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

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.

• You're counting multiple times the cases where all the objects go to the same bucket. First intentionally, and then twice unintentionally, since some of the times that objects are limited to either of two buckets, they are in fact limited to exactly one of the two buckets. (That is to say, if I say that eight objects go into either Bucket A or Bucket B, nothing prevents them from all going into Bucket A or all into Bucket B.) – Brian Tung Jun 24 at 20:40
• Where did you find this question , can you share your source please ? – Bulbasaur Jun 24 at 21:03
• @Bulbasaur This question is from a past exam paper for a first-year university statistics course. – Cara vdC Jun 24 at 23:17

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

• Should we always approach the indistinguishable objects as distinguishable in probability questions ? For example , what if the indistinguishable balls were distributed at the same time , would we treat like one by one and also distinguishable – Bulbasaur Jun 24 at 20:54
• @bulbasaur - I believe that is the physical reality. Flip two indistinguishable fair coins. Is the probability of one heads and one tails $\frac12$ or $\frac13$? – Henry Jun 24 at 20:56
• hımm elegant example... they are like die (ordered pairs) or coins as you said . Is there any book that may help me to learn this tricks ? Maybe other questions ? – Bulbasaur Jun 24 at 21:00

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,

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

• Why did you treat indistinguishable balls like distinguishable . Cant the answer be $$1-[C(6,1) \times \frac{1}{C(13,5)} + C(6,2) \times \frac{C(9,1)}{C(13,5)}]$$ – Bulbasaur Jun 24 at 20:47
• @Bulbasaur stars and bars method should not be used to calculate probability. All arrangements in stars and bars are not equally probable. – Math Lover Jun 24 at 20:50
• thanks , i see now – Bulbasaur Jun 24 at 20:52
• @CaravdC Even if you assume that all distributions of the indistinguishable objects are equally likely, the probability would work out to be $1-(6+\binom62\times 7)/\binom{8+6-1}{5}$, which is not equal to the book's answer. This is discussed in the second bullet point of Henry's answer. There is no conceivable way your source's answer is correct. – Mike Earnest Jun 25 at 0:33
• @CaravdC Mike Earnest's comments already answers your question. There is no way to get to that answer for the given question. Your first approach is optimal except a mistake that you made. – Math Lover Jun 25 at 9:49

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 $$P(X\geq 3)$$

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.

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.

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