Quick Statistics Question

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samshok

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Me and my friends are playing a game.
The game goes like this, 10 people sit in a circle, there are ten papers, one has a lucky number on it. The papers are folded so no one could see what's written inside, and all put in a jar.
The jar is passed around in a circle and everyone reaches to get one paper.
Question: The first one to reach out and the last one to reach out a paper have the same chance to get the lucky number, or does the last one have a better chance?
Thank-you very much.
 
samshok said:
Me and my friends are playing a game.
The game goes like this, 10 people sit in a circle, there are ten papers, one has a lucky number on it. The papers are folded so no one could see what's written inside, and all put in a jar.
The jar is passed around in a circle and everyone reaches to get one paper.
Question: The first one to reach out and the last one to reach out a paper have the same chance to get the lucky number, or does the last one have a better chance?
Thank-you very much.
Click to expand...
I am assuming that:

You want to calculate and show the reasoning behind your answer (instead of one "word" answer and​

once a paper is picked it is NOT replaced back in the jar.​

The first picker has 1/10 chance to pick the lucky paper and 9/10 chance to pick "unlucky" paper

For the 2nd picker to pick to pick the lucky paper

The first picker must choose an unlucky paper (9/10) and​
​
The second picker chooses the lucky paper (1/9)​

continue......
 
they have the same chance

the probability the first one to win is [MATH]\frac{1}{10}[/MATH]
the probability the last one to win is

[MATH]\frac{9}{10} \cdot \frac{8}{9} \cdot \frac{7}{8} \cdot \frac{6}{7} \cdot \frac{5}{6} \cdot \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} \cdot 1 = \frac{1}{10}[/MATH]
 
joshuaa said:
they have the same chance

the probability the first one to win is [MATH]\frac{1}{10}[/MATH]
the probability the last one to win is

[MATH]\frac{9}{10} \cdot \frac{8}{9} \cdot \frac{7}{8} \cdot \frac{6}{7} \cdot \frac{5}{6} \cdot \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} \cdot 1 = \frac{1}{10}[/MATH]
Click to expand...
Please let the student have a chance to DISCOVER the solution
 
Tha
joshuaa said:
they have the same chance

the probability the first one to win is [MATH]\frac{1}{10}[/MATH]
the probability the last one to win is

[MATH]\frac{9}{10} \cdot \frac{8}{9} \cdot \frac{7}{8} \cdot \frac{6}{7} \cdot \frac{5}{6} \cdot \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} \cdot 1 = \frac{1}{10}[/MATH]
Click to expand...

Thank-you very much
 
This is the same as putting a piece of paper in front of each person.
Each person has the same chance of having the winning paper in front of them.
As already stated, p( person i winning) = 1/10, where i= 1,2,3,...,10
 
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