

Subject:
Probability of pulling a #1 pingpong ball out of a bag without replacement
Category: Science > Math Asked by: justaguy1969ga List Price: $10.00 
Posted:
30 Aug 2006 11:06 PDT
Expires: 29 Sep 2006 11:06 PDT Question ID: 760847 

There is no answer at this time. 

Subject:
Re: Probability of pulling a #1 pingpong ball out of a bag without replacement
From: research_helpga on 30 Aug 2006 11:54 PDT 
By definition, everyone has the same odds of pulling a certain ball if 16 people each pick 1 ball out of 16 without replacement. What kind of answer are you looking for? 
Subject:
Re: Probability of pulling a #1 pingpong ball out of a bag without replacement
From: pinkfreudga on 30 Aug 2006 11:59 PDT 
Researchers and Commenters may want to read some of the comments posted to the earlier version of this question: http://www.answers.google.com/answers/threadview?id=759701 
Subject:
Re: Probability of pulling a #1 pingpong ball out of a bag without replacement
From: jack_of_few_tradesga on 30 Aug 2006 12:47 PDT 
Justaguy, I guess you're unhappy with the comments on the former question... Here is the math showing that everyone has exactly the same odds: 1st person chooses 1 ball from 16. There is a 100% chance that the winning ball is in the bag. 1 / 16 * 100% = 6.25% chance of winning 2nd person chooses 1 ball from 15. There is a 93.75% chance that the winning ball is in the bag. (100%  6.25% = 93.75%) 1 / 15 * 93.75% = 6.25% chance of winning 3rd person 1 / 14 * 87.5% = 6.25% 4th 1 / 13 * 81.25% = 6.25% 5th 1 / 12 * 75% = 6.25% 6th 1 / 11 * 68.75% = 6.25% 7th 1 / 10 * 62.5% = 6.25% 8th 1 / 9 * 56.25% = 6.25% 9th 1 / 8 * 50% = 6.25% 10th 1 / 7 * 43.75% = 6.25% 11th 1 / 6 * 37.5% = 6.25% 12th 1 / 5 * 31.25% = 6.25% 13th 1 / 4 * 25% = 6.25% 14th 1 / 3 * 18.75% = 6.25% 15th 1 / 2 * 12.5% = 6.25% 16th 1 / 1 * 6.25% = 6.25% Notice that the sooner you draw, the more likely the winning ball is still in the bag... but the sooner you draw, the more balls you have to choose from. These positive and negative influences exactly cancels eachother out and each person is left with exactly the same odds of winning. 
Subject:
Re: Probability of pulling a #1 pingpong ball out of a bag without replacement
From: myoaringa on 30 Aug 2006 12:48 PDT 
Of course, the odds are the same, but the moment the #1 ball has been drawn (greater than 50% chance this will happen by the time the 9th ball is drawn), the rest of the participants will feel cheated. This gets back to your idea about randomly choosing the participants' order of drawing, but that only eliminates the rush to draw at the start when all know the #1 ball is still available  and puts to question the randomness of that step. This whole problem is avoided by running your lottery in the traditional manner (I should have thought of this before): Each person draws a ball (or buys a ticket), AND THEN the winning ball or ticket is drawn. My suggestion to your other question does this, but a better alternative to that would be to have have two nonparticipants draw: the first drawing from one urn the color or symbol, and then the second, drawing the symbol or color. 
Subject:
Re: Probability of pulling a #1 pingpong ball out of a bag without replacement
From: saltytowersga on 31 Aug 2006 04:29 PDT 
If the #1 ball is chosen early, the probability for the rest is nil. If it is not chosen right down to the end, the probability is 100% for the final person. The maths for doing probabilites goes like this. If you use the word AND you use X in the calculation. If you use the word OR then you use + in the calculation. For example. To calculate on the 3rd person being the one to pull #1 ball. 15/16(any non #1) X 14/15(any non #1) X 1/14 (the #1)= 6.25% chance of that one person picking that one significant ball. It is also 6.25% for the first person to choose the one significant ball. So in effect (at the outset)each person has the SAME chance at pulling #1 ball regardless of where they stand in the line. 
Subject:
Re: Probability of pulling a #1 pingpong ball out of a bag without replacement
From: celinga on 04 Sep 2006 08:01 PDT 
There are two differnt answers to this question. First,everyone can know who get the exact ball Numble for each other, so the probability of each one people get the #1 ball is: the 1st people P=1/16 the 2nd people P=(15/16)*(1/15)=1/16,because if the second people pull the right #1 ball,the first people must get one of the #2#16 ball. so the probability of the first people don't get the #1 ball is 15/16; and the probability of the 3th people pull the #1 ball can be calculated,which also is (14/16)*(1/14)=1/16; ....., in the same way we can caculate the probability of the other 13 people who get the #1 ball are all 1/16. the second way is ......,to be continued. 
Subject:
Re: Probability of pulling a #1 pingpong ball out of a bag without replacement
From: jack_of_few_tradesga on 05 Sep 2006 04:58 PDT 
**on the edge of his seat** Celin, I can't wait for the exciting conclusion! 
Subject:
Re: Probability of pulling a #1 pingpong ball out of a bag without replacement
From: celinga on 05 Sep 2006 07:35 PDT 
continue,the second way we should suppose everyone can?t know who get the exact ball numble for each other, and the situation equal to the sixteen people get the ball in the same time, so we can calculate the probability of the each people get the #1 ball. Actually, the #1 ball must be pulled one of the sixteen people, let p(i) equal to the probability of the ith people get the #1 ball, so we can know p(1)+p(2)+?+p(16)=1. on the other hand, we can easily realize that p(i) is equal to each other according to the suppose condition, so p(i)=1/16. Actually, there have one exact answer, but we must realize the problem on different way. That is very important for solving problems. Over. 
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