Mixing
Chance of getting six in three dice
$begingroup$
I am having a hard time wrapping my head around this and am sure that my answers are wrong.
There are three dice.
A. Chance of getting exactly one six on the three dice.
(1/6) * 3 = 1/3
B. Chance of getting exactly two sixes.
(1/6 * 1/6) * 1.5 = 1/24
C. Chance of getting exactly 3 sixes.
1/6 * 1/6 * 1/6 = 1/216
D. Chance of any combination of A, B and C
1/3 + 1/24 + 1/216
72/216 + 9/216 + 1/216 = 82/216
probability dice
edited Oct 14 '13 at 10:08
azimut
16.5k1052101
asked Oct 14 '13 at 9:04
Peter BushnellPeter Bushnell
157126
$endgroup$
add a comment |
$begingroup$
I am having a hard time wrapping my head around this and am sure that my answers are wrong.
There are three dice.
A. Chance of getting exactly one six on the three dice.
(1/6) * 3 = 1/3
B. Chance of getting exactly two sixes.
(1/6 * 1/6) * 1.5 = 1/24
C. Chance of getting exactly 3 sixes.
1/6 * 1/6 * 1/6 = 1/216
D. Chance of any combination of A, B and C
1/3 + 1/24 + 1/216
72/216 + 9/216 + 1/216 = 82/216
probability dice
edited Oct 14 '13 at 10:08
azimut
16.5k1052101
asked Oct 14 '13 at 9:04
Peter BushnellPeter Bushnell
157126
$endgroup$
3
$begingroup$
The chance of getting a six on any of the three dice would be $frac{1}{6}times 6=1$ if $6$ dice were thrown. Certainty. Doesn't that make you suspicuous?
$endgroup$
– drhab
Oct 14 '13 at 9:12
add a comment |
$begingroup$
I am having a hard time wrapping my head around this and am sure that my answers are wrong.
There are three dice.
A. Chance of getting exactly one six on the three dice.
(1/6) * 3 = 1/3
B. Chance of getting exactly two sixes.
(1/6 * 1/6) * 1.5 = 1/24
C. Chance of getting exactly 3 sixes.
1/6 * 1/6 * 1/6 = 1/216
D. Chance of any combination of A, B and C
1/3 + 1/24 + 1/216
72/216 + 9/216 + 1/216 = 82/216
probability dice
edited Oct 14 '13 at 10:08
azimut
16.5k1052101
asked Oct 14 '13 at 9:04
Peter BushnellPeter Bushnell
157126
$endgroup$
I am having a hard time wrapping my head around this and am sure that my answers are wrong.
There are three dice.
A. Chance of getting exactly one six on the three dice.
(1/6) * 3 = 1/3
B. Chance of getting exactly two sixes.
(1/6 * 1/6) * 1.5 = 1/24
C. Chance of getting exactly 3 sixes.
1/6 * 1/6 * 1/6 = 1/216
D. Chance of any combination of A, B and C
1/3 + 1/24 + 1/216
72/216 + 9/216 + 1/216 = 82/216
probability dice
probability dice
edited Oct 14 '13 at 10:08
azimut
16.5k1052101
asked Oct 14 '13 at 9:04
Peter BushnellPeter Bushnell
157126
edited Oct 14 '13 at 10:08
azimut
16.5k1052101
asked Oct 14 '13 at 9:04
Peter BushnellPeter Bushnell
157126
edited Oct 14 '13 at 10:08
azimut
16.5k1052101
edited Oct 14 '13 at 10:08
azimut
16.5k1052101
edited Oct 14 '13 at 10:08
azimut
16.5k1052101
16.5k1052101
asked Oct 14 '13 at 9:04
Peter BushnellPeter Bushnell
157126
asked Oct 14 '13 at 9:04
Peter BushnellPeter Bushnell
157126
asked Oct 14 '13 at 9:04
Peter BushnellPeter Bushnell
157126
157126
3
$begingroup$
The chance of getting a six on any of the three dice would be $frac{1}{6}times 6=1$ if $6$ dice were thrown. Certainty. Doesn't that make you suspicuous?
