Let's say I have a bag with 8 numbered balls inside, and I'm drawing them 1 by 1 until I get ball #1. What's the chance of me drawing ball #1 as last one? I was under the impression that the chances for me to draw it last is

$$\frac18* \frac17*\frac16* \frac15* \frac14* \frac13*\frac12=\frac1{40320}$$

since I'm drawing them 1 by 1, so 1st time I draw, I have 1/8 chance to pull #1, next draw I have 1/7, and so on?

Is that wrong? A friend of mine says it's just 1/8, am I misunderstanding this?

  • 5
    $\begingroup$ One ball has to be last. Each has the same probability. $\endgroup$
    – shawnt00
    Jun 27 at 20:02
  • 1
    $\begingroup$ There are only 8 possible options for "last ball", so it is just plain 1/8. How you got to the last ball doesn't affect it. What you have describe is the probability of drawing the entire sequence of 8 balls in that specific order. $\endgroup$
    – Nelson
    Jun 28 at 2:16

3 Answers 3


Your friend is correct. The problem is identical to choosing the order of the balls coming out and from symmetry, ball #1 can be in any place with the same probability.

If you want to compute it in your way, you need to NOT DRAW the ball in the first 7 draws:

$$\tfrac{7}{8}\times \tfrac{6}{7} \times \ldots \tfrac{1}{2}=\tfrac{1}{8}$$

  • $\begingroup$ So If I wouldn't draw the ball, it'd be $1/8*1/8*1/8$ and so on until I draw it? $\endgroup$
    – Slava K.
    Jun 27 at 8:55
  • $\begingroup$ If you draw the balls with replacement, then it's like tossing a die. The probability to get ball #1 only on the 8th roll and not before is $(\tfrac{1}{8})^7 \cdot \tfrac{1}{8}$. $\endgroup$
    – YJT
    Jun 27 at 8:57
  • 2
    $\begingroup$ The term $1/8*1/8*.... $ can be used to compute the probability of getting ball #1 everytime. $\endgroup$
    – YJT
    Jun 27 at 8:58
  • $\begingroup$ Wait, now I'm confused I'm trying to get probability to get ball #1 only on last roll and not before, how is it different from what you typed about rolling the die? $\endgroup$
    – Slava K.
    Jun 27 at 10:27
  • 1
    $\begingroup$ If you have replacement, it is not different than rolling a die because you have the same set of options in each roll. This distribution is called geometric. If you do not replace the balls, then the problem is equivalent to arranging them in a line, and the location of ball #1 is uniform. $\endgroup$
    – YJT
    Jun 27 at 10:33

The probabilities that the last ball is $k$ are all equal, and their sum is $1$,

Therefore the probability is $\frac18$.


@YJT's answer is good, but it seems like it could use a little more explanation. In your proposed formula:

$$\frac18* \frac17*\frac16* \frac15* \frac14* \frac13*\frac12=\frac1{40320}$$

What do the terms stand for and why are you multiplying them? The first one is correct, you have a $\frac18$ chance of pulling your chosen ball first. But if you think about it, if you all 8 balls out then you have a $\frac11$ chance of getting your ball at some point, right? Think of it as if the balls are in a random order. The chance that the chosen ball will be in any of the 8 positions should be equal, so the odds should be $\frac18$ for each position. There is no reason it should be more favored to be in any position than another.

Going back to your formula... After the first ball is pulled, there is a 1/8 chance you got the chosen ball, and a 7/8 chance you didn't, so the odds of getting it on your second attempt is $\frac78*\frac17$. The sevens cancel so you have a $\frac18$ chance of getting it second. Therefore there is a $\frac68$ chance you haven't gotten it in the first two pulls. Since you have 6 balls left, there is a $\frac16$ chance you pull the right one, but that is only in the $\frac68$ remaining probability, so $\frac68 * \frac16$ is again $\frac18$

If you keep going to the end, there is a 1/8 chance you haven't pulled it in the first 7 attempts, and you have a 1/1 chance of pulling it with only the one ball remaining, so that's the final 1/8.


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