A friend posed a riddle. There are two boxes, one with ten tickets, and one with one hundred tickets. There is one lucky ticket in each. Do you choose one chance with the ten or ten chances with the hundred?

 

I reasoned that with the ten out of one hundred, a person's chances are better because as you're picking tickets out of the box, there are less tickets in the box. Therefore the probability is 1/100 + 1/99 + 1/98 + 1/97 +... + 1/91 for a 10.48% chance of getting the lucky ticket.

 

My friend says no, the probabilities are equal.

 

Which one of us is correct?

 

EDIT: I'm wrong because I know that this series doesn't add up to one, but I still have the feeling that since you're picking without replacement, your odds would be better.

Edited August 19, 201312 yr by kisai

Ok the way to look at this

 

chance of you losing in the ten ticket game - 9 tickets lose 1 wins 9/10 chance = 0.9

 

chance of you losing in the hundred ticket game

1st draw 99 tickets lose and 1 tickets wins ie lose = 99/100 AND

2nd draw 98/99 AND

3rd draw 97/98 AND...

10th draw 90/91

 

you need to lose each time to lose over-all so you multiply

 

chance of losing (100 ticket game) = 99/100 * 98/99 * 97/98...* 90/91

 

to make the arithmetic easier put all the tops together and all the bottoms

 

[latex]\frac{99*98*97*96*95*94*93*92*91*90}{100* 99*98*97*96*95*94*93*92*91}[/latex]

 

this makes it obvious you can do some cancelling - ie the 99 on the top cancels the 99 on the bottom. In fact all cancel except the 100 bottom left and 90 top right giving

 

chance of losing (100 ticket game) = 90/100 = 0.9

 

Your friend is quite correct.

 

Thank you very much for the explanation!