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Biased dice probability question [on hold]

-2

2

$begingroup$

A biased six sided dice is rolled twice. Show that the probability that the two results are the same is at least $frac{1}{6}$.
(Hint: $(p_1 − a)^2 + . . . + (p_6 − a)^2 ≥ 0$ and choose suitable
$p_1, . . . , p_6$, a.)

probability

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edited yesterday

mathpadawan

2,030422

asked yesterday

mandymandy

92

New contributor

mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by Arnaud D., Xander Henderson, Saad, TheSimpliFire, user21820 20 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Arnaud D., Xander Henderson, Saad, TheSimpliFire, user21820

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: First try to show that if a coin is flipped twice, the probability that the two results are the same is at least $1/2$. This will help you figure out what to choose as $a$.
  $endgroup$
  – Lorenzo
  yesterday
  

  
  
  
  

add a comment | 

-2

2

$begingroup$

A biased six sided dice is rolled twice. Show that the probability that the two results are the same is at least $frac{1}{6}$.
(Hint: $(p_1 − a)^2 + . . . + (p_6 − a)^2 ≥ 0$ and choose suitable
$p_1, . . . , p_6$, a.)

probability

share|cite|improve this question

edited yesterday

mathpadawan

2,030422

asked yesterday

mandymandy

92

New contributor

mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by Arnaud D., Xander Henderson, Saad, TheSimpliFire, user21820 20 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Arnaud D., Xander Henderson, Saad, TheSimpliFire, user21820

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: First try to show that if a coin is flipped twice, the probability that the two results are the same is at least $1/2$. This will help you figure out what to choose as $a$.
  $endgroup$
  – Lorenzo
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

A biased six sided dice is rolled twice. Show that the probability that the two results are the same is at least $frac{1}{6}$.
(Hint: $(p_1 − a)^2 + . . . + (p_6 − a)^2 ≥ 0$ and choose suitable
$p_1, . . . , p_6$, a.)

probability

share|cite|improve this question

edited yesterday

mathpadawan

2,030422

asked yesterday

mandymandy

92

New contributor

mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited yesterday

mathpadawan

2,030422

asked yesterday

mandymandy

92

New contributor

mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited yesterday

mathpadawan

2,030422

asked yesterday

mandymandy

92

New contributor

mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited yesterday

mathpadawan

2,030422

edited yesterday

mathpadawan

2,030422

asked yesterday

mandymandy

92

New contributor

mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

mandymandy

92

3 Answers
3

active

oldest

votes

7

$begingroup$

Let $p_i$ be the probability of rolling $i$. Then $sum_{i=1}^6 p_i = 1$.

By Cauchy-Schwarz inequality,

$$begin{align*}
left(sum_{i=1}^6 1^2right) left(sum_{i=1}^6 p_i^2right) &ge
left(sum_{i=1}^6 1p_iright)^2\
6left(sum_{i=1}^6 p_i^2right) &ge 1\
sum_{i=1}^6 p_i^2 &ge frac16end{align*}$$

Equality holds when all the $p_i$ are the same, i.e. when the die is unbiased.

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answered yesterday

peterwhypeterwhy

12.4k21229

$endgroup$

add a comment | 

2

$begingroup$

Using the hint: For any $ain mathbb{R}$

$$ sum (p_i − a)^2 ge 0$$
$$ sum (a^2 -2ap_i+p_i^2) ge 0$$
$$ 6 a^2 -2asum p_i + sum p_i^2 ge 0$$
$$ sum p_i^2 ge -2a(3a-1)$$

The RHS is a quadratic with roots at $a=0$, $a=1/3$, and maximum at $a_0=1/6$. At that point, the RHS value is $-2 frac 16 (3 frac 16 -1)=1/6$

Hence $$ sum p_i^2 ge frac16$$

Of course, the LHS is the probability of having the same result in both rolls.

