You throw a dice 3 times,probabilty whats the probability of the number k to be the highest?












3












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















3












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














3












3








3





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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 $P = frac {k^3 - (k-1)^3}{216}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!!!
    $endgroup$
    – David
    Dec 4 '18 at 17:53



















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 $P = frac {k^3 - (k-1)^3}{216}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!!!
    $endgroup$
    – David
    Dec 4 '18 at 17:53
















1












$begingroup$

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 $P = frac {k^3 - (k-1)^3}{216}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!!!
    $endgroup$
    – David
    Dec 4 '18 at 17:53














1












1








1





$begingroup$

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 $P = frac {k^3 - (k-1)^3}{216}$.






share|cite|improve this answer









$endgroup$



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 $P = frac {k^3 - (k-1)^3}{216}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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











2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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