8 Indistinguishable objects randomly sorted into six buckets - What is the probability of at least three buckets receiving the objects?

Question

Eight indistinguishable objects are to be randomly put into six buckets. What is the probability that at least three buckets will receive the objects?

My approach was to let X= the number of buckets that receive objects, and then the required probability is

$P(X≥3) = 1- P(X=1) - P(X=2)$

$= 1 - [(6C1)(1/6)^8 + (6C2)(2/6)^8]$ $= 0.99771$

My reasoning was that for all of the objects falling into one bucket, there are 6 possible buckets $(6C1)$, and the probability for eight objects to fall into the same bucket in a row is $(1/6)^8 $

By the same logic, for all of the objects to be divided into two of the six buckets, there are $(6C2)$ possible ways to choose those two buckets...And the probability is $(2/6)^8 $.

Then, the probability that at least three buckets will receive objects is 1- (the probability of only one bucket receiving all the objects + two buckets receiving all the objects)

However, the answer given to this question was 0.8904. Where am I going wrong, and is there a better way to approach questions like this? Does it fall into any specific discrete distribution?

Note:

I also tried $1- [(6C1)(1/6)^8] - (6C2)[(1/6)^7(5/6) + (1/6)^6(5/6)^2 + (1/6)^5(5/6)^3 + (1/6)^4(5/6)^4 + (1/6)^3(5/6)^5 + (1/6)^2(5/6)^6 + (1/6)(5/6)^7]$
And got $0.8721$. I am not sure if this approach is better.

Answer 1

I get $1 - (6C1)(1/6)^8 - (6C2)[(2/6)^8-2(1/6)^8] \approx 0.997728$, close to but not the same as your first calculation. This treats the objects as distinguishable, as I believe that is the physical reality of putting objects into buckets.

Another way of getting this answer is to say there are $6^8= 1679616$ ways of putting $8$ items into $6$ buckets, and $6$ of them have them all going into one bucket while $3810$ have them going into exactly $2$ buckets, leaving $1675800$ possibility using at least $3$ buckets, with $\frac{1675800}{1679616} \approx 0.997728$.

Bad approaches might be to say

• there are ${13 \choose 5}=1287$ distributions of $8$ indistinguishable objects among $6$ distinguishable buckets, of which $6$ use one bucket and $105$ use exactly $2$ buckets, suggesting a probability of $\frac{1176}{1287}\approx 0.913752$ as the probability
• there are $20$ distributions of $8$ indistinguishable objects among $6$ indistinguishable buckets, of which $1$ uses one bucket and $4$ use exactly $2$ buckets, suggesting a probability of $\frac{15}{20}=0.75$ as the probability

but I think these are wrong because the distributions are not equally likely

Answer 2

The official answer is incorrect. Coming to your approach, it is correct except that you have a small mistake when you are distributing objects to two buckets.

$\displaystyle P(X=1) = {6 \choose 1} \cdot \frac{1}{6^8}$

$\displaystyle P(X=2) = {6 \choose 2} \cdot \frac{2^8 - 2}{6^8}$

Please note that $2^8$ will also count the two arrangements where all objects go to one of the two chosen buckets. So we subtract $2$ from $2^8$.

Therefore,

$\displaystyle P(X \geq 3) = 1 - P(X=1) - P(X=2) = \frac{23275}{23328}$

Answer 3

I'm assuming that the buckets are distinguishable! And the indistinguishable objects are balls.

A hint to answer the question, assuming the balls are indistinguishable

$P(X=1)=(6C1)P(\text{all balls fall into bucket 1})$

In the case of indistinguishability, you will have to use conditioning to get the answer. The calculation remains exactly the same. The only difference is that you are using conditioning instead of combinatorics.

Let $A_i$ be the event that the $i^{th}$ ball chosen falls into bucket 1

$P(A_1\cap A_2 \cap \dots\cap A_8)=P(A_1)\cdot P(A_2|A_1)\cdot P(A_3|A_1\cap A_2)\dots$

There is one way to choose the first ball(since balls are indistinguishable!) and the probability that, that ball goes into the first bucket is $(1/6)$(because the buckets are distinguishable). so, $P(A_1)=1/6$

Now, given that the first ball goes into the first bucket, the number of ways to choose the second ball is also one(balls are indistinguishable!), and the probability that that ball goes into the first bucket is also $(1/6)$(because the buckets are distinguishable). so, $P(A_2|A_1)=1/6$

And so on...

hence $P(\text{all balls fall into bucket 1})=(1/6)^8$ and $P(X=1)=(6C1)P(\text{all balls fall into bucket 1})=(6C1)(1/6)^8$

I believe, it is straightforward to similarly calculate $P(X=2)$ and $P(X\geq 3)$

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I would like to quote @DavidK here(from this answer, you should check out that answer, it is an absolute gem),

A way I think of this intuitively is that we are modeling a world in which writing a number on a ball or erasing the number does not cause that ball to magically run away from you when you reach in the back nor jump into your hand. In fact, the distinguishing marks (or lack thereof) on the balls have no effect on the probability of drawing a ball each time. So a correct way to compute $P(X=k)$ with indistinguishable balls is to compute $P(X=k)$ with distinguishable balls and simply copy the final result. This yields the same formulas.

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I hope, it is okay to quote other users, if it is relevant. If it is not, let me know and I will remove the quote.

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Regarding your note:

the probability of dividing 6 objects into exactly 2 buckets is not $[(1/6)^7(5/6) + (1/6)^6(5/6)^2 + (1/6)^5(5/6)^3 + (1/6)^4(5/6)^4 + (1/6)^3(5/6)^5 + (1/6)^2(5/6)^6 + (1/6)(5/6)^7]$

it should be ${nCr}\left(8,1\right)\left(\frac{1}{6}\right)^{7}\left(\frac{1}{6}\right)+{nCr}\left(8,2\right)\left(\frac{1}{6}\right)^{6}\left(\frac{1}{6}\right)^{2}+{nCr}\left(8,3\right)\left(\frac{1}{6}\right)^{5}\left(\frac{1}{6}\right)^{3}+{nCr}\left(8,4\right)\left(\frac{1}{6}\right)^{4}\left(\frac{1}{6}\right)^{4}+{nCr}\left(8,5\right)\left(\frac{1}{6}\right)^{3}\left(\frac{1}{6}\right)^{5}+{nCr}\left(8,6\right)\left(\frac{1}{6}\right)^{2}\left(\frac{1}{6}\right)^{6}+{nCr}\left(8,7\right)\left(\frac{1}{6}\right)^{1}\left(\frac{1}{6}\right)^{7}$