Probability that 4/4, 2/4 Discrete Uniform RVs aren't equal

Question

If I have four discrete $uniform(1,32)$ RVs. I'm trying to figure out the probability that a) None are the same b) exactly two are the same

I thought that the probability that none are the same would be $\frac{1}{(1/31)^4} = 0.00000108281241$ although I might be over simplifying this

for exactly 2 I'm not sure where to start I was thinking maybe I should use either counting methods or summing the PMFs by convolving 4 uniform PMFs and dividing by 4 which would give me something like a triangle distribution?

Would someone please point me in the right direction?

Answer 1

Consider a sequence of four observations, one from each random variable: $$(x_1,x_2,x_3,x_4).$$ There are $32^4 = 1{,}048{,}576$ such sequences, each of which is equally likely.

For a), there are 32 choices for $x_1$, leaving 31 choices for $x_2$, leaving 30 choices for $x_3$, leaving 29 choices for $x_4$. By the multiplication principle, there are $32 \times 31 \times 30 \times 29 = \frac{32!}{28!} = 863{,}040$ sequences $(x_1,x_2,x_3,x_4)$ in which each $x_i$ is distinct from the others. Each of those sequences is equally likely. Therefore, the probability in question is $\frac{863{,}040}{1{,}048{,}576} = \frac{13{,}485}{16{,}384}$.

For (b), choose:

• two of the observations to be equal;
• a single integer in $[1,32]$ for those two observations;
• two integers (distinct from the previous choice and from each other) in $[1,32]$ for the remaining two observations;

There are $\binom{4}{2}$ ways to make the first choice, $32$ ways to make the second choice, leaving $31 \times 30$ ways to make the third, and final, choice. By the multiplication principle, there are $\binom{4}{2} \times 32 \times 31 \times 30 = 178{,}560$ sequences $(x_1,x_2,x_3,x_4)$ in which exactly two observations are the same. Each of those sequences is equally likely. Therefore, the probability in question is $\frac{178{,}560}{1{,}048{,}576} = \frac{1{,}395}{8{,}192}$.

Answer 2

There are $32^4$ possible outcomes for $(a,b,c,d)$, since each one has $32$ outcomes.

a) If none of them are equal, then outcomes for $(a,b,c,d)$ are $32\times 31\times 30\times 29$. Thus, the probability is $\frac{32\times 31\times 30\times 29}{32^4}=\frac{13485}{16384}=0.823$

b) Select the common number: $32$ outcomes. Select the places: $\binom{4}{2}$. Select the other two outcomes: $\binom{31}{2}$. For each of the last, "we can put one of the two choices in one place and the other in the other or viceversa". This means that we have multiply by two. So, we have $32\times\binom{4}{2}\times\binom{31}{2}\times 2$ total outcomes, so the probability is $\frac{32\times\binom{4}{2}\times\binom{31}{2}\times 2}{32^4}=\frac{1395}{8192}=0.170$