# Probability In Multi-Throw Dice Game

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.

More questions:

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?

• This is complicated. The strategy used by the player will effect the probability of any particular outcome. For instance, I assume that this game is played with more than one player. If I go second, I just need to beat the first player's score to win. So, I don't need to shoot for the maximum possible score. This will result in a different strategy than if I am going first. Also, it is not clear when one should keep a 2 or 4. If you keep them early, then you decrease the number of dice you are throwing on all subsequent throws; this might not be the best strategy for maximizing scores. – Matthew Conroy Jan 7 '14 at 20:55
• I'm not following your counting. For instance, where you say "there are 6 ways to roll a 6". If you roll 6 dice, there are $6^6$ possible rolls, and $5^6$ of them have no 6, so there are $6^6-5^6=31031$ ways to roll at least one 6. There are an identical number of ways to roll at least one 2 (or 4), and here we are counting rolls that have both a 6 and a 2 (or a 4) in both counts. – Matthew Conroy Jan 7 '14 at 21:01
• You might like to take a look at problem 38 in this pdf: madandmoonly.com/doctormatt/mathematics/dice1.pdf Solution starts on page 38. The game of Drop Dead is similar in some ways to your game, but simpler in that there are no strategy issues. – Matthew Conroy Jan 7 '14 at 21:07
• Let's see of I can be a bit clearer. I don't care about the strategy of two people playing the game, I am more interested in a probability table such that I can have one player play against the house and pay passed on his score. In order to do this I want to calculate the probability of a particular score. If you throw 6 dice let's call them A,B,C,D,E,F there are 6 ways A=6 or B=6 or C=6 or D=6 or E=6 or F=6 that you can throw one 6. I don't believe the player's strategy should have any effect on the probabilities. I am just trying to come up with how many ways it can be accomplished / 46656 – Bradley M. Small Jan 7 '14 at 21:09
• Thanks Matthew I will take a look at that. – Bradley M. Small Jan 7 '14 at 21:13

The strategy really does matter. Using dynamic programming, you can show that the value of the game (the expected score when played optimally) is about $18.6759$. To see this, compute the value of holding any given set $S$ of dice as follows: If $|S|=6$, the value is just the score of the dice. If $|S|<6$, you average, over all possible throws of $6-|S|$ dice, the maximum value of $S\cup T$ over all nonempty subsets $T$ of the throw. This is an inductive formula, and you work your way back to the value of the empty set of dice -- i.e. the expected value before you start playing.
In contrast, Monte Carlo simulation indicates that the average score using (a very slight variant of) the obvious strategy indicated by @RossMillikan is between $15.8$ and $15.9$.
You need to be clear about the rules before you ask us to start calculating probabilities. As you have stated it, I can get a perfect throw with probability $1$. I keep a $2$ or $4$ and any $6$'s I roll and reroll the rest. If I don't hav any of those I keep some random die. I can keep rerolling until I have a perfect roll. I suspect that once you keep a die you are stuck with it and you need to keep at least one new die each time. You did not say that. Then I suspect that if you fail to get a 2 and 4 by the end you get zero.