$endgroup$
– drhab
Oct 14 '13 at 9:12
add a comment |
3
$begingroup$
The chance of getting a six on any of the three dice would be $frac{1}{6}times 6=1$ if $6$ dice were thrown. Certainty. Doesn't that make you suspicuous?
$endgroup$
– drhab
Oct 14 '13 at 9:12
3
3
$begingroup$
The chance of getting a six on any of the three dice would be $frac{1}{6}times 6=1$ if $6$ dice were thrown. Certainty. Doesn't that make you suspicuous?
$endgroup$
– drhab
Oct 14 '13 at 9:12
$begingroup$
The chance of getting a six on any of the three dice would be $frac{1}{6}times 6=1$ if $6$ dice were thrown. Certainty. Doesn't that make you suspicuous?
$endgroup$
– drhab
Oct 14 '13 at 9:12
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
A. There is a total of 6^3=216 combinations if you roll 3 dice. There are 5^2x3=75 combinations that you will get one 6. Thus there is a 75/216=25/72 chance of getting only one 6 when rolling 3 dice.
B. There are 5x3 combinations that you will get 2 6s. Thus there is a 15/216=5/72 chance of getting a 2 6s when rolling 3 dice.
C. There is 1 combination where you will get 3 6s. Thus there is a 1/216 chance you will get 3 6s when rolling 3 dice. (Good job you got this correct)
D. There is a 75+15+1/216=91/216 chance of any of them happening.
answered Oct 14 '13 at 9:41
EpicGuyEpicGuy
21519
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@DanielFischer Thx I always get the 2 mixed up :)
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– EpicGuy
Oct 14 '13 at 9:56
2
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Where this number came from (5^2)*3?
$endgroup$
– Misaki
May 9 '16 at 9:36
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Seems like if you don't care about order, rolling a six once would give you 25 possibilities, not 75.
$endgroup$
– PitaJ
Nov 13 '17 at 19:02
1
$begingroup$
@Misaki I figured it out. There are three "cases" where there is exactly one six on three dice:6,x,x/x,6,x/x,x,6. Thexstands for anything but 6, so a number in [1, 5] (5 different sides). When dice #1 is 6 (the first case), there are 5*5 = 5^2 = 25 (the twox) combinations where dice #2 and #3 are anything but 6. The same is true for the other two cases, and thus: 5^2 + 5^2 + 5^2 = (5^2)*3.
$endgroup$
– Joschua
Mar 13 '18 at 21:05
add a comment |
$begingroup$
Hint for A: what is the chance that no sixes appear? If you know that chance then you automatically know the chance that sixes do appear.
answered Oct 14 '13 at 9:13
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
The chance of getting a 6 with three dice is 91/216 because if you subtract 125/216 (the probability of rolling three dice without getting a 6) from 216/216 (the probability of any combination of numbers), you get 91/216.
answered Nov 26 '13 at 16:09
user111236user111236
4914
$endgroup$
$begingroup$
Yes! I earned my first up vote!
$endgroup$
– user111236
Nov 26 '13 at 16:45
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
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$begingroup$
A. There is a total of 6^3=216 combinations if you roll 3 dice. There are 5^2x3=75 combinations that you will get one 6. Thus there is a 75/216=25/72 chance of getting only one 6 when rolling 3 dice.
B. There are 5x3 combinations that you will get 2 6s. Thus there is a 15/216=5/72 chance of getting a 2 6s when rolling 3 dice.
C. There is 1 combination where you will get 3 6s. Thus there is a 1/216 chance you will get 3 6s when rolling 3 dice. (Good job you got this correct)
D. There is a 75+15+1/216=91/216 chance of any of them happening.
answered Oct 14 '13 at 9:41
EpicGuyEpicGuy
21519
$endgroup$
$begingroup$
@DanielFischer Thx I always get the 2 mixed up :)
$endgroup$
– EpicGuy
Oct 14 '13 at 9:56
2
$begingroup$
Where this number came from (5^2)*3?
