# Probability-Equally likely or not?

Probability

Question. "$11$ identical balls are distributed in $4$ distinct boxes randomly. Then the probability that any $3$ boxes will together get a greater number of balls than the remaining one is:"

I am confused about the distribution of balls. Are all the cases i.e. $\{5,5,0,1\}$ and $\{2,3,4,2\}$ equally likely or not?

• No, those two cases are not equally likely (according to the most straightforward reading of the problem). Each ball is equally likely to be put into any of the four boxes, but it is not equally likely for all eleven to be put into box 1, for example, than it is for two to be put into box 1, and three each to be put into the other boxes. The question is asking you to identify what the distribution of number in boxes is, and to use that in your answer. – Brian Tung Sep 17 '17 at 20:04
• By the way, "i.e." means "that is"; this would imply that your two examples are all the cases. You probably mean "e.g.," which means "for example." (It's a trivial issue, but sometimes mathematics turns on trivialities!) – Brian Tung Sep 17 '17 at 20:05

The point is to find all combinations of natural numbers $(a,b,c,d)$ such that $\max\{a,b,c,d\}\le 5$ and $a+b+c+d=11$. Without loss of generality, assume $a\ge b\ge c\ge d$. Notice that $\max\{a,b,c,d\}$ can only be $3$, or $4$, or $5$.

• If $\max\{a,b,c,d\}=3$, then the only possible combination is $(3,3,3,2)$;
• If $\max\{a,b,c,d\}=4$, then the possible combinations are $(4,4,3,0)$, $(4,4,2,1)$, $(4,3,3,1)$,$(4,3,2,2)$;
• If $\max\{a,b,c,d\}=5$, then possible combinations include $(5,5,1,0)$,$(5,4,2,0)$,$(5,4,1,1)$,$(5,3,3,0)$, $(5,3,2,1)$,$(5,2,2,2)$.

Together, since four boxes are distinct, for each combination above, the number of ways to rearrange is

• For $(5,4,2,0)$ and $(5,3,2,1)$: $A^4_4$;
• For $(4,4,3,0)$, $(4,4,2,1)$,$(4,3,3,1)$,$(4,3,2,2)$,$(5,5,1,0)$,$(5,4,1,1)$, $(5,3,3,0)$: $A^4_4/A^2_2$;
• For $(3,3,3,2)$ and $(5,2,2,2)$:$A^4_4/A^3_3$

Since the total number of events are $4^{11}$, the probability is $$\frac{2A^4_4+7A^4_4/A^2_2+2A^4_4/A^3_3}{4^{11}}.$$

• That number seems too low. I think you must also multiply each box configuration by the ways the balls can end up in that configuration. For example, for $(5,4,2,0)$ the first $5$ balls could go to box 1, or they could be the $4$ first ones and the sixth, and so on. Simulation gives the answer somewhere around $0.86$. – ploosu2 Sep 18 '17 at 9:09

The solution given by OnoL has the correct configurations for the ball amounts in the boxes but it is missing the possible ways the balls can be placed to the boxes. Because we are considering all the ways $11$ balls can be placed to $4$ boxes ($4^{11}$) these multipliers must be considered. They are also different for different configuration (this answers your second question: the different cases have different probabilities). Luckily it doesn't depend on the order of the boxes so we can still only consider the ordered configurations (and then multiply by the number of permutations of that configuration).

The way $11$ balls can be placed to form $(a, b, c, d)$ is

$${{11}\choose{a}} {{11-a}\choose{b}} {{11-a-b}\choose{c}}$$

(The balls for the first box can be chosen freely, then choose from the rest $11-a$ balls $b$ balls for the second and so on. The last (fourth) box has to have the remaining balls so that coefficient ${{11-a-b-c}\choose{d}} = {{d}\choose{d}} = 1$, since $a+b+c+d=11$.)

The number of ball placings that satisfy the claim "any $3$ boxes will together get a greater number of balls than the remaining one" is given by the sum over all possible configurations (these were found by OnoL)

$(3,3,3,2), (4,4,3,0), (4,4,2,1), (4,3,3,1), (4,3,2,2), (5,5,1,0), (5,4,2,0), (5,4,1,1), (5,3,3,0), (5,3,2,1), (5,2,2,2)$

where the summand is the product

$($number of permutations of that configuration$)$ $\times$ $($the number of ways to place the balls to form that configuration$)$.

I calculated this with a computer and got $3618384$. So the wanted probability is

$$\frac{3618384}{4^{11}} = \frac{226149}{262144} \approx 0.86269.$$