the total is 9*(0.2^8)*0.8 =0.000018432. Since you asked for a max of one success,you add the two results together to get 0.000018944. The best way to do these problems is learn the binomial distribution. There are free calculators on the Internet. You can also use Excel’s BINOMDIST.

Quote:
 If you are looking to show a site is rigged disregard my answer
Haha no Just curiosity

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