$endgroup$
– Misaki
May 9 '16 at 9:36
$begingroup$
Seems like if you don't care about order, rolling a six once would give you 25 possibilities, not 75.
$endgroup$
– PitaJ
Nov 13 '17 at 19:02
1
$begingroup$
@Misaki I figured it out. There are three "cases" where there is exactly one six on three dice:6,x,x/x,6,x/x,x,6. Thexstands for anything but 6, so a number in [1, 5] (5 different sides). When dice #1 is 6 (the first case), there are 5*5 = 5^2 = 25 (the twox) combinations where dice #2 and #3 are anything but 6. The same is true for the other two cases, and thus: 5^2 + 5^2 + 5^2 = (5^2)*3.
$endgroup$
– Joschua
Mar 13 '18 at 21:05
add a comment |
$begingroup$
A. There is a total of 6^3=216 combinations if you roll 3 dice. There are 5^2x3=75 combinations that you will get one 6. Thus there is a 75/216=25/72 chance of getting only one 6 when rolling 3 dice.
B. There are 5x3 combinations that you will get 2 6s. Thus there is a 15/216=5/72 chance of getting a 2 6s when rolling 3 dice.
C. There is 1 combination where you will get 3 6s. Thus there is a 1/216 chance you will get 3 6s when rolling 3 dice. (Good job you got this correct)
D. There is a 75+15+1/216=91/216 chance of any of them happening.
answered Oct 14 '13 at 9:41
EpicGuyEpicGuy
21519
$endgroup$
$begingroup$
@DanielFischer Thx I always get the 2 mixed up :)
$endgroup$
– EpicGuy
Oct 14 '13 at 9:56
2
$begingroup$
Where this number came from (5^2)*3?
$endgroup$
– Misaki
May 9 '16 at 9:36
$begingroup$
Seems like if you don't care about order, rolling a six once would give you 25 possibilities, not 75.
$endgroup$
– PitaJ
Nov 13 '17 at 19:02
1
$begingroup$
@Misaki I figured it out. There are three "cases" where there is exactly one six on three dice:6,x,x/x,6,x/x,x,6. Thexstands for anything but 6, so a number in [1, 5] (5 different sides). When dice #1 is 6 (the first case), there are 5*5 = 5^2 = 25 (the twox) combinations where dice #2 and #3 are anything but 6. The same is true for the other two cases, and thus: 5^2 + 5^2 + 5^2 = (5^2)*3.
$endgroup$
– Joschua
Mar 13 '18 at 21:05
add a comment |
$begingroup$
A. There is a total of 6^3=216 combinations if you roll 3 dice. There are 5^2x3=75 combinations that you will get one 6. Thus there is a 75/216=25/72 chance of getting only one 6 when rolling 3 dice.
B. There are 5x3 combinations that you will get 2 6s. Thus there is a 15/216=5/72 chance of getting a 2 6s when rolling 3 dice.
C. There is 1 combination where you will get 3 6s. Thus there is a 1/216 chance you will get 3 6s when rolling 3 dice. (Good job you got this correct)
D. There is a 75+15+1/216=91/216 chance of any of them happening.
answered Oct 14 '13 at 9:41
EpicGuyEpicGuy
21519
$endgroup$
A. There is a total of 6^3=216 combinations if you roll 3 dice. There are 5^2x3=75 combinations that you will get one 6. Thus there is a 75/216=25/72 chance of getting only one 6 when rolling 3 dice.
B. There are 5x3 combinations that you will get 2 6s. Thus there is a 15/216=5/72 chance of getting a 2 6s when rolling 3 dice.
