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Thread: Permutations involving additional chances

  1. #1
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    Permutations involving additional chances




    I am attempting to calculate the probabilities of a simple dice game. I have actually calculated the perfect score, and the additional permutations involved because it gives additional chances. Though I think I came up with a good formula, I can't prove it and would love a little help so that I can calculate subsequent scores (not perfect) without having to manually work each and every one out.

    Here is how the game works:

    Toss 6 dice. Each toss you can keep one or more up to the total number of dice tossed. You continue to toss until you have no dice remaining. You may not reintroduce a die once it has been kept.

    In order to get a score in this game you must have retained both a 2 and a 4. You get no score for them but without them you get no score at all (0). Your score is the sum of the remaining 4 dice. So a perfect score would be 2,4,6,6,6,6 which would get you 24 (also the name of the game).

    To start, I determined that the permutation of 6 dice having 4 numbers the same and 2 numbers different is 6!/4! (or 30) and given that there are 6^6 possible throws the odds of a perfect score in one throw of the dice is (n(6^6)-(6!/4!)) : 6!/4!

    That comes out to 1554.2 : 1 ... Pretty tough odds.

    Now the game allows you to keep one of more die and then re-toss the rest. If I add up all the different combinations of keeping and tossing I get a total of 900 different ways from "throwing them all and keeping them all", to "throwing them 6 times keeping 1 each time" and every where in between. So now that gives me odds on a perfect game of 6^6 - 900 : 900

    That comes out to 50.84 : 1 ... Not as tough odds.

    Where I am having trouble is understanding the calculations, because I simply worked them out by hand instead of coming up with a calculation. I would like to ultimately come up with the odds of score 23 (2,4,6,6,6,5) and so on down to score 4 (2,4,1,1,1,1)

    When I came up with the 900 different ways, I liked that it appears to be the square of the permutations to get the perfect score in one toss. However, I can't come up with a way to prove that so that I can apply it. The other thing that I noticed was that there were the same number of ways to get the perfect score by specifically retaining 1 die each time as there were getting the perfect score in one throw. I seem to feel they are related some way as well. But, that also is just intuition.

    I really feel I am close to a breakthrough on this, but I am just missing one piece of understanding. Can someone please push me in the right direction?

    Oh, yeah... if this is not 100% clear please ask me, I have tried to make the explanation as clear as possible, and would hate to miss out on a good answer just because the problem was not presented clearly.

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    Re: Permutations involving additional chances

    You said one thing, that makes me want to make sure about the rules of the game; do you have to keep at least one tossed die? You said you want to come up with odds for scores all the way down to 4, but isn't it possible to get 0 if you don't get a 2 and a 4?

    I'm trying to calculate the probably of getting a 24, and that's a beast calculation. I'm trying to think of the probability of getting a 24, using {# of ways to get 24} / {# of all possibilities}, and the # of all possibilities isn't so easy to get.

    Let's say you throw six 6's first. Do you keep four of the 6's, and take a chance that you'll get a 2 or 4 (or, 2 and 4 if you're extra lucky) out of just two dice on the next throw, and then the last required throw? Or keep just one of the 6's, so you have more chance to get the 2 and 4 you need in order to score anything? To get the # of all possibilities, we'd have to count all these different strategies. And that's just if you get six 6's first.

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    Re: Permutations involving additional chances

    Yes, you must keep at least one die each toss. The least number of tosses is 1 and the max is 6. And you are correct that you will have a score of 0 in any series of tosses that fails to result in at least one 2 and one 4.

    The number of possibilities is 6^6 = 46656
    The number of ways to get 24 = 900 or (30 if you have to do it in one toss)

    The 30 is from 6! / 4! .. that is a simple calculation for permutations where some number of the outcome are the same.
    The 900 comes from writing out all the possible responses to each possible toss and counting them.

    If you throw and keep 4 6's you then have 2 chances (2,4 or 4,2) out of 36 possibilities. If you throw 6,6,2,4 first then you have 1 chance (6,6) out of 36 possibilities. But that is merely a discussion of strategy. Which, in fact, means more considering how you play. (e.g. if you are playing only to get 24 or if you are playing against an opponent that has some score less than 24 which you are trying to beat.)

    However, the strategy does not affect the odds. Rather, understanding the odds will affect how someone should devise a strategy.

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    Re: Permutations involving additional chances

    Just to add some more twist, I just looked through the spreadsheet and found that I hadn't completed the series totally. So there will be more than 900 ways to get a perfect score. That does, however, eliminate the question of the squaring of the first number.

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    Re: Permutations involving additional chances

    It has been some time, but I am getting back to looking at this again, here is what I have determined:


    As I understand this to throw a perfect roll (2,4,6,6,6,6) (score of 24) is calculated as 6!/4! which gives me 30 possibilities out of 6^6 (46656). Which 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 rethrow 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 thow 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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    Re: Permutations involving additional chances

    I guess you better think of a dynamic prgramming recursion along the lines of the solution of the dice game of Threes in http://www.madandmoonly.com/doctorma...tics/dice1.pdf

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    Re: Permutations involving additional chances


    The last post was caught by the spam filter but I just let it through - this post is mainly to bump the thread.
    I don't have emotions and sometimes that makes me very sad.

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