Dice Probabilities with 3 dice
April 3rd, 2013 at 6:37:31 PM
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Is my math correct with the following probabilities using 3 dice?
Chance of: Odds Decimal Percentage
Rolling a specific single No. 75/216 0.347222222 35.0%
Rolling a specific Double No. 15/216 0.069444444 7.0%
Rolling a specific Triple No. 1/216 0.004629629 0.5%
Rolling a double or triple for a specific No. 16/216 0.074074074 7.5%
Rolling a specific No. at all 91/216 0.421296296 42.0%
Rolling a triple of any No. 6/216 0.027777777 3.0%
Rolling a double of any No. 90/216 0.416666666 41.5%
Rolling a double or triple of any No. 96/216 0.444444444 44.5%
Many thanks
Anton
Chance of: Odds Decimal Percentage
Rolling a specific single No. 75/216 0.347222222 35.0%
Rolling a specific Double No. 15/216 0.069444444 7.0%
Rolling a specific Triple No. 1/216 0.004629629 0.5%
Rolling a double or triple for a specific No. 16/216 0.074074074 7.5%
Rolling a specific No. at all 91/216 0.421296296 42.0%
Rolling a triple of any No. 6/216 0.027777777 3.0%
Rolling a double of any No. 90/216 0.416666666 41.5%
Rolling a double or triple of any No. 96/216 0.444444444 44.5%
Many thanks
Anton
April 3rd, 2013 at 7:22:19 PM
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Yes they are correct for standard 6 sided dice.
“Man Babes” #AxelFabulous
April 3rd, 2013 at 10:02:28 PM
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miplet,
Thank you for your help.
Cheers
Anton
Thank you for your help.
Cheers
Anton
April 3rd, 2013 at 10:05:30 PM
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10 and 11 is the 7 of 3 dice.
April 4th, 2013 at 1:04:57 AM
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Sure and you have some wild rounding for your percentages.Quote: antonicalIs my math correct with the following probabilities using 3 dice?
The Wizard has a few pages also on this
https://wizardofodds.com/games/sic-bo/appendix/1/
http://www.mathproblems.info/gam470/games/sicbo/
I do just a few
"Rolling a specific single No. 75/216"
This would be the pattern 1AB
The # of choices for A = 5 (can not be the 1)
The # of choices for B = 5 (can not be the 1)
The 1 can be in 3 different positions
5*5*3 = 75
"Rolling a double of any No. 90/216"
This would be pattern AAB
The # of choices would be 6 (for the first A) * 5 (for the second A) * 3 for the B
6*5*3 = 90
The pattern ABC is simply 6*5*4 = 120
AAA is easy at 6
more of this method here
http://blog.plover.com/math/yahtzee.html
winsome johnny (not Win some johnny)
April 4th, 2013 at 1:16:14 AM
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For the sums, yes. 7 for 2d6 and 10 and 11 for 3d6Quote: nezbit10 and 11 is the 7 of 3 dice.
Can you show proof for those interested?
here are the combinations
I need to show the # if ways for each.
hint: 3 different #s is just 3!
For 2 different #s it is 3C2 (3!/2!)
9,10,11,12 have the same number of combinations but different # of ways
3 ; 1 + 1 + 1 = 1 way
4 ; 1 + 1 + 2 = 3 ways
5 ; 1 + 1 + 3 ; 2 + 2 + 1 = 6 ways (3+3)
6 ; 1 + 1 + 4 ; 1 + 2 + 3 ; 2 + 2 + 2 = 10 ways (3+6+1)
7 ; 1 + 1 + 5 ; 2 + 2 + 3 ; 3 + 3 + 1 ; 1 + 2 + 4 = 15 ways (3+3+3+6)
8 ; 1 + 1 + 6 ; 2 + 3 + 3 ; 4 + 3 + 1 ; 1 + 2 + 5 ; 2 + 2 + 4 = 21 ways (3+3+6+6+3)
9 ; 6 + 2 + 1 ; 4 + 3 + 2 ; 3 + 3 + 3 ; 2 + 2 + 5 ; 1 + 3 + 5 ; 1 + 4 + 4 = 25 ways (6+6+1+3+6+3)
10 ; 6 + 3 + 1 ; 6 + 2 + 2 ; 5 + 3 + 2 ; 4 + 4 + 2 ; 4 + 3 + 3 ; 1 + 4 + 5 = 27 ways (6+3+6+3+3+6)
11 ; 6 + 4 + 1 ; 1 + 5 + 5 ; 5 + 4 + 2 ; 3 + 3 + 5 ; 4 + 3 + 4 ; 6 + 3 + 2
12 ; 6 + 5 + 1 ; 4 + 3 + 5 ; 4 + 4 + 4 ; 5 + 2 + 5 ; 6 + 4 + 2 ; 6 + 3 + 3
13 ; 6 + 6 + 1 ; 5 + 4 + 4 ; 3 + 4 + 6 ; 6 + 5 + 2 ; 5 + 5 + 3
14 ; 6 + 6 + 2 ; 5 + 5 + 4 ; 4 + 4 + 6 ; 6 + 5 + 3
15 ; 6 + 6 + 3 ; 6 + 5 + 4 ; 5 + 5 + 5
16 ; 6 + 6 + 4 ; 5 + 5 + 6
17 ; 6 + 6 + 5
18 ; 6 + 6 + 6
the rest is easy
just select the first 10 rows with your pointer
3 ; 1 + 1 + 1 = 1 way
4 ; 1 + 1 + 2 = 3 ways
5 ; 1 + 1 + 3 ; 2 + 2 + 1 = 6 ways (3+3)
6 ; 1 + 1 + 4 ; 1 + 2 + 3 ; 2 + 2 + 2 = 10 ways (3+6+1)
7 ; 1 + 1 + 5 ; 2 + 2 + 3 ; 3 + 3 + 1 ; 1 + 2 + 4 = 15 ways (3+3+3+6)
8 ; 1 + 1 + 6 ; 2 + 3 + 3 ; 4 + 3 + 1 ; 1 + 2 + 5 ; 2 + 2 + 4 = 21 ways (3+3+6+6+3)
9 ; 6 + 2 + 1 ; 4 + 3 + 2 ; 3 + 3 + 3 ; 2 + 2 + 5 ; 1 + 3 + 5 ; 1 + 4 + 4 = 25 ways (6+6+1+3+6+3)
10 ; 6 + 3 + 1 ; 6 + 2 + 2 ; 5 + 3 + 2 ; 4 + 4 + 2 ; 4 + 3 + 3 ; 1 + 4 + 5 = 27 ways (6+3+6+3+3+6)
11 ; 6 + 4 + 1 ; 1 + 5 + 5 ; 5 + 4 + 2 ; 3 + 3 + 5 ; 4 + 3 + 4 ; 6 + 3 + 2
12 ; 6 + 5 + 1 ; 4 + 3 + 5 ; 4 + 4 + 4 ; 5 + 2 + 5 ; 6 + 4 + 2 ; 6 + 3 + 3
13 ; 6 + 6 + 1 ; 5 + 4 + 4 ; 3 + 4 + 6 ; 6 + 5 + 2 ; 5 + 5 + 3
14 ; 6 + 6 + 2 ; 5 + 5 + 4 ; 4 + 4 + 6 ; 6 + 5 + 3
15 ; 6 + 6 + 3 ; 6 + 5 + 4 ; 5 + 5 + 5
16 ; 6 + 6 + 4 ; 5 + 5 + 6
17 ; 6 + 6 + 5
18 ; 6 + 6 + 6
the rest is easy
just select the first 10 rows with your pointer
winsome johnny (not Win some johnny)
