If I placed sixteen numbered ping-pong balls in a bag, and sixteen
people pulled one ball out of the bag.  Is the probability for getting
the #1 ball the same for everybody?

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Clarification of Question by justaguy1969-ga on 30 Aug 2006 11:11 PDT

Is there an advantage of pulling the first ball or early, compared to late or last?

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Clarification of Question by justaguy1969-ga on 30 Aug 2006 12:49 PDT

This is ultimately for a fantasy football draft.  I posted a question
about this very recently.  There are individuals in this group who
believe there is an advantage to drawing first or early, because once
the #1 ball is gone, then there is no chance to get that ball.  My
understanding, along with most in the group, is that before any ball
is drawn, everyone has a one in sixteen chance to get any ball. 
Obviously the odds would change as balls were taken out of the bag.  I
just wanted to get a second opinion on this to show them.

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?

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

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.

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 non-participants draw:
the first drawing from one urn the color or symbol, and then the
second, drawing the symbol or color.

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.

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.

**on the edge of his seat**

Celin, I can't wait for the exciting conclusion!

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.