C. There is 1 combination where you will get 3 6s. Thus there is a 1/216 chance you will get 3 6s when rolling 3 dice. (Good job you got this correct)
D. There is a 75+15+1/216=91/216 chance of any of them happening.
answered Oct 14 '13 at 9:41
EpicGuyEpicGuy
21519
edited Oct 14 '13 at 10:17
answered Oct 14 '13 at 9:41
EpicGuyEpicGuy
21519
answered Oct 14 '13 at 9:41
EpicGuyEpicGuy
21519
answered Oct 14 '13 at 9:41
EpicGuyEpicGuy
21519
21519
$begingroup$
@DanielFischer Thx I always get the 2 mixed up :)
$endgroup$
– EpicGuy
Oct 14 '13 at 9:56
2
$begingroup$
Where this number came from (5^2)*3?
$endgroup$
– Misaki
May 9 '16 at 9:36
$begingroup$
Seems like if you don't care about order, rolling a six once would give you 25 possibilities, not 75.
$endgroup$
– PitaJ
Nov 13 '17 at 19:02
1
$begingroup$
@Misaki I figured it out. There are three "cases" where there is exactly one six on three dice:6,x,x/x,6,x/x,x,6. Thexstands for anything but 6, so a number in [1, 5] (5 different sides). When dice #1 is 6 (the first case), there are 5*5 = 5^2 = 25 (the twox) combinations where dice #2 and #3 are anything but 6. The same is true for the other two cases, and thus: 5^2 + 5^2 + 5^2 = (5^2)*3.
$endgroup$
– Joschua
Mar 13 '18 at 21:05
add a comment |
$begingroup$
@DanielFischer Thx I always get the 2 mixed up :)
$endgroup$
– EpicGuy
Oct 14 '13 at 9:56
2
$begingroup$
Where this number came from (5^2)*3?
$endgroup$
– Misaki
May 9 '16 at 9:36
$begingroup$
Seems like if you don't care about order, rolling a six once would give you 25 possibilities, not 75.
$endgroup$
– PitaJ
Nov 13 '17 at 19:02
1
$begingroup$
@Misaki I figured it out. There are three "cases" where there is exactly one six on three dice:6,x,x/x,6,x/x,x,6. Thexstands for anything but 6, so a number in [1, 5] (5 different sides). When dice #1 is 6 (the first case), there are 5*5 = 5^2 = 25 (the twox) combinations where dice #2 and #3 are anything but 6. The same is true for the other two cases, and thus: 5^2 + 5^2 + 5^2 = (5^2)*3.
$endgroup$
– Joschua
Mar 13 '18 at 21:05
$begingroup$
@DanielFischer Thx I always get the 2 mixed up :)
$endgroup$
– EpicGuy
Oct 14 '13 at 9:56
$begingroup$
@DanielFischer Thx I always get the 2 mixed up :)
$endgroup$
– EpicGuy
Oct 14 '13 at 9:56
2
2
$begingroup$
Where this number came from (5^2)*3?
$endgroup$
– Misaki
May 9 '16 at 9:36
$begingroup$
Where this number came from (5^2)*3?
$endgroup$
– Misaki
May 9 '16 at 9:36
$begingroup$
Seems like if you don't care about order, rolling a six once would give you 25 possibilities, not 75.
$endgroup$
– PitaJ
Nov 13 '17 at 19:02
$begingroup$
Seems like if you don't care about order, rolling a six once would give you 25 possibilities, not 75.