For the equality to hold, the first inequality must be an equality, hence $p_i=a=frac16$.

share|cite|improve this answer

edited yesterday

answered yesterday

leonbloyleonbloy

42.4k647108

$endgroup$

add a comment | 

1

$begingroup$

Here’s another way to solve the problem, just for fun. Let $p_i$ be the probability of rolling $i$ when the die is rolled once. Also for simplicity, suppose the faces of the die are numbered $0$ through $5$. This won’t change the probability in question.

For two-roll sequences, consider the event $E_d$ that the second roll is $d$ bigger than the first roll (modulo $6$). So the sequence $3,5$ would be part of event $E_2$; $3,3$ would be part of $E_0$; and $5,3$ would be part of $E_color{red}4$, because $5+color{red}4equiv3$ (mod $6$).

The probability $P(E_d)$ is easy to calculate. It’s $sum_{i=0}^5 p_ip_{(i+d)(!!!!mod!!6)}$, which is the dot product of the vector $vec{p}=langle p_0,p_1,p_2,p_3,p_4,p_5rangle$ and a vector $vec{p_d}$ equal in length to $vec{p}$ having the same coordinates as $vec{p}$ but cyclically permuted. Also, when the die is rolled twice, exactly one of the events $E_d$ for $0le dle5$ occurs, so $sum_{d=0}^5P(E_d)=1$.

Then $P(E_d)=vec vcdotvec v_d={|vec v|^2overcostheta}$, where $theta$ is the angle between $vec v$ and $vec v_d$. The die is biased, so $vec v parallel vec v_d$ only when $d=0$, and $P(E_d)$ has a unique maximum when $d=0$, and $P(E_0)$ is the probability of the die showing the same number on both throws.

The unique maximum value among $6$ numbers whose sum is $1$ must be greater than $1over 6$.

share|cite|improve this answer

answered yesterday

Steve KassSteve Kass

11.4k11530

$endgroup$

add a comment | 

7

$begingroup$

Let $p_i$ be the probability of rolling $i$. Then $sum_{i=1}^6 p_i = 1$.

By Cauchy-Schwarz inequality,

$$begin{align*}
left(sum_{i=1}^6 1^2right) left(sum_{i=1}^6 p_i^2right) &ge
left(sum_{i=1}^6 1p_iright)^2\
6left(sum_{i=1}^6 p_i^2right) &ge 1\
sum_{i=1}^6 p_i^2 &ge frac16end{align*}$$

Equality holds when all the $p_i$ are the same, i.e. when the die is unbiased.

share|cite|improve this answer

answered yesterday

peterwhypeterwhy

12.4k21229

$endgroup$

add a comment | 

7

$begingroup$

Let $p_i$ be the probability of rolling $i$. Then $sum_{i=1}^6 p_i = 1$.

By Cauchy-Schwarz inequality,

$$begin{align*}
left(sum_{i=1}^6 1^2right) left(sum_{i=1}^6 p_i^2right) &ge
left(sum_{i=1}^6 1p_iright)^2\
6left(sum_{i=1}^6 p_i^2right) &ge 1\
sum_{i=1}^6 p_i^2 &ge frac16end{align*}$$

Equality holds when all the $p_i$ are the same, i.e. when the die is unbiased.

share|cite|improve this answer

answered yesterday

peterwhypeterwhy

12.4k21229

$endgroup$

add a comment | 

$begingroup$

Let $p_i$ be the probability of rolling $i$. Then $sum_{i=1}^6 p_i = 1$.

By Cauchy-Schwarz inequality,

$$begin{align*}
left(sum_{i=1}^6 1^2right) left(sum_{i=1}^6 p_i^2right) &ge
left(sum_{i=1}^6 1p_iright)^2\
6left(sum_{i=1}^6 p_i^2right) &ge 1\
sum_{i=1}^6 p_i^2 &ge frac16end{align*}$$

Equality holds when all the $p_i$ are the same, i.e. when the die is unbiased.

