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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?

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  • $\begingroup$ 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. $\endgroup$ – Matthew Conroy Jan 7 '14 at 20:55
  • $\begingroup$ 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. $\endgroup$ – Matthew Conroy Jan 7 '14 at 21:01
  • $\begingroup$ 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. $\endgroup$ – Matthew Conroy Jan 7 '14 at 21:07
  • $\begingroup$ 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 $\endgroup$ – Bradley M. Small Jan 7 '14 at 21:09
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    $\begingroup$ Thanks Matthew I will take a look at that. $\endgroup$ – Bradley M. Small Jan 7 '14 at 21:13
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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 $15.9$.

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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.

Then an obvious strategy is to keep the first 2 and 4 you see, any 6's, and if none are available the highest die. If you have to add one more fixed die each roll, clearly when you are rolling only two dice if both are 4 or above you should keep them both.

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