More dice questions
July 21st, 2014 at 1:40:37 PM
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Somebody smarter than me please help. What are odds with 3d6 of the lowest two dice thrown being a pair of 1s, 2s, or 3s? I have already bumbled my way into figuring that the chance of there being a pair of 1s,2s, or 3s is roughly 22%, but I cant figure the odds of that pair being on the lowest 2 dice. Thanks in advance for any help.
July 21st, 2014 at 2:57:04 PM
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Not including triple 1's, 2's, or 3's, I get 24 instances where the pair is the lowest value of the 3 dice out of 216 possible rolls, for an incidence of 11.1111% of the time. Including triples, for 27 instances, would be 12.5%. YMMV.
If the House lost every hand, they wouldn't deal the game.
July 21st, 2014 at 3:16:31 PM
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Quote: beachbumbabsNot including triple 1's, 2's, or 3's, I get 24 36 instances where the pair is the lowest value of the 3 dice out of 216 possible rolls, for an incidence of 11.1111% 16.667% of the time. Including triples, for 27 39 instances, would be 12.5% 18.056% . YMMV.
Fixed that for you BBB.
“Man Babes” #AxelFabulous
July 21st, 2014 at 3:22:52 PM
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Quote: mipletQuote: beachbumbabsNot including triple 1's, 2's, or 3's, I get 24 36 instances where the pair is the lowest value of the 3 dice out of 216 possible rolls, for an incidence of 11.1111% 16.667% of the time. Including triples, for 27 39 instances, would be 12.5% 18.056% . YMMV.
Fixed that for you BBB.
Ah, you're right that I made an error. But I think it's 33 instances (9 I left out) rather than 12. Lemme look at my spreadsheet again.
Edit: you're right, I left out 3-1, 2-1, and 3-2. 36 it is.
If the House lost every hand, they wouldn't deal the game.
July 21st, 2014 at 3:45:19 PM
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how about some math
second math will be for you if your take the challenge
I think I agree with the other two replies
So 111 or 222 does not count
There are 90 ways for the 3 dice to roll a pair
AAB (one pair and one single)
A = any 6 values
B = any 5 values and any of the 3 dice
6*5*3 = 90 ways out of 6^3 total ways (216)
90/6=15 pairs per each value
you want 3 values so 45 ways are what you want for a pair of 1s, 2s or 3s
45/216 = 20.83%
one could even just list all the possible sequences too
that saves on hurting any math
see if you can figure out using math only
I could keep on going but that might take out the fun for you I think
have fun!
Sally
second math will be for you if your take the challenge
I think I agree with the other two replies
some bees bumble tooQuote: DonkI have already bumbled my way into figuring that the chance of there being a pair of 1s,2s, or 3s is roughly 22%, but I cant figure the odds of that pair being on the lowest 2 dice.
So 111 or 222 does not count
There are 90 ways for the 3 dice to roll a pair
AAB (one pair and one single)
A = any 6 values
B = any 5 values and any of the 3 dice
6*5*3 = 90 ways out of 6^3 total ways (216)
90/6=15 pairs per each value
you want 3 values so 45 ways are what you want for a pair of 1s, 2s or 3s
45/216 = 20.83%
one could even just list all the possible sequences too
that saves on hurting any math
we can still use math...Quote: DonkWhat are odds with 3d6 of the lowest two dice thrown being a pair of 1s, 2s, or 3s?
see if you can figure out using math only
I could keep on going but that might take out the fun for you I think
my answer too = 1/6
I did a simple list method here
where the pair = 1
all smaller! That =15 ways
1 1 2
1 1 3
1 1 4
1 1 5
1 1 6
1 2 1
1 3 1
1 4 1
1 5 1
1 6 1
2 1 1
3 1 1
4 1 1
5 1 1
6 1 1
not all smaller
2 2 1
2 2 3<yes
2 2 4<yes
2 2 5<yes
2 2 6<yes
2 1 2
2 3 2<yes
2 4 2<yes
2 5 2<yes
2 6 2<yes
1 2 2
3 2 2<yes
4 2 2<yes
5 2 2<yes
6 2 2<yes
only 12 ways
3 3 1
3 3 2
3 3 4<yes
3 3 5<yes
3 3 6<yes
3 1 3
3 2 3
3 4 3<yes
3 5 3<yes
3 6 3<yes
1 3 3
2 3 3
4 3 3<yes
5 3 3<yes
6 3 3<yes
only 9 ways
15+12+9/216 = 36/216 should be your answer that you have been seeking
now try with just using math. see some patterns already?
it gets easier the more times you do try ;)
I did a simple list method here
where the pair = 1
all smaller! That =15 ways
1 1 2
1 1 3
1 1 4
1 1 5
1 1 6
1 2 1
1 3 1
1 4 1
1 5 1
1 6 1
2 1 1
3 1 1
4 1 1
5 1 1
6 1 1
not all smaller
2 2 1
2 2 3<yes
2 2 4<yes
2 2 5<yes
2 2 6<yes
2 1 2
2 3 2<yes
2 4 2<yes
2 5 2<yes
2 6 2<yes
1 2 2
3 2 2<yes
4 2 2<yes
5 2 2<yes
6 2 2<yes
only 12 ways
3 3 1
3 3 2
3 3 4<yes
3 3 5<yes
3 3 6<yes
3 1 3
3 2 3
3 4 3<yes
3 5 3<yes
3 6 3<yes
1 3 3
2 3 3
4 3 3<yes
5 3 3<yes
6 3 3<yes
only 9 ways
15+12+9/216 = 36/216 should be your answer that you have been seeking
now try with just using math. see some patterns already?
it gets easier the more times you do try ;)
have fun!
Sally
I Heart Vi Hart
July 21st, 2014 at 6:00:07 PM
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Here's my answer:
Of the 216 possible ways to throw 3d6:
There is one way to throw 1-1-1
1-1-(something higher than 1) can be thrown 5 (the "other" die) x 3 (the number of dice that can be the "other" die) = 15 times
2-2-2 can be thrown once
2-2-(something higher than 2) can be thrown 4 x 3 = 12 times
3-3-3 can be thrown once
3-3-(something higher than 3) can be thrown 3 x 3 = 9 times
The total number of throws that meet the condition is 1 + 15 + 1 + 12 + 1 + 9 = 39.
Thus the probability is 39 / 216 = 13 / 72.
This assumes that, for example, 3-3-3 meets the condition that "the lowest two dice are a pair of 1s, 2s, or 3s" - otherwise, the three triplets are ignored, and the probability is 36 / 216 = 1 / 6.
Of the 216 possible ways to throw 3d6:
There is one way to throw 1-1-1
1-1-(something higher than 1) can be thrown 5 (the "other" die) x 3 (the number of dice that can be the "other" die) = 15 times
2-2-2 can be thrown once
2-2-(something higher than 2) can be thrown 4 x 3 = 12 times
3-3-3 can be thrown once
3-3-(something higher than 3) can be thrown 3 x 3 = 9 times
The total number of throws that meet the condition is 1 + 15 + 1 + 12 + 1 + 9 = 39.
Thus the probability is 39 / 216 = 13 / 72.
This assumes that, for example, 3-3-3 meets the condition that "the lowest two dice are a pair of 1s, 2s, or 3s" - otherwise, the three triplets are ignored, and the probability is 36 / 216 = 1 / 6.