$endgroup$
– PitaJ
Nov 13 '17 at 19:02
1
1
$begingroup$
@Misaki I figured it out. There are three "cases" where there is exactly one six on three dice:
6,x,x / x,6,x / x,x,6. The x stands for anything but 6, so a number in [1, 5] (5 different sides). When dice #1 is 6 (the first case), there are 5*5 = 5^2 = 25 (the two x) combinations where dice #2 and #3 are anything but 6. The same is true for the other two cases, and thus: 5^2 + 5^2 + 5^2 = (5^2)*3.$endgroup$
– Joschua
Mar 13 '18 at 21:05
$begingroup$
@Misaki I figured it out. There are three "cases" where there is exactly one six on three dice:
6,x,x / x,6,x / x,x,6. The x stands for anything but 6, so a number in [1, 5] (5 different sides). When dice #1 is 6 (the first case), there are 5*5 = 5^2 = 25 (the two x) combinations where dice #2 and #3 are anything but 6. The same is true for the other two cases, and thus: 5^2 + 5^2 + 5^2 = (5^2)*3.$endgroup$
– Joschua
Mar 13 '18 at 21:05
add a comment |
$begingroup$
Hint for A: what is the chance that no sixes appear? If you know that chance then you automatically know the chance that sixes do appear.
answered Oct 14 '13 at 9:13
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Hint for A: what is the chance that no sixes appear? If you know that chance then you automatically know the chance that sixes do appear.
answered Oct 14 '13 at 9:13
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Hint for A: what is the chance that no sixes appear? If you know that chance then you automatically know the chance that sixes do appear.
answered Oct 14 '13 at 9:13
drhabdrhab
103k545136
$endgroup$
Hint for A: what is the chance that no sixes appear? If you know that chance then you automatically know the chance that sixes do appear.
answered Oct 14 '13 at 9:13
drhabdrhab
103k545136
answered Oct 14 '13 at 9:13
drhabdrhab
103k545136
answered Oct 14 '13 at 9:13
drhabdrhab
103k545136
answered Oct 14 '13 at 9:13
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
$begingroup$
The chance of getting a 6 with three dice is 91/216 because if you subtract 125/216 (the probability of rolling three dice without getting a 6) from 216/216 (the probability of any combination of numbers), you get 91/216.
answered Nov 26 '13 at 16:09
user111236user111236
4914
$endgroup$
$begingroup$
Yes! I earned my first up vote!
$endgroup$
– user111236
Nov 26 '13 at 16:45
add a comment |
$begingroup$
The chance of getting a 6 with three dice is 91/216 because if you subtract 125/216 (the probability of rolling three dice without getting a 6) from 216/216 (the probability of any combination of numbers), you get 91/216.
answered Nov 26 '13 at 16:09
user111236user111236
4914
$endgroup$
$begingroup$
Yes! I earned my first up vote!
$endgroup$
– user111236
Nov 26 '13 at 16:45
add a comment |
$begingroup$
The chance of getting a 6 with three dice is 91/216 because if you subtract 125/216 (the probability of rolling three dice without getting a 6) from 216/216 (the probability of any combination of numbers), you get 91/216.
answered Nov 26 '13 at 16:09
user111236user111236
4914
$endgroup$
The chance of getting a 6 with three dice is 91/216 because if you subtract 125/216 (the probability of rolling three dice without getting a 6) from 216/216 (the probability of any combination of numbers), you get 91/216.
answered Nov 26 '13 at 16:09
user111236user111236
4914
answered Nov 26 '13 at 16:09
user111236user111236
4914
answered Nov 26 '13 at 16:09
user111236user111236
4914
answered Nov 26 '13 at 16:09
user111236user111236
4914
4914
$begingroup$
Yes! I earned my first up vote!
$endgroup$
– user111236
Nov 26 '13 at 16:45
add a comment |
$begingroup$
Yes! I earned my first up vote!
$endgroup$
– user111236
Nov 26 '13 at 16:45
$begingroup$
Yes! I earned my first up vote!
$endgroup$
– user111236
Nov 26 '13 at 16:45
$begingroup$
Yes! I earned my first up vote!
$endgroup$
– user111236
Nov 26 '13 at 16:45
add a comment |
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The chance of getting a six on any of the three dice would be $frac{1}{6}times 6=1$ if $6$ dice were thrown. Certainty. Doesn't that make you suspicuous?
$endgroup$
– drhab
Oct 14 '13 at 9:12