Tetrahedral dice game

This week’s Riddler Classic is a game of four-sided dice:

You have four fair tetrahedral dice whose four sides are numbered 1 through 4.

You play a game in which you roll them all and divide them into two groups: those whose values are unique, and those which are duplicates. For example, if you roll a 1, 2, 2 and 4, then the 1 and 4 will go into the “unique” group, while the 2s will go into the “duplicate” group.

Next, you reroll all the dice in the duplicate pool and sort all the dice again. Continuing the previous example, that would mean you reroll the 2s. If the result happens to be 1 and 3, then the “unique” group will now consist of 3 and 4, while the “duplicate” group will have two 1s.

You continue rerolling the duplicate pool and sorting all the dice until all the dice are members of the same group. If all four dice are in the “unique” group, you win. If all four are in the “duplicate” group, you lose.

What is your probability of winning the game?

My solution:
[Show Solution]

4 thoughts on “Tetrahedral dice game”

  1. I got slightly different coefficients.
    For the three duplicates self transition i got 3/16. And the transition from three to two duplicates is correspondingly smaller.

    The way I did the 3 duplicates was with 4 throws: select the duplicate value (4 options) select the throw positions (4 options) select the non duplicate (3 options).
    p = 3*4*4/4^4 = 3/16

    1. I agree with Tlk – imagine my dismay when Laurent Lessard posted a solution from mine!

      Confirming the probability from three duplicates to two: select the location of the duplicates (6 options), then the identity of the duplicates (4 options), then the identity of the two singletons where order matters (3*2 options). P=(6*4*3*2)/(4^4)=9/16.

      Aha! I see you just made the edit. Well done!

      1. Thank you both for your comments! Somehow in my mind, 12/64 simplified to 3/32. Oops! I corrected my mistake and updated my solution.

  2. A faster way to get to the final result is to notice that A^infinity (call it L for limit) must be of the form L=
    1 1-x 1-y 0
    0 0 0 0
    0 0 0 0
    0 x y 1
    where x is the probability of winning starting from 3 duplicates and y is the probability of winning starting from 2 duplicates (which we don’t care about). Then we must have L.A = L, where L.A means matrix multiplication. This yields two equations for x and y, whose solution is x=9/20 and y=29/60.

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