I am having some trouble coming up with a probability table for a game that I wrote.
The game uses 6 dice. The player can throw the dice multiple times, but must retain at least 1 on each throw. In order to complete a game with a score the player must retain one '2' and one '4'. The score is equal to the sum of the other 4 retained dice.
As I understand this to throw a perfect roll (2,4,6,6,6,6) (score of 24) is calculated as 6!/4! (30) possibilities out of 6^6 (46656) single throws. This should happen ~ 1 in every 1555.2 6 dice throws.
However, because the player can toss all 6 dice and then retain either 1,2,3,4,5 of them and re-throw the rest. The play is much easier than 1 : 1555.
So the first question is, How do I calculate each of the other possible methods as well as other scores besides 24 perfect score.
I started out thinking about breaking down each combination of rolls for example 2 throws. 1,5 ; 2,4; 3,3; 4,2; 5,1
I assumed there are 6 ways to throw 6 dice and retain 1 of them. Then depending on which value was retained there are either 5!/3! (if the one kept was a 6) or 5!/4! (if the one kept was a 2 or a 4).
So if my thinking is correct there are 6 ways to roll a 6, 6 ways to roll a 2 and 6 ways to roll a 4.
Given a 6 then there are 20 ways to roll 6,6,6,2,4 6 * 20 = 120 ways to roll this pair of two rolls
Given a 4 then there are 5 ways to roll 6,6,6,6,2 6 * 5 = 30 ways to roll this pair of two rolls
Given a 2 then there are 5 ways to roll 6,6,6,6,4 6 * 5 = 30 ways to roll this pair of two rolls
This would give me 30 ways to get the score with 1 roll retaining all the dice on a single throw.
This would give me 180 ways to get the score with two rolls retaining 1 die and rethrowing 5 dice.
To take this a step further if the player retains 2 dice from the first toss and rethrows the remaining 4 then:
I have 6 choose 2 (15) ways to throw each of the following:
6,6 6,2 6,4 2,4
Given them there are either 4!/2! or 4!/3! or 4!/4! ways to throw the remaining winning throw (depending on the number of 6's retained in the initial throw). If I do the math like on the first one I get 315 ways to get the perfect score with 2 rolls where 2 dice are held, and the remaining 4 are thrown and kept.
Am I doing this correctly? Are these the correct assumptions to do this calculation? How does this change when I get to 3,4,5,6 rolls? Is there formula for doing what I am attempting to do? Do any of these gyrations have any affect on the 46656 (6^6) that I am using for the total possibilities?