Permutations with additional chances (like draw poker I guess)
July 8th, 2012 at 4:53:13 PM
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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?
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?
July 11th, 2012 at 3:41:35 AM
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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.
