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Problem Statement

We have a bag with 20 balls, numbered from 1 through 20. We pick a ball at random, observe its number and put it back. We do the same operation two more times. (This is called sampling with replacement since we replace the object that we pick – it may be chosen again.) In the end, we have three numbers. Let X denote the largest of these. What is the probability that X ≥ 17?

Solution

$ P \{ X=i \} = { {3 \choose 1} (i-1)^2 +{3 \choose 2} (i-1) +{3 \choose 3} \over { 20^3 }} $

$ P \{ X \geq 17 \} \approx. 0.4880$

If the balls were not replaced then the solution would be:

$ P\{X=i \} = { {i-1 \choose 2} \over { 20 \choose 3} } $

Where $ P \{ X \geq 17 \} \approx. 0.508$

Question

Should it not be more likely to find a ball that is 17 or greater when the balls are replaced?

What exactly is wrong with the following solution:

$ P\{X=i \} = { {3 \choose 1}{1 \choose 1}{2 \choose 2}{i \choose 1}{i \choose 1} \over { 20^3 } }$

$ = {3i^2 \over 20^3}$

that is choose one of the 3 balls, give it the value of i. Now out of the remaining 2 balls each can take on one of i values.

So that$ P \{ X \geq 17 \} \approx. 0.515$

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    $\begingroup$ Suppose we pulled 17 balls. With replacement the probability that one ball is greater than or equal to 17 is high, but not certain. But without replacement it is guaranteed! Your first two results are correct. However, I would tackle this problem looking at the probability no ball is $17^+.$ i.e. with replacement $P(x\ge 17) = 1 - \left(\frac {16}{20}\right)^3.$ And without $P(x\ge 17) = 1 - \frac {16\choose 3}{20\choose 3}$. I am not following your 3rd equation. $\endgroup$
    – Doug M
    Jan 2, 2018 at 16:16
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    $\begingroup$ When $i=20$, the third formula gives $3/20 = 0.15$. But that's too high, since it should be $1 - (19/20)^3 = 0.1426...$ i.e. complement of the set of triples that don't contain a "20". $\endgroup$
    – Ned
    Jan 2, 2018 at 18:26
  • $\begingroup$ Thanks @DougM it helps looking at it from the perspective of the compliment also your example puts me more at ease. I think part of my problem lies with defining the minimum vs maximum i.e What would the probability be let's say to draw the number 20 if only one of the balls may be 20, with replacement? $\endgroup$
    – Johnny M
    Jan 3, 2018 at 10:29

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