My Math Forum Probability problem
 User Name Remember Me? Password

 Probability and Statistics Basic Probability and Statistics Math Forum

 March 14th, 2017, 09:46 AM #1 Newbie   Joined: Mar 2017 From: Netherlands Posts: 1 Thanks: 0 Probability problem Hello everyone, This is my first post here. I'm wondering how to solve the following problem without bruteforcing my way out of it. Let's assume we have a box with 16 balls, numbered 1 to 16. We're going to draw balls out of the box, write down the number and put it back in. I want to know what the odds are of getting 3 or less unique balls, which can be calculated by odds(3orlessuniqueballs) = (1 - odds(5uniqueballs) - odds(4uniqueballs)) 5 unique ones if ofcourse 1*15/16*14/16*13/16*12/16=0.5 for 4 unique ones I thought there was a way to calculate the odds of for example unique/unique/unique/unique/copy and multiply that by the amount of possible orders you can draw 4 unique balls The problem is the different orders don't have the same odds, which makes sense since it's less likely you draw the copy on draw 2 than on draw 5. odds(unique/u/u/u/copy) =1*15/16*14/16*13/16*4/16 = 0.17 odds(unique/u/u/copy/u) =1*15/16*14/16*3/16*13/16 = 0.13 This makes me think the only way to calculate odds(4uniqueballs) is to manually calculate this for all 4 orders, am I correct? (Not very much work but just doesn't feel very smart) Thanks, VanLunturu
 March 15th, 2017, 05:06 PM #2 Newbie   Joined: Mar 2017 From: Usa Posts: 5 Thanks: 0 Are you drawing 5 balls each time?
March 15th, 2017, 06:18 PM   #3
Senior Member

Joined: May 2016
From: USA

Posts: 1,151
Thanks: 479

Quote:
 Originally Posted by VanLunturu Hello everyone, This is my first post here. I'm wondering how to solve the following problem without bruteforcing my way out of it. Let's assume we have a box with 16 balls, numbered 1 to 16. We're going to draw balls out of the box, write down the number and put it back in. I want to know what the odds are of getting 3 or less unique balls, which can be calculated by odds(3orlessuniqueballs) = (1 - odds(5uniqueballs) - odds(4uniqueballs)) 5 unique ones if ofcourse 1*15/16*14/16*13/16*12/16=0.5 for 4 unique ones I thought there was a way to calculate the odds of for example unique/unique/unique/unique/copy and multiply that by the amount of possible orders you can draw 4 unique balls The problem is the different orders don't have the same odds, which makes sense since it's less likely you draw the copy on draw 2 than on draw 5. odds(unique/u/u/u/copy) =1*15/16*14/16*13/16*4/16 = 0.17 odds(unique/u/u/copy/u) =1*15/16*14/16*3/16*13/16 = 0.13 This makes me think the only way to calculate odds(4uniqueballs) is to manually calculate this for all 4 orders, am I correct? (Not very much work but just doesn't feel very smart) Thanks, VanLunturu
I am guessing that you are drawing FIVE balls (you do not say so).

If a ball with the same number appears 5 times, then there is no ball that appears only once. Failure.

If a ball with one number appears exactly thrice and a ball with a different number appears exactly twice, then there is no ball that appears only once. Failure.

Every other case is a success.

If a ball with the same number appears exactly four times, then the remaining ball has a number that appears but once.

If a ball with the same number appears exactly thrice and the other two balls differ then two different numbers appear just once.

If a ball with the same number appears exactly twice and the other three draws show the same number, that was our second case of failure and has already been counted. Otherwise, there is either one number that appears just once or three numbers that appear just once, which are both successes.

If a ball with the same number appears exactly once, that is a success.

So we need to compute

1 - P(five identical balls) - P(three identical balls and two identical balls).

$1 - \left \{ \dbinom{16}{1} \left ( \dfrac{1}{16} \right )^5 \right \} - \left \{ \dbinom{16}{2} \left ( \dfrac{1}{16} \right )^3 \left ( \dfrac{1}{15} \right )^2 \right \} =$

$1 - \left ( \dfrac{16!}{(1! * (15!)} * \dfrac{1}{16^5} \right ) - \left ( \dfrac{16!}{(2!) * (14!)} * \dfrac{1}{16^3} * \dfrac{1}{15^2} \right ) =$

$1 - \left ( \dfrac{16}{16^5} \right ) - \left ( \dfrac{1}{2} * \dfrac{16}{16^3} * \dfrac{15}{15^2} \right ) = 1 - \dfrac{1}{16^2} * \left \{ \dfrac{1}{16^2} + \dfrac{1}{2} * \dfrac{1}{15} \right \} =$

$1 - \dfrac{1}{256} * \left \{ \dfrac{1}{256} + \dfrac{1}{30} \right \} = 1 - \dfrac{1}{256} * \left \{ \dfrac{30}{7680} + \dfrac{256}{7680} \right \} =$

$1 - \dfrac{1}{256} * \dfrac{286}{7680} \approx 99.985\%.$

Please check my work; I am very sleepy. But it appears that it is practically a certainty that at least one number will appear only once.

 Tags probability, problem

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post szz Probability and Statistics 3 March 6th, 2015 11:50 AM bsjmath Advanced Statistics 2 March 1st, 2014 02:50 PM ystiang Advanced Statistics 1 February 13th, 2014 04:27 PM najaa Probability and Statistics 3 April 19th, 2012 08:46 PM power-of-2 Probability and Statistics 0 October 5th, 2011 11:43 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2018 My Math Forum. All rights reserved.