share|cite|improve this answer

answered yesterday

peterwhypeterwhy

12.4k21229

$endgroup$

share|cite|improve this answer

answered yesterday

peterwhypeterwhy

12.4k21229

answered yesterday

peterwhypeterwhy

12.4k21229

answered yesterday

peterwhypeterwhy

12.4k21229

2

$begingroup$

Using the hint: For any $ain mathbb{R}$

$$ sum (p_i − a)^2 ge 0$$
$$ sum (a^2 -2ap_i+p_i^2) ge 0$$
$$ 6 a^2 -2asum p_i + sum p_i^2 ge 0$$
$$ sum p_i^2 ge -2a(3a-1)$$

The RHS is a quadratic with roots at $a=0$, $a=1/3$, and maximum at $a_0=1/6$. At that point, the RHS value is $-2 frac 16 (3 frac 16 -1)=1/6$

Hence $$ sum p_i^2 ge frac16$$

Of course, the LHS is the probability of having the same result in both rolls.

For the equality to hold, the first inequality must be an equality, hence $p_i=a=frac16$.

share|cite|improve this answer

edited yesterday

answered yesterday

leonbloyleonbloy

42.4k647108

$endgroup$

add a comment | 

2

$begingroup$

Using the hint: For any $ain mathbb{R}$

$$ sum (p_i − a)^2 ge 0$$
$$ sum (a^2 -2ap_i+p_i^2) ge 0$$
$$ 6 a^2 -2asum p_i + sum p_i^2 ge 0$$
$$ sum p_i^2 ge -2a(3a-1)$$

The RHS is a quadratic with roots at $a=0$, $a=1/3$, and maximum at $a_0=1/6$. At that point, the RHS value is $-2 frac 16 (3 frac 16 -1)=1/6$

Hence $$ sum p_i^2 ge frac16$$

Of course, the LHS is the probability of having the same result in both rolls.

For the equality to hold, the first inequality must be an equality, hence $p_i=a=frac16$.

share|cite|improve this answer

edited yesterday

answered yesterday

leonbloyleonbloy

42.4k647108

$endgroup$

add a comment | 

$begingroup$

Using the hint: For any $ain mathbb{R}$

$$ sum (p_i − a)^2 ge 0$$
$$ sum (a^2 -2ap_i+p_i^2) ge 0$$
$$ 6 a^2 -2asum p_i + sum p_i^2 ge 0$$
$$ sum p_i^2 ge -2a(3a-1)$$

The RHS is a quadratic with roots at $a=0$, $a=1/3$, and maximum at $a_0=1/6$. At that point, the RHS value is $-2 frac 16 (3 frac 16 -1)=1/6$

Hence $$ sum p_i^2 ge frac16$$

Of course, the LHS is the probability of having the same result in both rolls.

For the equality to hold, the first inequality must be an equality, hence $p_i=a=frac16$.

share|cite|improve this answer

edited yesterday

answered yesterday

leonbloyleonbloy

42.4k647108

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

leonbloyleonbloy

42.4k647108

answered yesterday

leonbloyleonbloy

42.4k647108

answered yesterday

leonbloyleonbloy

42.4k647108

1

$begingroup$

Here’s another way to solve the problem, just for fun. Let $p_i$ be the probability of rolling $i$ when the die is rolled once. Also for simplicity, suppose the faces of the die are numbered $0$ through $5$. This won’t change the probability in question.

For two-roll sequences, consider the event $E_d$ that the second roll is $d$ bigger than the first roll (modulo $6$). So the sequence $3,5$ would be part of event $E_2$; $3,3$ would be part of $E_0$; and $5,3$ would be part of $E_color{red}4$, because $5+color{red}4equiv3$ (mod $6$).

The probability $P(E_d)$ is easy to calculate. It’s $sum_{i=0}^5 p_ip_{(i+d)(!!!!mod!!6)}$, which is the dot product of the vector $vec{p}=langle p_0,p_1,p_2,p_3,p_4,p_5rangle$ and a vector $vec{p_d}$ equal in length to $vec{p}$ having the same coordinates as $vec{p}$ but cyclically permuted. Also, when the die is rolled twice, exactly one of the events $E_d$ for $0le dle5$ occurs, so $sum_{d=0}^5P(E_d)=1$.

