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There are 100 numbered balls in an urn. We make 100 random draws with replacement. Of course, we can not expect to draw every number exactly once, there will be multiples. What is the expected value of multiply drawn numbers?

I asked this question before (expect number of multipe draws), but the obtained answer does not seem to be correct.

For a deeper understanding think of just 2 balls and 2 draws: Then the probability of multiples is 50%. The answers given so far focussed on the probability that a fixed number is drawn at least twice, which is 25%.

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  • $\begingroup$ Not sure I see any problem with the other solution, could you elaborate? Each ball has a rough $25\%$ chance of being chosen multiply...so I expect to see roughly $25$ balls more than once. In the two ball case, that means I expect to see $\frac 14\times 2=\frac 12$ a ball more than once. this is true! With equal odds I see no ball twice or one ball twice, so the expectation is $\frac 12$ a ball. $\endgroup$ – lulu May 29 '16 at 14:26
  • $\begingroup$ @lulu: Now do you know the general formula to answer the case of 100 balls? $\endgroup$ – scenario May 31 '16 at 17:53
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    $\begingroup$ Judging from your response to the posted solution, I believe there is a confusion in meaning. What I, and I believe the other solvers, are counting is the expected number of values which will be drawn more than once. Thus, in your two ball case, if you draw $\#2$ twice you have seen exactly one value multiply. Accordingly, still in the two ball case, you see $0$ multiple-values half the time and $1$ multiple value half the time, so the answer is that you expect to see $\frac 12$ a ball multiple times. If you mean something else, you need to clarify it. $\endgroup$ – lulu May 31 '16 at 18:01
  • $\begingroup$ to be clear: surely you aren't actually just after the probability that there will be multiple draws? That is effectively $1$ for large $n$. Easy to compute...it's just $1-\frac {n!}{n^n}$. $\endgroup$ – lulu May 31 '16 at 18:09
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    $\begingroup$ Well, I don't see anything wrong with with the other solvers have said. The expected number of balls that are seen exactly $i$ times is $e_i=100\times \binom {100}i\times \left( \frac 1{100}\right)^i\times \left(\frac {99}{100}\right)^{100-i}$. So, the expected number that are seen multiple times is $\sum_{i=2}^{100}e_i\sim 26.4328$ $\endgroup$ – lulu Jun 1 '16 at 19:26
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The chance that a particular number is not drawn at all is $(\frac{99}{100})^{100}$.

The chance that a particular number is drawn exactly once is $\binom{100}{1}\frac{1}{100}(\frac{99}{100})^{99}$.

Now you can find the chance that a particular number is multiply drawn. (Interestingly, this never differs by much from $\frac14$, which is the value in the case of just $2$ balls).

Multiply by $100$ to get the expected count of such numbers (thanks to additivity of expectation).

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  • $\begingroup$ Indeed, let "p" be the probability that any given number is drawn more than once. Then, letting I_n be the indicator function of the event "the n-th number is drawn more than once" (for n=1,...,100) then the total number of balls that are drawn more than once is I_1+I_2+...+I_100. Using the linearity of the expectation, this sum has expected value p*100. $\endgroup$ – karlahrnndz May 31 '16 at 2:48
  • $\begingroup$ @Henning Makholm: The answer for 2 balls is 1/2, not 1/4. Your answer is a mere repetition of the answer given to my first attempt. Nobody seems to understand what I mean .... See the third paragraph of my question, we have the outcomes 1 & 1, 1 & 2, 2 & 1, 2 & 2, so there is a 50% chance for multiply drawn balls. Your answer does not adress this situation. $\endgroup$ – scenario May 31 '16 at 17:50
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    $\begingroup$ @scenario: $\frac14$ is the the chance that a particular ball is drawn more than once is $\frac14$ -- that is, the chance that ball number 1 is drawn more than once is $\frac14$, and the chance that ball number 2 is drawn more than once is $\frac14$. $\endgroup$ – Henning Makholm May 31 '16 at 18:27
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    $\begingroup$ @scenario: As written, your question is not about the probability that there is some ball being drawn more than once, but about the expected number of multiply-drawn balls. To find that, you should add the probabilities that each ball is multiply drawn -- and since these probabilities are all equal this is the same as multiplying the common probability by the number of balls. $\endgroup$ – Henning Makholm May 31 '16 at 18:28
  • $\begingroup$ @Henning Makholm: I agree with your first comment and don't understand your second comment. $\endgroup$ – scenario Jun 1 '16 at 18:55

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