Probability that Amy, Bill and Poor pete win money 6.042 MIT I was doing the following problem:

Problem 2. [12 points] Amy, Bill, and Poor Pete play a game:
  
  
*
  
*Each player puts \$2 on the table.
  
*Each player secretly writes a number between 1 and 4.
  
*They roll a fair, four-sided die with faces numbered 1, 2, 3, and 4.
  
*The money on the table is divided among the players that guessed correctly. If no one guessed correctly, then everyone gets their
  money back and Poor Pete is paid \$0.25 in “service fees”.
Suppose that, Amy and Bill cheat by picking a pair of distinct
  numbers uniformly at random.

How do you calculate the following Probability:

Pete guesses right AND either Amy or Bill guesses right?

I think it should be a simple question but I can't seem to get the answer $\frac{1}{8}$

This is what I have tried:
Let P denote the event that Pete wins, A and B that Amy and Bill will respectively.
So we want $Pr[P=wins \cap (A=wins \cup B=wins)]$ which is equal to:
$Pr[P=wins]Pr[(A=wins \cup B=wins)]$
So I tried calculating each one by writing down the tree corresponding to each Pr and then multiplying the probability.
Pr[P = wins] = Pr[P chooses the same as the number in the real die] = Pr[1 came up]Pr[Pete chooses 1]+ Pr[2 came up]Pr[Pete chooses 2] + Pr[3 came up]Pr[Pete chooses 3] + Pr[4 came up]Pr[Pete chooses 4] = $\frac{4}{16} = \frac{1}{4}$
The other probability $Pr[(A=wins \cup B=wins)]$ I was simply going to compute via inclusion exclusion. i.e.
$$Pr[(A=wins \cup B=wins)] = Pr[A=wins] + Pr[B = wins] - Pr[(A=wins \cap B=wins)]$$
is this approach correct? Can't seem to make it work and give me $\frac{1}{8}$.
I also had to compute:

Pete guesses right AND both Amy and Bill guess wrong
Pete guesses wrong AND either Amy or Bill guesses right
Pete guesses wrong AND both Amy and Bill guess wrong

and was wondering if this general approach was correct.

You've understood Pete's behavior well, as well as the independence of Pete's behavior and that of Amy & Bill's. But you haven't explored Amy & Bill's behavior.
The key is to understand what a pair of distinct numbers at random means.
Amy and Bill, as a pair, select one of:
$\begin{array}{cc}
\text{Amy} & \text{Bill} \\
1 & 2 \\
1 & 3 \\
1 & 4 \\
2 & 1 \\
2 & 3 \\
2 & 4 \\
3 & 1 \\
3 & 2 \\
3 & 4 \\
4 & 1 \\
4 & 2 \\
4 & 3
\end{array}$
And they, as a pair, select each with equal probability. There are twelve, so the selection given in each row has probability $1/12$.
Then $P[\text{A wins}] = 1/4$ (sum over the four possibilities as you did for Pete: $P[A = 1] = 3/12 = 1/4$, etc) and $P[\text{B wins}] = 1/4$, but $P[\text{A wins and B wins}] = 0$.