Then $P(E_d)=vec vcdotvec v_d={|vec v|^2overcostheta}$, where $theta$ is the angle between $vec v$ and $vec v_d$. The die is biased, so $vec v parallel vec v_d$ only when $d=0$, and $P(E_d)$ has a unique maximum when $d=0$, and $P(E_0)$ is the probability of the die showing the same number on both throws.

The unique maximum value among $6$ numbers whose sum is $1$ must be greater than $1over 6$.

share|cite|improve this answer

answered yesterday

Steve KassSteve Kass

11.4k11530

$endgroup$

add a comment | 

1

$begingroup$

Here’s another way to solve the problem, just for fun. Let $p_i$ be the probability of rolling $i$ when the die is rolled once. Also for simplicity, suppose the faces of the die are numbered $0$ through $5$. This won’t change the probability in question.

For two-roll sequences, consider the event $E_d$ that the second roll is $d$ bigger than the first roll (modulo $6$). So the sequence $3,5$ would be part of event $E_2$; $3,3$ would be part of $E_0$; and $5,3$ would be part of $E_color{red}4$, because $5+color{red}4equiv3$ (mod $6$).

The probability $P(E_d)$ is easy to calculate. It’s $sum_{i=0}^5 p_ip_{(i+d)(!!!!mod!!6)}$, which is the dot product of the vector $vec{p}=langle p_0,p_1,p_2,p_3,p_4,p_5rangle$ and a vector $vec{p_d}$ equal in length to $vec{p}$ having the same coordinates as $vec{p}$ but cyclically permuted. Also, when the die is rolled twice, exactly one of the events $E_d$ for $0le dle5$ occurs, so $sum_{d=0}^5P(E_d)=1$.

Then $P(E_d)=vec vcdotvec v_d={|vec v|^2overcostheta}$, where $theta$ is the angle between $vec v$ and $vec v_d$. The die is biased, so $vec v parallel vec v_d$ only when $d=0$, and $P(E_d)$ has a unique maximum when $d=0$, and $P(E_0)$ is the probability of the die showing the same number on both throws.

The unique maximum value among $6$ numbers whose sum is $1$ must be greater than $1over 6$.

share|cite|improve this answer

answered yesterday

Steve KassSteve Kass

11.4k11530

$endgroup$

add a comment | 

$begingroup$

Here’s another way to solve the problem, just for fun. Let $p_i$ be the probability of rolling $i$ when the die is rolled once. Also for simplicity, suppose the faces of the die are numbered $0$ through $5$. This won’t change the probability in question.

For two-roll sequences, consider the event $E_d$ that the second roll is $d$ bigger than the first roll (modulo $6$). So the sequence $3,5$ would be part of event $E_2$; $3,3$ would be part of $E_0$; and $5,3$ would be part of $E_color{red}4$, because $5+color{red}4equiv3$ (mod $6$).

The probability $P(E_d)$ is easy to calculate. It’s $sum_{i=0}^5 p_ip_{(i+d)(!!!!mod!!6)}$, which is the dot product of the vector $vec{p}=langle p_0,p_1,p_2,p_3,p_4,p_5rangle$ and a vector $vec{p_d}$ equal in length to $vec{p}$ having the same coordinates as $vec{p}$ but cyclically permuted. Also, when the die is rolled twice, exactly one of the events $E_d$ for $0le dle5$ occurs, so $sum_{d=0}^5P(E_d)=1$.

Then $P(E_d)=vec vcdotvec v_d={|vec v|^2overcostheta}$, where $theta$ is the angle between $vec v$ and $vec v_d$. The die is biased, so $vec v parallel vec v_d$ only when $d=0$, and $P(E_d)$ has a unique maximum when $d=0$, and $P(E_0)$ is the probability of the die showing the same number on both throws.

The unique maximum value among $6$ numbers whose sum is $1$ must be greater than $1over 6$.

share|cite|improve this answer

answered yesterday

Steve KassSteve Kass

11.4k11530

$endgroup$

share|cite|improve this answer

answered yesterday

Steve KassSteve Kass

11.4k11530

answered yesterday

Steve KassSteve Kass

11.4k11530

answered yesterday

Steve KassSteve Kass

11.4k11530