Can You Roll Unique Dice?

Riddler Classic

From Ross O’Brien comes a game of nonconformist 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?

Solution

If there are four dice, there are $4^4 = 256$ outcomes.

I looked at the combinations of unique and duplicate dice, as well as further explored the ones where rerolls were possible.

1. All Four Unique

Each die should be different than the previously rolled. $\dfrac{4}{4} \cdot \dfrac{3}{4} \cdot \dfrac{2}{4} \cdot \dfrac{1}{4} = \dfrac{24}{256} = \dfrac{3}{32}$.

2. All Four Duplicate or Two Pairs of Two Duplicates

• All four dice could be the same, and it can be any of the four numbers. $\dfrac{4}{256} = \dfrac{1}{64}$.
  
  
• Two pairs of two dice could be the same, and there are 3 ways to pair the dice. $\left( \dfrac{4}{4}\cdot\dfrac{1}{4} \cdot \dfrac{3}{4} \cdot \dfrac{1}{4} \right) \cdot 3 = \dfrac{36}{256} = \dfrac{9}{64} $.
  
  
• $\dfrac{1}{64} + \dfrac{9}{64} = \dfrac{5}{32}$.

3. Two Unique, Two Duplicate

Two dice are the unique, two dice are the same, and there are six ways to choose the duplicated pair. $\left( \dfrac{4}{4}\cdot\dfrac{3}{4} \cdot \dfrac{2}{4} \cdot \dfrac{1}{4} \right) \cdot 6 = \dfrac{144}{256} = \dfrac{9}{16}$.

These two duplicate dice must be rerolled. Here are the outcomes.

• Result in All Four Unique - The two dice must be different from each other and the unique dice. $\dfrac{2}{4}\cdot\dfrac{1}{4} = \dfrac{1}{8}$.
  
  
• Result in Two Pairs of Two Duplicate - Each duplicated die must be be the same as a unique die. There are two ways to paire up. $\left(\dfrac{1}{4}\cdot\dfrac{1}{4}\right) \cdot 2 = \dfrac{1}{8}$.
  
  
• Result in Two Unique, Two Duplicate, again
  
  
  • The duplicate dice can again be duplicate of each other. $\dfrac{2}{4}\cdot\dfrac{1}{4} = \dfrac{1}{8}$.
    
    
  • A duplicate die and unique die could now be the same. Two ways to choose one of the original unique die, two ways to choose one of the original duplicated die, and the remaining duplicated die has two possibilities left to now be unique. $\left(\dfrac{1}{4}\cdot\dfrac{2}{4}\right)\cdot 2 \cdot 2 = \dfrac{1}{2}$.
    
    
  • $\dfrac{1}{8} + \dfrac{1}{2} = \dfrac{5}{8}$.
    
    
• Result in One Unique, Three Duplicate - One unique remains unique, the other unique is duplicated by the original duplicated dice. There are two ways to choose the new unique $\left(\dfrac{1}{4}\cdot\dfrac{1}{4}\right) \cdot 2 = \dfrac{1}{8}$.

4. One Unique, Three Duplicate

One die is unique, three dice are the same, and there are four ways to choose the unique. $\left( \dfrac{4}{4}\cdot\dfrac{3}{4} \cdot \dfrac{1}{4} \cdot \dfrac{1}{4} \right) \cdot 4 = \dfrac{48}{256} = \dfrac{3}{16}.$

These three duplicate dice must be rerolled. Here are the outcomes.

• Result in All Four Unique - The three dice must be different from each other and the unique dice. $\dfrac{3}{4}\cdot\dfrac{2}{4}\cdot\dfrac{1}{4} = \dfrac{3}{32}$.
  
  
• Result in All Duplicate or Two Pairs of Two Duplicate
  
  
  • The duplicate dice can again the same as the unique. $\dfrac{1}{4}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4} = \dfrac{1}{64}$.
    
    
  • A duplicate die and the unique die are the same, and the remaining duplicate dice are the same. Three ways to choose one of the original duplicate dice. $\left(\dfrac{1}{4}\cdot\dfrac{3}{4}\cdot\dfrac{1}{4}\right)\cdot 3 = \dfrac{9}{64}$.
    
    
  • $\dfrac{1}{64} + \dfrac{9}{64} = \dfrac{5}{32}$
    
    
• Result in Two Unique, Two Duplicate
  
  
  • The unique die remains unique. One duplicated dice is also now unique, and the two remaining duplicate dice are again duplicate. There are three ways to choose the duplicate die that becomes unique. $\left(\dfrac{3}{4}\cdot\dfrac{2}{4}\cdot\dfrac{1}{4}\right) \cdot 3 = \dfrac{18}{64}$.
    
    
  • Two duplicate dice are now unique and the remaining duplicate die is the same as the unique die could now be the same. There are three ways to choose the duplicate die that becomes unique. $\left(\dfrac{3}{4}\cdot\dfrac{2}{4}\cdot\dfrac{1}{4}\right)\cdot 3 = \dfrac{18}{64}$.
    
    
  • $\dfrac{18}{64} + \dfrac{18}{64} = \dfrac{9}{16}$
    
    
• Result in One Unique, Three Duplicate, again.
  
  
  • Unique remains unique, duplicate remain duplicate. $\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}\right) \cdot 2 = \dfrac{3}{64}$.
  • Unique becomes duplicated by two of the duplicates, and the remaining duplicate becomes unique. There are three ways to choose the duplicate die that becomes unique. $\left(\dfrac{1}{4}\cdot\dfrac{1}{4}\cdot\dfrac{3}{4}\right) \cdot 3 = \dfrac{9}{64}$.
  • $\dfrac{3}{64} + \dfrac{9}{64} = \dfrac{3}{16}$.

Answer

Looking at the Markov Chain the initial distribution can be represented as,

$\pi_0 = \left[\begin{matrix} \dfrac{3}{32} \\ \dfrac{5}{32} \\ \dfrac{9}{16} \\ \dfrac{3}{16} \end{matrix} \right]$

and transition matrix can be represented as,

$P = \left[\begin{matrix} 1 & 0 & \dfrac{1}{8} & \dfrac{3}{32} \\ 0 & 1 & \dfrac{1}{8} & \dfrac{5}{32} \\ 0 & 0 & \dfrac{5}{8} & \dfrac{9}{16} \\ 0 & 0 & \dfrac{1}{8} & \dfrac{3}{16}\end{matrix} \right]$

The terminating states are the first, which is all unique, and the second, which is all duplicate.

Solving $P\pi_n = \pi_n$ yields,

$$\pi_n = \left[\begin{matrix} \dfrac{9}{20} \\ \dfrac{11}{20} \\ 0 \\ 0 \end{matrix} \right]$$

The probability of winning the game is 45%.

I also ran a simulation of 10 million games. I received 4,496,805 wins and 5,503,195 losses.

Rohan Lewis

2022.05.16

Code can be found here.