You have an urn with 6 numbered balls and you pull out 3 (with replacement). What is the chance of getting 3 different numbers when the order doesn't matter?

I have two solutions that both seem reasonable.

  1. Looking at chances:
    Chance to pick the first ball: 6/6
    Chance to pick the second ball: 5/6
    Chance to pick the third ball: 4/6
    i.e. 20/36

  2. Looking at possibilities (combinatorics):
    To get 3 different balls out of a urn with 6 when no number can be repeated and the order doesn't matter is the same as a lottery.

Therefore there are ${6 \choose 3}$ = 20 possibilities.
Looking at all possibilities, I have 6*6*6 = 216. i.e. 20/216

Which one is wrong and why?

  • $\begingroup$ Hint: how would the original question be different if order did matter? $\endgroup$ – Marc van Leeuwen Dec 17 '11 at 4:47

The $20$ possibilities are without regard to order, while the $6*6*6$ considers the order you picked the balls. This is exactly the factor $6$ between the answers. (First answer: 36; multiply by 6, equals the second answer: 216) Your $20/36$ is correct.

  • $\begingroup$ But then it should be 56 possibilities, shouldn't it? (6+3-1) over 3 or adding 36 possibilities with double or tripple numbers to the 20 I have already. Therefore I would end up at 20/56. $\endgroup$ – Sai Dec 16 '11 at 22:28
  • $\begingroup$ Each of the $20$ sets can be ordered in $6$ ways. So there are $6*20=120$ ways to pick three different balls of $6$ with replacement, giving a chance of $120/216$ $\endgroup$ – Ross Millikan Dec 16 '11 at 22:39
  • $\begingroup$ Okay, I understand that now, thank you! But still I wonder why it is not 56 possibilities for 3 balls out of 6 without order with replacement and double/triple numbers allowed (i.e. my denomninator) $\endgroup$ – Sai Dec 16 '11 at 22:47
  • 1
    $\begingroup$ There are $20$ combinations with the three distinct, $30$ with two of one and one of another, and $6$ with all the same. But the chance of drawing each is not the same. Because the $20$ with three distinct balls can come in six orders they are more likely. $20*6+30*3+6=216$ $\endgroup$ – Ross Millikan Dec 16 '11 at 22:51
  • $\begingroup$ Aaah, that is my problem! :) Thank you very much! $\endgroup$ – Sai Dec 16 '11 at 22:53

First you pick one ball.

Then you pick a second ball. With 5/6 probability it is different from the first ball.

Then you pick the third ball. With 4/6 probability it is different from the first two.

5/6 * 4/6 = 20/36.


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