You throw a dice 3 times,probabilty whats the probability of the number k to be the highest?
$begingroup$
You throw a fair dice 3 times,
let X be the highest number you will get.
What is the PDF of $P(X=K), K=1,...6$ ?
The answer is : $$frac{k^3-(k-1)^3}{216}$$
However, I can't understand it, could help me to understand why is this the right answer?
Thank you
probability statistics dice
$endgroup$
add a comment |
$begingroup$
You throw a fair dice 3 times,
let X be the highest number you will get.
What is the PDF of $P(X=K), K=1,...6$ ?
The answer is : $$frac{k^3-(k-1)^3}{216}$$
However, I can't understand it, could help me to understand why is this the right answer?
Thank you
probability statistics dice
$endgroup$
$begingroup$
See en.m.wikipedia.org/wiki/…. There’s a somewhat simpler derivation for the PDF of a max or min of a set of i.i.d. random variables than that for the general case.
$endgroup$
– amd
Dec 4 '18 at 17:18
$begingroup$
Where did you get this problem and its answer? Somewhere in there I’d expect that there’s an explanation.
$endgroup$
– amd
Dec 4 '18 at 17:19
$begingroup$
This is from a book of questions of basic probability, this specific problem doesn't have a detailed explanation.
$endgroup$
– David
Dec 4 '18 at 17:22
$begingroup$
Have you tried to work this out for yourself? Add your attempts, if any, to your question.
$endgroup$
– amd
Dec 4 '18 at 17:23
$begingroup$
I have been trying to answer the question (without knowing the answer) for 2 hours. Then I tried to figure out why the answer is correct for the last hour
$endgroup$
– David
Dec 4 '18 at 17:28
add a comment |
$begingroup$
You throw a fair dice 3 times,
let X be the highest number you will get.
What is the PDF of $P(X=K), K=1,...6$ ?
The answer is : $$frac{k^3-(k-1)^3}{216}$$
However, I can't understand it, could help me to understand why is this the right answer?
Thank you
probability statistics dice
$endgroup$
You throw a fair dice 3 times,
let X be the highest number you will get.
What is the PDF of $P(X=K), K=1,...6$ ?
The answer is : $$frac{k^3-(k-1)^3}{216}$$
However, I can't understand it, could help me to understand why is this the right answer?
Thank you
probability statistics dice
probability statistics dice
edited Dec 4 '18 at 17:19
David
asked Dec 4 '18 at 17:15
DavidDavid
184
184
$begingroup$
See en.m.wikipedia.org/wiki/…. There’s a somewhat simpler derivation for the PDF of a max or min of a set of i.i.d. random variables than that for the general case.
$endgroup$
– amd
Dec 4 '18 at 17:18
$begingroup$
Where did you get this problem and its answer? Somewhere in there I’d expect that there’s an explanation.
$endgroup$
– amd
Dec 4 '18 at 17:19
$begingroup$
This is from a book of questions of basic probability, this specific problem doesn't have a detailed explanation.
$endgroup$
– David
Dec 4 '18 at 17:22
$begingroup$
Have you tried to work this out for yourself? Add your attempts, if any, to your question.
$endgroup$
– amd
Dec 4 '18 at 17:23
$begingroup$
I have been trying to answer the question (without knowing the answer) for 2 hours. Then I tried to figure out why the answer is correct for the last hour
$endgroup$
– David
Dec 4 '18 at 17:28
add a comment |
$begingroup$
See en.m.wikipedia.org/wiki/…. There’s a somewhat simpler derivation for the PDF of a max or min of a set of i.i.d. random variables than that for the general case.
$endgroup$
– amd
Dec 4 '18 at 17:18
$begingroup$
Where did you get this problem and its answer? Somewhere in there I’d expect that there’s an explanation.
$endgroup$
– amd
Dec 4 '18 at 17:19
$begingroup$
This is from a book of questions of basic probability, this specific problem doesn't have a detailed explanation.
$endgroup$
– David
Dec 4 '18 at 17:22
$begingroup$
Have you tried to work this out for yourself? Add your attempts, if any, to your question.
$endgroup$
– amd
Dec 4 '18 at 17:23
$begingroup$
I have been trying to answer the question (without knowing the answer) for 2 hours. Then I tried to figure out why the answer is correct for the last hour
$endgroup$
– David
Dec 4 '18 at 17:28
$begingroup$
See en.m.wikipedia.org/wiki/…. There’s a somewhat simpler derivation for the PDF of a max or min of a set of i.i.d. random variables than that for the general case.
