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April 6th, 2017, 03:48 PM  #1 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 590 Thanks: 81  Probability Question With Different People Knowing Different Information
People A, B, C, and D make drawings A, B, C, D, respectively. Each person makes one drawing and knows that each person made one drawing. The drawings are collected, and everybody guesses who made each drawing. All four people guess without knowing how the other people guessed. Everybody knows their own drawing and randomly guesses for the other three. 1. Since each person has a 1/3 chance of guessing each of the three drawing he didn't make, does that mean the average number of correct guesses per person is 1 (2 including the person knowing their own drawing)? Is it more complicated than that? 2. There are 3! = 6 permutations of 3 drawings, so I'm assuming each person has a 1/6 chance at correctly guessing all 3 drawings he didn't make. How well each person guesses is independent of how well the other people guess, so are these correct? P(everybody guesses everything correct) = (1/6)^4 = 1/1,296 P(at least one person guesses everything correct) = 1 (5/6)^4 = 671/1,296 = about 51.8% 
April 6th, 2017, 07:49 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,785 Thanks: 920 
1) It's a bit more complicated than that. Each person is faced with matching 3 painters to 3 paintings. If you look at how correct guesses are distributed you get $P[\text{k correct guesses}] = \begin{cases} \dfrac 1 3 &k=0 \\ \dfrac 1 2 &k=1 \\ 0 &k=2 \\ \dfrac 1 6 &k=3 \end{cases}$ The expected value does happen to equal 1. If each person painted 3 paintings, and out of those 3 1 was selected, and you had to select the painter that would be the situation you described. Here we "select w/o replacement". We'll never select the same painter twice. Looking at (2) You have $P[\text{everyone guesses all correctly}]=(1/6)^4$ correct You have the second one correct too though you left out a minus sign. well done 

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