Chance of getting $3$ same faces and $3$ different faces when $6$ dice are thrown
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What is the probability of showing $3$ faces same and $3$ faces different when $6$ dice are thrown simultaneously?
Here's my approach:
$6$ dice thrown. Total number of outcomes that first $3$ dice show the same number is $6$, actually ${(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)}$.
Total number of outcomes that next $3$ dice show all different numbers ( other than the one appeared in first $3$ dice ) are $5C3$ .
Total number of outcomes $= 6^6$. So according to me it should be $6times 5C3 / 6^6$ .
Please guide me through the correct answer.
probability dice
edited Nov 17 at 11:49
drhab
94.9k543125
asked Nov 17 at 7:05
Vaibhav Sachdeva
11
add a comment |
up vote
0
down vote
favorite
What is the probability of showing $3$ faces same and $3$ faces different when $6$ dice are thrown simultaneously?
Here's my approach:
$6$ dice thrown. Total number of outcomes that first $3$ dice show the same number is $6$, actually ${(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)}$.
Total number of outcomes that next $3$ dice show all different numbers ( other than the one appeared in first $3$ dice ) are $5C3$ .
Total number of outcomes $= 6^6$. So according to me it should be $6times 5C3 / 6^6$ .
Please guide me through the correct answer.
probability dice
edited Nov 17 at 11:49
drhab
94.9k543125
asked Nov 17 at 7:05
Vaibhav Sachdeva
11
1
Please post your attempt with the question to indicate how you're thinking about this question.
– OnceUponACrinoid
Nov 17 at 7:09
6 dices thrown. Total number of outcomes that first 3 dices show the same number is 6 {(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)} . Total number of outcomes that next 3 dices show all different numbers ( other than the one appeared in first 3 dices ) are 5C3 . Total number of outcomes = 6^6. So according to me it should be 6 x 5C3 / 6^6
– Vaibhav Sachdeva
Nov 17 at 7:59
1
That calculation counts the number of ways, when it's the first three dice that show the same number.
– Gerry Myerson
Nov 17 at 8:29
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What is the probability of showing $3$ faces same and $3$ faces different when $6$ dice are thrown simultaneously?
Here's my approach:
$6$ dice thrown. Total number of outcomes that first $3$ dice show the same number is $6$, actually ${(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)}$.
Total number of outcomes that next $3$ dice show all different numbers ( other than the one appeared in first $3$ dice ) are $5C3$ .
Total number of outcomes $= 6^6$. So according to me it should be $6times 5C3 / 6^6$ .
Please guide me through the correct answer.
probability dice
edited Nov 17 at 11:49
drhab
94.9k543125
asked Nov 17 at 7:05
Vaibhav Sachdeva
11
What is the probability of showing $3$ faces same and $3$ faces different when $6$ dice are thrown simultaneously?
Here's my approach:
$6$ dice thrown. Total number of outcomes that first $3$ dice show the same number is $6$, actually ${(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)}$.
Total number of outcomes that next $3$ dice show all different numbers ( other than the one appeared in first $3$ dice ) are $5C3$ .
Total number of outcomes $= 6^6$. So according to me it should be $6times 5C3 / 6^6$ .
Please guide me through the correct answer.
probability dice
probability dice
edited Nov 17 at 11:49
drhab
94.9k543125
asked Nov 17 at 7:05
Vaibhav Sachdeva
11
edited Nov 17 at 11:49
drhab
94.9k543125
asked Nov 17 at 7:05
Vaibhav Sachdeva
11
edited Nov 17 at 11:49
drhab
94.9k543125
edited Nov 17 at 11:49
drhab
94.9k543125
edited Nov 17 at 11:49
drhab
94.9k543125
94.9k543125
asked Nov 17 at 7:05
Vaibhav Sachdeva
11
asked Nov 17 at 7:05
Vaibhav Sachdeva
11
asked Nov 17 at 7:05
Vaibhav Sachdeva
11
11
1
Please post your attempt with the question to indicate how you're thinking about this question.
– OnceUponACrinoid
Nov 17 at 7:09
6 dices thrown. Total number of outcomes that first 3 dices show the same number is 6 {(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)} . Total number of outcomes that next 3 dices show all different numbers ( other than the one appeared in first 3 dices ) are 5C3 . Total number of outcomes = 6^6. So according to me it should be 6 x 5C3 / 6^6
– Vaibhav Sachdeva
Nov 17 at 7:59
1
That calculation counts the number of ways, when it's the first three dice that show the same number.
– Gerry Myerson
Nov 17 at 8:29
add a comment |
1
Please post your attempt with the question to indicate how you're thinking about this question.
– OnceUponACrinoid
Nov 17 at 7:09
6 dices thrown. Total number of outcomes that first 3 dices show the same number is 6 {(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)} . Total number of outcomes that next 3 dices show all different numbers ( other than the one appeared in first 3 dices ) are 5C3 . Total number of outcomes = 6^6. So according to me it should be 6 x 5C3 / 6^6
– Vaibhav Sachdeva
Nov 17 at 7:59
1
That calculation counts the number of ways, when it's the first three dice that show the same number.
– Gerry Myerson
Nov 17 at 8:29
1
1
Please post your attempt with the question to indicate how you're thinking about this question.