$endgroup$
– amd
Dec 4 '18 at 17:18
$begingroup$
See en.m.wikipedia.org/wiki/…. There’s a somewhat simpler derivation for the PDF of a max or min of a set of i.i.d. random variables than that for the general case.
$endgroup$
– amd
Dec 4 '18 at 17:18
$begingroup$
Where did you get this problem and its answer? Somewhere in there I’d expect that there’s an explanation.
$endgroup$
– amd
Dec 4 '18 at 17:19
$begingroup$
Where did you get this problem and its answer? Somewhere in there I’d expect that there’s an explanation.
$endgroup$
– amd
Dec 4 '18 at 17:19
$begingroup$
This is from a book of questions of basic probability, this specific problem doesn't have a detailed explanation.
$endgroup$
– David
Dec 4 '18 at 17:22
$begingroup$
This is from a book of questions of basic probability, this specific problem doesn't have a detailed explanation.
$endgroup$
– David
Dec 4 '18 at 17:22
$begingroup$
Have you tried to work this out for yourself? Add your attempts, if any, to your question.
$endgroup$
– amd
Dec 4 '18 at 17:23
$begingroup$
Have you tried to work this out for yourself? Add your attempts, if any, to your question.
$endgroup$
– amd
Dec 4 '18 at 17:23
$begingroup$
I have been trying to answer the question (without knowing the answer) for 2 hours. Then I tried to figure out why the answer is correct for the last hour
$endgroup$
– David
Dec 4 '18 at 17:28
$begingroup$
I have been trying to answer the question (without knowing the answer) for 2 hours. Then I tried to figure out why the answer is correct for the last hour
$endgroup$
– David
Dec 4 '18 at 17:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For $k$ to be the highest at least one must be $k$ and the other two must be at less or equal to $k$.
If there is only one that has $k$ there are $3$ dice which may be the $k$, and the other two must be less than $k$, there are $k-1$ options for each. So there are $3(k-1)^2$ ways for there to be exactly one $k$ and it being the highest.
If there are two that have $k$ then there are are ${3 choose 2} =3$ dice that might have the $k$s and $1$ that has the $k-1$ option to not be $k$. So there are $3(k-1)$ ways for that to occur.
And finally there is exactly $1$ way for there to be all three showing $k$.
So $3(k-1)^2 + 3(k-1) + 1 = $
$[(k-1)^3 +3(k-1)^2 + 3(k-1) + 1] - (k-1)^3 = $
$[(k-1) + 1]^3 - (k-1)^3 =$
$k^3 - (k-1)^3$ ways to do this.
And there are $6^3=216$ ways to roll three dice total.
So $P = frac {k^3 - (k-1)^3}{216}$.
$endgroup$
$begingroup$
Thank you very much!!!
$endgroup$
– David
Dec 4 '18 at 17:53
add a comment |
$begingroup$
The probability that the biggest roll will be $K$ is the probability that all three rolls are less than or equal to $K$, minus the case where all the rolls are in fact strictly less than $K$.
Let $X_1,X_2,X_3$ be the outcome of our three rolls (so $X = max(X_1,X_2,X_3)$). Then $$P(X = K) = P(X_1 leq K)P(X_2 leq K)P(X_3 leq K) - P(X_1 leq K - 1)P(X_2 leq K - 1)P(X_3 leq K - 1)$$
Now $P(X_i leq K) = frac{K}{6}$. Hence $$P(X = K) = (frac{K}{6})^3 - (frac{K-1}{6})^3$$ which is the answer you are looking for.
$endgroup$
$begingroup$
Oh wow this is much simpler than I thought. Thank you very much for helping me and spending time to clarify the answer for me!I really do appreciate it!
$endgroup$
– David
Dec 4 '18 at 17:32
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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$begingroup$
For $k$ to be the highest at least one must be $k$ and the other two must be at less or equal to $k$.
If there is only one that has $k$ there are $3$ dice which may be the $k$, and the other two must be less than $k$, there are $k-1$ options for each. So there are $3(k-1)^2$ ways for there to be exactly one $k$ and it being the highest.
If there are two that have $k$ then there are are ${3 choose 2} =3$ dice that might have the $k$s and $1$ that has the $k-1$ option to not be $k$. So there are $3(k-1)$ ways for that to occur.
And finally there is exactly $1$ way for there to be all three showing $k$.
So $3(k-1)^2 + 3(k-1) + 1 = $
$[(k-1)^3 +3(k-1)^2 + 3(k-1) + 1] - (k-1)^3 = $
$[(k-1) + 1]^3 - (k-1)^3 =$
$k^3 - (k-1)^3$ ways to do this.
