I would like to ask a simple question that happened to me on an interview. The question was very simple and it was as follows:

We have 80 balls.
We have 4 random groups (each group will have 20 balls).

What is the probability of having the first 2 balls in the same group?

My answer was: 1/4 (0.25) because the first ball will go with probability 1 to whichever group. And the 2nd ball has 1/4 chances to join the same group as the 1st ball.

The answer they gave me was: 19/79 (=0.24)

Was my logic so wrong? I'm not sure how 19/79 really answers that the first 2 balls are in the same group.

  • $\begingroup$ You cannot choose the first ball twice without replacement $\endgroup$ – Henry Oct 8 at 10:34
  • $\begingroup$ I'm not choosing the first ball twice. Just once and then i choose for the second ball. $\endgroup$ – mlds1337 Oct 8 at 10:45
  • $\begingroup$ How many possible choices are there for the second ball? $\endgroup$ – Henry Oct 8 at 10:46
  • $\begingroup$ 4 choices, because we have 4 groups $\endgroup$ – mlds1337 Oct 8 at 10:47
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    $\begingroup$ The four groups are no longer equally likely as there are $79$ gaps for the second ball, making the first ball's group less likely $\endgroup$ – Henry Oct 8 at 10:49

As you say, the first ball goes to some group with probability 1. Now, the bit where you are going wrong is the assumption that there is a 1 in 4 chance of the ball going to that same group.

Imagine that we are going to assign a random number between 1-80 to each of the 80 balls, and if the number is between 1-20 the ball goes in group 1, 21-40 group 2, etc. And assume now, without loss of generality, that the first ball was assigned the number 1. Now the available numbers to be assigned to the second ball are the integers from 2-80. To be assigned to group 1 you have to be assigned a number between 2-20, and the rest of the groups are unchanged. Because we know that assignments are made uniformly at random we can make the following observation: there are 79 different possible numbers that we can have, and only 19 of those numbers would assign the second ball to the same group as the first, hence there is a $\frac{19}{79}$ probability of the first and second ball ending up the same group.

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  • 1
    $\begingroup$ Thanks for the explanation. I think you described it perfectly. $\endgroup$ – mlds1337 Oct 9 at 12:53

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