– OnceUponACrinoid
Nov 17 at 7:09
Please post your attempt with the question to indicate how you're thinking about this question.
– OnceUponACrinoid
Nov 17 at 7:09
6 dices thrown. Total number of outcomes that first 3 dices show the same number is 6 {(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)} . Total number of outcomes that next 3 dices show all different numbers ( other than the one appeared in first 3 dices ) are 5C3 . Total number of outcomes = 6^6. So according to me it should be 6 x 5C3 / 6^6
– Vaibhav Sachdeva
Nov 17 at 7:59
6 dices thrown. Total number of outcomes that first 3 dices show the same number is 6 {(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)} . Total number of outcomes that next 3 dices show all different numbers ( other than the one appeared in first 3 dices ) are 5C3 . Total number of outcomes = 6^6. So according to me it should be 6 x 5C3 / 6^6
– Vaibhav Sachdeva
Nov 17 at 7:59
1
1
That calculation counts the number of ways, when it's the first three dice that show the same number.
– Gerry Myerson
Nov 17 at 8:29
That calculation counts the number of ways, when it's the first three dice that show the same number.
– Gerry Myerson
Nov 17 at 8:29
add a comment |
1 Answer
1
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votes
up vote
1
down vote
Guide
- Work in probability space ${1,2,3,4,5,6}^6$ (so the dice are ordered/numbered) have $6^6$ equiprobable outcomes.
- Choose one face out of $6$ to be the one that is shown $3$ times.
- Choose $3$ spots/numbers out of $6$ for the chosen face.
- Now $3$ spots are left to be filled up with other faces that are moreover distinct. For the first of these $3$ spots there are $5$ choices (of face), for the second $4$ and for the third $3$.
$$6^{-6}cdot6cdotbinom63cdot5cdot4cdot3$$
answered Nov 17 at 8:37
drhab
94.9k543125
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Guide
- Work in probability space ${1,2,3,4,5,6}^6$ (so the dice are ordered/numbered) have $6^6$ equiprobable outcomes.
- Choose one face out of $6$ to be the one that is shown $3$ times.
- Choose $3$ spots/numbers out of $6$ for the chosen face.
- Now $3$ spots are left to be filled up with other faces that are moreover distinct. For the first of these $3$ spots there are $5$ choices (of face), for the second $4$ and for the third $3$.
$$6^{-6}cdot6cdotbinom63cdot5cdot4cdot3$$
answered Nov 17 at 8:37
drhab
94.9k543125
add a comment |
up vote
1
down vote
Guide
- Work in probability space ${1,2,3,4,5,6}^6$ (so the dice are ordered/numbered) have $6^6$ equiprobable outcomes.
- Choose one face out of $6$ to be the one that is shown $3$ times.
- Choose $3$ spots/numbers out of $6$ for the chosen face.
- Now $3$ spots are left to be filled up with other faces that are moreover distinct. For the first of these $3$ spots there are $5$ choices (of face), for the second $4$ and for the third $3$.
$$6^{-6}cdot6cdotbinom63cdot5cdot4cdot3$$
answered Nov 17 at 8:37
drhab
94.9k543125
add a comment |
up vote
1
down vote
up vote
1
down vote
Guide
- Work in probability space ${1,2,3,4,5,6}^6$ (so the dice are ordered/numbered) have $6^6$ equiprobable outcomes.
- Choose one face out of $6$ to be the one that is shown $3$ times.
- Choose $3$ spots/numbers out of $6$ for the chosen face.
- Now $3$ spots are left to be filled up with other faces that are moreover distinct. For the first of these $3$ spots there are $5$ choices (of face), for the second $4$ and for the third $3$.
$$6^{-6}cdot6cdotbinom63cdot5cdot4cdot3$$
answered Nov 17 at 8:37
drhab
94.9k543125
Guide
- Work in probability space ${1,2,3,4,5,6}^6$ (so the dice are ordered/numbered) have $6^6$ equiprobable outcomes.
- Choose one face out of $6$ to be the one that is shown $3$ times.
- Choose $3$ spots/numbers out of $6$ for the chosen face.
- Now $3$ spots are left to be filled up with other faces that are moreover distinct. For the first of these $3$ spots there are $5$ choices (of face), for the second $4$ and for the third $3$.
$$6^{-6}cdot6cdotbinom63cdot5cdot4cdot3$$
answered Nov 17 at 8:37
drhab
94.9k543125
edited Nov 17 at 8:46
answered Nov 17 at 8:37
drhab
94.9k543125
answered Nov 17 at 8:37
drhab
94.9k543125
answered Nov 17 at 8:37
drhab
94.9k543125
94.9k543125
add a comment |
add a comment |
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Please post your attempt with the question to indicate how you're thinking about this question.
– OnceUponACrinoid
Nov 17 at 7:09
6 dices thrown. Total number of outcomes that first 3 dices show the same number is 6 {(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)} . Total number of outcomes that next 3 dices show all different numbers ( other than the one appeared in first 3 dices ) are 5C3 . Total number of outcomes = 6^6. So according to me it should be 6 x 5C3 / 6^6
– Vaibhav Sachdeva
Nov 17 at 7:59
1
That calculation counts the number of ways, when it's the first three dice that show the same number.
– Gerry Myerson
Nov 17 at 8:29