And there are $6^3=216$ ways to roll three dice total.
So $P = frac {k^3 - (k-1)^3}{216}$.
$endgroup$
$begingroup$
Thank you very much!!!
$endgroup$
– David
Dec 4 '18 at 17:53
add a comment |
$begingroup$
For $k$ to be the highest at least one must be $k$ and the other two must be at less or equal to $k$.
If there is only one that has $k$ there are $3$ dice which may be the $k$, and the other two must be less than $k$, there are $k-1$ options for each. So there are $3(k-1)^2$ ways for there to be exactly one $k$ and it being the highest.
If there are two that have $k$ then there are are ${3 choose 2} =3$ dice that might have the $k$s and $1$ that has the $k-1$ option to not be $k$. So there are $3(k-1)$ ways for that to occur.
And finally there is exactly $1$ way for there to be all three showing $k$.
So $3(k-1)^2 + 3(k-1) + 1 = $
$[(k-1)^3 +3(k-1)^2 + 3(k-1) + 1] - (k-1)^3 = $
$[(k-1) + 1]^3 - (k-1)^3 =$
$k^3 - (k-1)^3$ ways to do this.
And there are $6^3=216$ ways to roll three dice total.
So $P = frac {k^3 - (k-1)^3}{216}$.
$endgroup$
$begingroup$
Thank you very much!!!
$endgroup$
– David
Dec 4 '18 at 17:53
add a comment |
$begingroup$
For $k$ to be the highest at least one must be $k$ and the other two must be at less or equal to $k$.
If there is only one that has $k$ there are $3$ dice which may be the $k$, and the other two must be less than $k$, there are $k-1$ options for each. So there are $3(k-1)^2$ ways for there to be exactly one $k$ and it being the highest.
If there are two that have $k$ then there are are ${3 choose 2} =3$ dice that might have the $k$s and $1$ that has the $k-1$ option to not be $k$. So there are $3(k-1)$ ways for that to occur.
And finally there is exactly $1$ way for there to be all three showing $k$.
So $3(k-1)^2 + 3(k-1) + 1 = $
$[(k-1)^3 +3(k-1)^2 + 3(k-1) + 1] - (k-1)^3 = $
$[(k-1) + 1]^3 - (k-1)^3 =$
$k^3 - (k-1)^3$ ways to do this.
And there are $6^3=216$ ways to roll three dice total.
So $P = frac {k^3 - (k-1)^3}{216}$.
$endgroup$
For $k$ to be the highest at least one must be $k$ and the other two must be at less or equal to $k$.
If there is only one that has $k$ there are $3$ dice which may be the $k$, and the other two must be less than $k$, there are $k-1$ options for each. So there are $3(k-1)^2$ ways for there to be exactly one $k$ and it being the highest.
If there are two that have $k$ then there are are ${3 choose 2} =3$ dice that might have the $k$s and $1$ that has the $k-1$ option to not be $k$. So there are $3(k-1)$ ways for that to occur.
And finally there is exactly $1$ way for there to be all three showing $k$.
So $3(k-1)^2 + 3(k-1) + 1 = $
$[(k-1)^3 +3(k-1)^2 + 3(k-1) + 1] - (k-1)^3 = $
$[(k-1) + 1]^3 - (k-1)^3 =$
$k^3 - (k-1)^3$ ways to do this.
And there are $6^3=216$ ways to roll three dice total.
So $P = frac {k^3 - (k-1)^3}{216}$.
answered Dec 4 '18 at 17:34
fleabloodfleablood
69.7k22685
69.7k22685
$begingroup$
Thank you very much!!!
$endgroup$
– David
Dec 4 '18 at 17:53
add a comment |
$begingroup$
Thank you very much!!!
$endgroup$
– David
Dec 4 '18 at 17:53
$begingroup$
Thank you very much!!!
$endgroup$
– David
Dec 4 '18 at 17:53
$begingroup$
Thank you very much!!!
$endgroup$
– David
Dec 4 '18 at 17:53
add a comment |
$begingroup$
The probability that the biggest roll will be $K$ is the probability that all three rolls are less than or equal to $K$, minus the case where all the rolls are in fact strictly less than $K$.
Let $X_1,X_2,X_3$ be the outcome of our three rolls (so $X = max(X_1,X_2,X_3)$). Then $$P(X = K) = P(X_1 leq K)P(X_2 leq K)P(X_3 leq K) - P(X_1 leq K - 1)P(X_2 leq K - 1)P(X_3 leq K - 1)$$
Now $P(X_i leq K) = frac{K}{6}$. Hence $$P(X = K) = (frac{K}{6})^3 - (frac{K-1}{6})^3$$ which is the answer you are looking for.
