Calculating Expectation - die, 3 players.

Question

Three players are playing a game and have a fair six-sided die. This is an arbitrary game conditioned on the following rule:

Player 1 rolls first, Player 2 roles until he has a different number to 1, Player 3 rolls until he has a different number to players 1 and 2.

$\underline{Question}$: How do I go about calculating the expected value of each of the players rolls?

Let $X_i =$ number appearing for player $i=1,2,3$. I can get the first one but after I get stuck in setting up the equation for the next:

$\mathbb{E}[X_1] = \frac{1+2+...+6}{6} = 3.5$,

$\mathbb{E}[X_2 | X_2 \ne X_1]$ =? is this what I am looking for and if so any help calculating it would be appreciate.

Best wishes, I.

Answer 1

Since the numerical values of the rolls are of no relevance there could as well be six different animals on the faces of the die.

This should make it clear that the expected value is $3.5$ for each of the three players, by symmetry.

Answer 2

You might show that the distribution of numbers that the players receive is the same as if they each (one at a time) chose a numbered ball from a bag without replacement, where the bag originally contained six balls numbered $1$ through $6.$

Then consider whether there are any two balls such that one is more likely than the other to be selected second. Also consider the same question with "third" instead of "second."