Probability of voting in a group of four I have the following problem, but I'm not sure about my solutions.  

There is a reality show, with four people in it. They're {A, B, C, D}.
  They vote by giving each other a sign. The rules of voting are:  
  
  
*
  
*They can't vote for themselves
  
*They can vote for anyone from the other three
  

Q1. How many different ways can they vote?
I think the answer is 81, because everyone can vote for 3 other people, which means $3^4$.
Q2. What is the probability of everyone receiving a vote?
I got $$\frac{3*2*2*1}{81} = 0.1481$$ however when the person who just received a vote votes next, and we want 1 vote for each, they can choose like this: $3*3*1*1$ Should I add them together, or does the first one include the second?
Q3. What is the probability of 3 players receiving votes?
The preferred scenarios are either $3*3*2*2$ or $3*3*3*2$.
Q4. What is the probability of 2 players receiving 2-2 votes
The preferred scenarios are either $3*3*1*1$ or $3*3*2*1$.
I'd also like to know what's the correct way of solving these kind of problems (without a graph), because my approach (ie. at Q1, draw 4 lines, and think like: "Well, the 'A' can vote at anyone, so he gets a 3, lets vote for 'B', now 'B' can vote for anyone as well, he gets a 3 too, lets vote for 'C', 'C' can vote only for 'D', and 'D' can only vote for 'A', and now we have $3*3*1*1$)  is clearly wrong, because it matters to whom do I give these hypothetical votes.
 A: They can indeed vote in $3^4=81$ different ways.
The fourth question is perhaps the easiest of the rest. There are $\binom42=6$ ways to pick the two players getting votes. If those two players are $X$ and $Y$, and the other two are $W$ and $Z$, then $X$ and $Y$ must vote for each other, so $W$ and $Z$ must split their votes between $X$ and $Y$. They can do this in two ways: $W$ can vote for $X$ and $Z$ for $Y$, or the other way around. Thus, there are $\binom42\cdot2=12$ ways for two of the players to split the vote evenly, and the probability of this occurrence is $\frac{12}{81}=\frac4{27}$.
Now suppose that exactly $3$ of the players receive votes. Let $X$ be the unlucky player who gets no votes; there are $4$ ways to pick $X$. Say $X$ votes for $Y$; $Y$ must be one of the other $3$ players, so there are $3$ choices for $Y$. Say $Y$ votes for $Z$; $Z$ cannot be $X$ or $Y$, so there are $2$ choices for $Z$. Somebody has to vote for the remaining person, whom I’ll call $W$, and it isn’t $X,Y$, or $W$, so it must be $Z$. Finally, $W$ can vote for either $Y$ or $Z$. Thus, there are $4\cdot3\cdot2\cdot2=48$ ways for exactly $3$ players to receive votes, and the probability of this occurrence is $\frac{48}{81}=\frac{16}{27}$.
Finally, suppose that everyone receives a vote. There are $3$ ways to choose the recipient of $A$’s vote; let $X$ be that recipient. Now there are two possibilities.


*

*$X$ votes for $A$. In that case the other two players must vote for each other.  

*Someone else votes for $A$; call that person $Y$. There are $2$ choices for $Y$, and the remaining person must vote for $Y$ and be voted for by $X$.


In other words, once $X$ has been chosen, there are $1+2=3$ ways to finish the voting so that everyone receives a vote, and since there are $3$ ways to choose $X$, there are $3\cdot3=9$ ways to give everyone a vote. The probability of this event is therefore $\frac9{81}=\frac19$.
As a check, these four probabilities sum to $\frac{23}{27}$, so the probability of the one remaining distribution should be $\frac4{27}$. That’s the distribution in which one player gets $3$ votes. There are $4$ ways to choose the lucky player, and there are then $3$ ways to choose the recipient of that player’s vote, so there are $12$ ways to achieve this outcome, and its probability is indeed $\frac{12}{81}=\frac4{27}$.