$endgroup$
$begingroup$
Oh wow this is much simpler than I thought. Thank you very much for helping me and spending time to clarify the answer for me!I really do appreciate it!
$endgroup$
– David
Dec 4 '18 at 17:32
add a comment |
$begingroup$
The probability that the biggest roll will be $K$ is the probability that all three rolls are less than or equal to $K$, minus the case where all the rolls are in fact strictly less than $K$.
Let $X_1,X_2,X_3$ be the outcome of our three rolls (so $X = max(X_1,X_2,X_3)$). Then $$P(X = K) = P(X_1 leq K)P(X_2 leq K)P(X_3 leq K) - P(X_1 leq K - 1)P(X_2 leq K - 1)P(X_3 leq K - 1)$$
Now $P(X_i leq K) = frac{K}{6}$. Hence $$P(X = K) = (frac{K}{6})^3 - (frac{K-1}{6})^3$$ which is the answer you are looking for.
$endgroup$
$begingroup$
Oh wow this is much simpler than I thought. Thank you very much for helping me and spending time to clarify the answer for me!I really do appreciate it!
$endgroup$
– David
Dec 4 '18 at 17:32
add a comment |
$begingroup$
The probability that the biggest roll will be $K$ is the probability that all three rolls are less than or equal to $K$, minus the case where all the rolls are in fact strictly less than $K$.
Let $X_1,X_2,X_3$ be the outcome of our three rolls (so $X = max(X_1,X_2,X_3)$). Then $$P(X = K) = P(X_1 leq K)P(X_2 leq K)P(X_3 leq K) - P(X_1 leq K - 1)P(X_2 leq K - 1)P(X_3 leq K - 1)$$
Now $P(X_i leq K) = frac{K}{6}$. Hence $$P(X = K) = (frac{K}{6})^3 - (frac{K-1}{6})^3$$ which is the answer you are looking for.
$endgroup$
The probability that the biggest roll will be $K$ is the probability that all three rolls are less than or equal to $K$, minus the case where all the rolls are in fact strictly less than $K$.
Let $X_1,X_2,X_3$ be the outcome of our three rolls (so $X = max(X_1,X_2,X_3)$). Then $$P(X = K) = P(X_1 leq K)P(X_2 leq K)P(X_3 leq K) - P(X_1 leq K - 1)P(X_2 leq K - 1)P(X_3 leq K - 1)$$
Now $P(X_i leq K) = frac{K}{6}$. Hence $$P(X = K) = (frac{K}{6})^3 - (frac{K-1}{6})^3$$ which is the answer you are looking for.
answered Dec 4 '18 at 17:28
ODFODF
1,476510
1,476510
$begingroup$
Oh wow this is much simpler than I thought. Thank you very much for helping me and spending time to clarify the answer for me!I really do appreciate it!
$endgroup$
– David
Dec 4 '18 at 17:32
add a comment |
$begingroup$
Oh wow this is much simpler than I thought. Thank you very much for helping me and spending time to clarify the answer for me!I really do appreciate it!
$endgroup$
– David
Dec 4 '18 at 17:32
$begingroup$
Oh wow this is much simpler than I thought. Thank you very much for helping me and spending time to clarify the answer for me!I really do appreciate it!
$endgroup$
– David
Dec 4 '18 at 17:32
$begingroup$
Oh wow this is much simpler than I thought. Thank you very much for helping me and spending time to clarify the answer for me!I really do appreciate it!
$endgroup$
– David
Dec 4 '18 at 17:32
add a comment |
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$begingroup$
See en.m.wikipedia.org/wiki/…. There’s a somewhat simpler derivation for the PDF of a max or min of a set of i.i.d. random variables than that for the general case.
$endgroup$
– amd
Dec 4 '18 at 17:18
$begingroup$
Where did you get this problem and its answer? Somewhere in there I’d expect that there’s an explanation.
$endgroup$
– amd
Dec 4 '18 at 17:19
$begingroup$
This is from a book of questions of basic probability, this specific problem doesn't have a detailed explanation.
$endgroup$
– David
Dec 4 '18 at 17:22
$begingroup$
Have you tried to work this out for yourself? Add your attempts, if any, to your question.
$endgroup$
– amd
Dec 4 '18 at 17:23
$begingroup$
I have been trying to answer the question (without knowing the answer) for 2 hours. Then I tried to figure out why the answer is correct for the last hour
$endgroup$
– David
Dec 4 '18 at 17:28