1. ## Dice Pair Matching

4 players roll a die. They score 1 point if there is a matching pair. What are the probabilities of getting the possible scores.

For example: for the case where 2 people get 1, and the other two roll a 3 and a 4 respectively, then the teams score would be 1. Similarly, for the case that 3 people throw a 5 and the other one does not, the score would be 3, since there are 3 diﬀerent pairs of players that have the same number.

Proposed solution:
Imagine players as slots to fill with possibilities of rolling dice.

Score 0:
All 4 players roll different numbers. So 6*5*4*3 = 360 ways of getting score 0.

Score 1:
2 Players roll the same, other 2 are different from each other and the matching pair. So 6*1*5*4 = 120 ways of getting score 0.

Score 2:
2 pairs of players match. So 6*1*5*1 = 30 ways. The ones are the corresponding matches to the preceding rolls.

Score 3:
3 matches and 1 different so 6*1*1*5 = 30 ways again. The 4th person can have any one of the 5 choices left over.

Score 4:
All roll the same number so. 6*1*1*1 = 6 ways of getting score 4.

My problem is these all add up to 546, whereas I would imagine they'd have to add up to $1296 = 6^4$ possibilities.

Any suggestions?

2. Lets look at the case score 2. That could happen if the players roll $\boxed{2}\boxed{2}\;\boxed{6}\boxed{6}$.
But that can happen is $\frac{4!}{2^2}$ ways.
Moreover, there are ${6 \choose 2}$ ways to have two different pairs.
It seems to me that have under-counted this case.

3. Hello, utopiaNow!

You're not considering WHO gets the pairs, etc.

Four players roll a die.
They score 1 point if there is a matching pair.
What are the probabilities of getting the possible scores?

For example:
2 people get 1, and the other two roll a 3 and a 4 resp., then the score would be 1.
Similarly, that 3 people throw a 5 and the other one does not, the score would be 3.

Score 0
Four different numbers: $abcd$
Number of ways: . $6\cdot5\cdot4\cdot3 \:=\:{\color{blue}360}$

Score 1
A pair and two singles: $aabc$
There are: . ${4\choose2} = 6$ distributions . . . $\{aabc, abac, abca, baac, baca, bcaa\}$
Number of ways: . $6\cdot(6\cdot5\cdot4) \:=\:{\color{blue}720}$

Score 2
Two pairs: $aabb$
There are 3 distributions.
. . If the four players are $A,B,C,D$, they can be paired in 3 ways:
. . . . $\{AB,\,CD\},\;\{AC,\,BD\},\:\{AD,\,BC\}$
Number of ways: . $3\cdot(6\cdot5) \;=\;{\color{blue}90}$

Score 3
A triple and a single: $aaab$
There are 4 choices of who gets the triple.
Number of ways: . $4\cdot(6\cdot5) \;=\;{\color{blue}120}$

Score 4
A quadruple: $aaaa$
The only choice is the value of the quadruple.
Number of ways: . ${\color{blue}6}$

Check: . $360 + 720 + 90 + 120 + 6 \;=\;{\bf1296}$

4. Originally Posted by Soroban
Hello, utopiaNow!

You're not considering WHO gets the pairs, etc.

Score 0
Four different numbers: $abcd$
Number of ways: . $6\cdot5\cdot4\cdot3 \:=\:{\color{blue}360}$

Score 1
A pair and two singles: $aabc$
There are: . ${4\choose2} = 6$ distributions . . . $\{aabc, abac, abca, baac, baca, bcaa\}$
Number of ways: . $6\cdot(6\cdot5\cdot4) \:=\:{\color{blue}720}$

Score 2
Two pairs: $aabb$
There are 3 distributions.
. . If the four players are $A,B,C,D$, they can be paired in 3 ways:
. . . . $\{AB,\,CD\},\;\{AC,\,BD\},\:\{AD,\,BC\}$
Number of ways: . $3\cdot(6\cdot5) \;=\;{\color{blue}90}$

Score 3
A triple and a single: $aaab$
There are 4 choices of who gets the triple.
Number of ways: . $4\cdot(6\cdot5) \;=\;{\color{blue}120}$

Score 4
A quadruple: $aaaa$
The only choice is the value of the quadruple.
Number of ways: . ${\color{blue}6}$

Check: . $360 + 720 + 90 + 120 + 6 \;=\;{\bf1296}$

Ah, yes that's it. Thanks Soroban for your help and explanation.